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	<title>Chris Dean’s Personal Website &#187; Science</title>
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		<title>Office Experiment Fodder</title>
		<link>http://www.christopher-dean.co.uk/2008/02/21/office-experiment-fodder/</link>
		<comments>http://www.christopher-dean.co.uk/2008/02/21/office-experiment-fodder/#comments</comments>
		<pubDate>Wed, 20 Feb 2008 23:26:17 +0000</pubDate>
		<dc:creator>Chris</dc:creator>
				<category><![CDATA[Science]]></category>
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		<description><![CDATA[When I saw this on a website recently I just had to investigate further: Blow up your last Rolo. Observe how the chewy toffee centre fails to act in a chewy manner when exposed to internally concentrated explosive forces. Be Afraid Funtley&#8230;. Be Very Afraid&#8230;. Sphere: Related Content]]></description>
			<content:encoded><![CDATA[<!-- sphereit start --><p>When I saw this on a website recently I just had to investigate further:</p>
<p><a href="http://www.emanator.demon.co.uk/bigclive/rolo.htm">Blow up your last Rolo.</a><br />
Observe how the chewy toffee centre fails to act in a chewy manner when exposed to internally concentrated explosive forces.</p>
<p>Be Afraid Funtley&#8230;. Be Very Afraid&#8230;.</p>
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		<title>Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation</title>
		<link>http://www.christopher-dean.co.uk/2001/05/28/determination-of-the-boyle-temperature-of-a-gas-using-the-redlich-and-kwong-gas-equation/</link>
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		<pubDate>Mon, 28 May 2001 11:28:51 +0000</pubDate>
		<dc:creator>Chris</dc:creator>
				<category><![CDATA[Essays]]></category>
		<category><![CDATA[Science]]></category>

		<guid isPermaLink="false">http://www.christopher-dean.co.uk/2007/06/06/determination-of-the-boyle-temperature-of-a-gas-using-the-redlich-and-kwong-gas-equation/</guid>
		<description><![CDATA[In this project I shall be trying to determine the Boyle temperature for a set gas based around the Redlich and Kwong gas equation that has been written in a form analogous to the van der Waals equation. The equation is:                 Which can be re-arranged to:         First We Find The First Derivative Of This Equation: [...]]]></description>
			<content:encoded><![CDATA[<!-- sphereit start --><p><span style="font-size: 10pt; font-family: Times New Roman">In this project I shall be trying to determine the Boyle temperature for a set gas based around the Redlich and Kwong gas equation that has been written in a form analogous to the van der Waals equation.<br />
</span></p>
<p>The equation is:                 <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati1.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati1 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p>Which can be re-arranged to:        <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati2.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati2 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p><span style="font-family: Arial"><strong><em>First We Find The First Derivative Of This Equation:<br />
</em></strong></span></p>
<p>From a standard result, and the fact that the derivative of a constant times a variable is the same as the constant times the derivative of the variable:</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati3.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati3 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p>Next, treating the constant in the same way as above, and using the quotient rule we get:</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati4.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati4 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p><strong>So overall:<br />
</strong></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati5.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati5 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p><span style="font-family: Arial"><strong><em>Next, We Find The Second Derivative:<br />
</em></strong></span></p>
<p>Using the same method again we first have:</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati6.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati6 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p>And then for the second term we do the same but we also use the product rule, giving:</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati7.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati7 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p><strong>So overall we have:<br />
</strong></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati8.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati8 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p><strong>Next we want to find the critical constants, V<sub>c</sub>, T<sub>c</sub>, and P<sub>c</sub>.<br />
</strong></p>
<p>At the critical point <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati9.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati9 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /> are zero, so we can equate them to each other in order to find V<sub>c</sub>:</p>
<p><strong>Firstly:<br />
</strong></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati10.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati10 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p><strong>Secondly:<br />
</strong></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati11.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati11 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p><strong>Therefore:<br />
</strong></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati12.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati12 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p>This has 3 solutions, 2 are imaginary, and are therefore of no interest to us, the third is real however and gives us:            </p>
<p style="text-align: center">V<sub>c</sub> = (2<span style="font-size: 12pt"><sup>(⅔)</sup></span> + 2<span style="font-size: 12pt"><sup>(⅓)</sup></span> + 1)b = 3.8473b</p>
<p><strong>Next we use this result to find T<sub>c</sub><br />
<sup>3/2</sup>:</strong></p>
<p>We do this by rearranging one of the differential equations and substituting this value of V<sub>c</sub> for V:</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati13.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati13 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p><strong>Now that we have an expression for T<sub>c</sub> and V<sub>c</sub> we can determine P<sub>c</sub>, using the original equation:<br />
</strong></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati14.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati14 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p>As we only have an expression for T<sub>c </sub><sup>3/2</sup>, we can&#8217;t form an equation with just P<sub>c</sub>, a and b in it, but we can derive the following:</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati15.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati15 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p>If we now take units of atm for P, l mol<sup>-1</sup> for V and K for T, we have R = 0.08206 l atm K<sup>-1</sup>. Also as I have chosen to study the H<sub>2</sub> molecule, we can look up the values of P<sub>c</sub> and T<sub>c</sub>.</p>
<p><strong>We can use the values to find a, b and V<sub>c</sub> as follows:<br />
</strong></p>
<p style="text-align: center">P<sub>c</sub> = 12.8 atm, T<sub>c</sub> = 33.23 K</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati16.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati16 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p>b is a term that accounts for the discrepancy in volume due to the gas being non-ideal, and is directly related to the volume of the molecule in question – namely H<sub>2</sub>. We can use b to give a better value for the volume of 1 mole of H<sub>2</sub> by noting that: V = V<sub>m</sub> – b, where V<sub>m</sub> is the ideal molar volume. Then once we have the true volume of 1 mole of H<sub>2</sub> molecules, we can calculate the effective spherical volume occupied by just 1 molecule.</p>
<p>The ideal molar volume for H<sub>2</sub> at 1atm and 300 K is: V = <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati17.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati17 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " />= 0.08206 <span style="font-family: Arial">x</span> 300 = 24.618 l</p>
<p>This means that a better value for the true volume is = 24.618 – 0.018 = 24.6 l</p>
<p>So 24.6 l contains 6.02 <span style="font-family: Arial">x</span> 10<sup>23</sup> molecules, which implies that one molecule has an effective spherical volume of 4.086 <span style="font-family: Arial">x</span> 10<sup>-23</sup> l.</p>
<p>The next thing to do would be to expand P in powers of V<sup>-1</sup>, and show that B, the second virial coefficient is given by:</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati18.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati18 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p>The easiest way to do this is to apparently to expand the expression for a perfect gas, and compare this expression to it. The expression for a perfect gas is:</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati19.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati19 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p>And our equation for P is:</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati20.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati20 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p>So far I have obtained an expression for B that is almost true to this, but at the time of writing, I am having trouble getting an exact match. As a result I will leave this part of the analysis to be included as an appendix at the end of the project, and continue, assuming the above expression for B to be true.</p>
<p><strong>We can use a plot B for different values of T to determine the Boyle temperature for the gas:<br />
</strong></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati21.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati21 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati22.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati22 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p>At T = Boyle Temperature (T<sub>B</sub>), B = 0. As the graph is approximately linear around this point, a straight line fitted to this range of values yields the equation:</p>
<p style="text-align: center"><strong>B = 0.0003T – 0.0316</strong>.</p>
<p>Using this equation we can predict T<sub>B</sub> as follows:</p>
<p style="text-align: center">B = 0.0003T – 0.0316</p>
<p style="text-align: center">0 = 0.0003T – 0.0316 at T<sub>B<br />
</sub></p>
<p style="text-align: center"><span style="font-family: Symbol">Þ</span><strong><br />
<img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati23.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati23 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /><br />
</strong></p>
<p>Therefore the Boyle Temperature for H<sub>2</sub> is 105.33 K, however the book states that it is 110.0K my result is a little out, but this is probably due to both rounding errors, and approximating the region of the graph where the T axis is crossed as linear instead of fitting an exact curve to the data points.</p>
<p>Whilst doing this project I also tried a couple of other scenarios; one using Xe as the basis &#8216;molecule&#8217;, and the other using NH<sub>3</sub>. For Xe the predicted value for TB, was approximately <sup>1</sup>/<sub>7</sub> of the value stated in the book, and the predicted value for NH<sub>3</sub> is 1350 K. There is no value for T<sub>B</sub> of NH<sub>3</sub> in the book, this is most probably because NH<sub>3</sub> dissociates below this predicted temperature. Also the value for Xe stated in the book is about 7 times as much as I predicted (about 130 – 170 K). This is probably due to Xe being an atom and not considerable as a molecule, and therefore this expression doesn&#8217;t apply, or perhaps Xe could be considered as a molecule but in a way that I&#8217;m not yet familiar with.</p>
<h1><span style="font-size: 12pt">Appendix:<br />
</span></h1>
<h2><span style="font-size: 11pt">Expanding P In Powers Of V<sup>-1</sup> To Show That The Second Virial Co-Efficient Is:<br />
</span></h2>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati24.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati24 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p>Well, I&#8217;ve reached the project deadline and I still can&#8217;t get this quite right, but rather than avoid the question as others have, I have included my best attempt at reaching the correct expression.</p>
<p>The perfect gas expansion is:</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati25.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati25 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p>Our equation is:</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati26.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati26 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p>So re-writing our expression in the form of the ideal gas expression gives:</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati27.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati27 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
<p>However assuming that the first term is always 1 (which I believe it is) then the equation should be modified slightly, and so should B. However this just leads to an even messier expression, which no more correct than the first, as simplification still doesn&#8217;t remove the &#8220;V&#8217;s&#8221;:</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2032-determinati28.png" title="Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " alt="060607 2032 determinati28 Determination of the Boyle Temperature of a Gas using the Redlich and Kwong Gas Equation   " /></p>
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		<title>My Autobiography as a Scientist</title>
		<link>http://www.christopher-dean.co.uk/2001/05/13/my-autobiography-as-a-scientist/</link>
		<comments>http://www.christopher-dean.co.uk/2001/05/13/my-autobiography-as-a-scientist/#comments</comments>
		<pubDate>Sun, 13 May 2001 10:00:30 +0000</pubDate>
		<dc:creator>Chris</dc:creator>
				<category><![CDATA[Essays]]></category>
		<category><![CDATA[Science]]></category>

		<guid isPermaLink="false">http://www.christopher-dean.co.uk/2007/06/06/my-autobiography-as-a-scientist/</guid>
		<description><![CDATA[From a very early age, in fact ever since my dad started to teach me maths at the age of three or four, I have been interested in solving problems. This was partially due to my inquisitive nature, and my mum&#8217;s interest in crosswords, and other such puzzles that I tried to help her solve [...]]]></description>
			<content:encoded><![CDATA[<!-- sphereit start --><p><span style="font-size: 9pt">From a very early age, in fact ever since my dad started to teach me maths at the age of three or four, I have been interested in solving problems. This was partially due to my inquisitive nature, and my mum&#8217;s interest in crosswords, and other such puzzles that I tried to help her solve when I was a little older. My dad was a blacksmith at Chatham Dockyard, where he made springs. Conveniently, he is very creative and very good at metalwork and his main hobby was (and still is) wrought iron work, especially making gates.<br />
</span></p>
<p><span style="font-size: 9pt">We have a large workshop in our back garden, and a 10 ft high forge that he built to aid in his metalwork. Ever since the age of about five I was encouraged to be outside with him, and learn what he was doing as a start in learning a trade. There are many technical and creative aspects to his work, these started me thinking in this way too, and I have found this to be very useful in many areas of my life.<br />
</span></p>
<p><span style="font-size: 9pt">Aside to metalwork my dad had a passion for trying to make broken things work. This is probably due to the fact that he had never been, and we have never been financially &#8216;rich&#8217;, and as a result, if he saw a broken TV set or old stereo system dumped by the side of the road for example, then he would stop his car, pick it up, bring it home and try to fix it, and most of the time he succeeded. However, there were times when he couldn&#8217;t fix these things and would pass them onto me (and my younger brother) to play with (in terms of dismantling and destroying them).I must admit that that I didn&#8217;t learn very much in dismantling these old pieces of electrical equipment, but it did start an interest in electronics.<br />
</span></p>
<p><span style="font-size: 9pt">Dad also liked science fiction series on TV and in the movies (as did mum, although to a lesser extent), these programmes were always on around teatime (six o&#8217;clock) and I can remember watching them whilst eating my dinner, from a very early age. Dad&#8217;s like for things like Star Trek and The Invaders, also influenced me and started my interest in Science Fiction, although nowadays I prefer to read the older stories like The Time Machine or The War of the Worlds. I find that modern novels are more suited to the cinema, and I enjoy them there, but otherwise they seem quite bland and &#8216;samey&#8217;.<br />
</span></p>
<p><span style="font-size: 9pt">When I was a bit older, about six or seven I was bought a &#8216;junior tool kit&#8217;, which was a real but not very practical, woodworking set, that I tried to use to make different things, just like my dad, however these were not very successful. Things carried on like this, me working with dad, playing with the TV&#8217;s, trying unsuccessful woodworking projects, watching Science fiction TV, and the like until a few years before I started secondary school.<br />
</span></p>
<p><span style="font-size: 9pt">When I was about nine or ten, I asked for an electronics kit for Christmas. It was nothing big, just a small pre-wired circuit board, with springs for connection terminals and a series of wires that you could configure in different ways to produce about fifty different things. Most of these things didn&#8217;t interest me at all, electronic counters and the like, but a few really caught my imagination. One was a burglar alarm, which really worked, but only for about five minutes, and another was a transistor radio, which was much more successful.<br />
</span></p>
<p><span style="font-size: 9pt">The radio was great because it had the one key thing that meant that no-one ever knew that I was listening to it when I should have been asleep – an ear piece. Mum never liked the idea of things like Walkman&#8217;s as she thought that noise playing directly into our ears would damage our hearing (as a young child I had a little trouble with my hearing, amongst other things), but she never realised what I could be doing with this ear piece, and as a result, for the next two years or so, I used to secretly listen to the radio late at night under the bed sheets.<br />
</span></p>
<p><span style="font-size: 9pt">Being a normal child I also liked the idea of making a mess and blowing things up. Having graduated past mud pies and the like, I started to think that chemistry was a pretty neat idea, and wanted a chemistry set. Mum didn&#8217;t like the idea of this too much either, as she thought I may hurt myself, but when I was eleven she reluctantly bought me one for Christmas (or rather Father Christmas brought me one). This was fine the set was OK and I tried a couple of the experiments, but these soon bored me, so I began just mixing things up. I also knew that vinegar was acidic, so I put this into my test tube with the other chemicals, and then corked it. Bad move! The vinegar reacted with one of the chemicals and popped the cork off of the tube and a sticky black gunge, started to bubble out over the worktop in the kitchen (where I was &#8216;experimenting&#8217;). I soon cleaned this mess up before mum saw it, and everything was fine until a few days later.<br />
</span></p>
<p><span style="font-size: 9pt">A friend of mine wanted to come over and play, which he did, and I showed him my new chemistry set. Mum had also just bought some nice peach coloured curtains and blinds for the kitchen that she was particularly fond of. Another bad move! I told my friend about what happens when you mix everything together and add vinegar, but he wanted to see it for himself of course. So he proceeded to mix everything<br />
</span></p>
<p><span style="font-size: 9pt">together and cork the test tube, and just as I was about to warn him not to, he started shaking the tube rather quickly. Well in no time at all, the cork and the rest of the top of the test tube actually blew away, and the kitchen ceiling, our clothes, but worst of all mum&#8217;s new curtains and blinds were covered in, you&#8217;ve guessed it, sticky black gunge. Great. My friend was sent home, me to my room, dad had to paint the kitchen ceiling, the curtains became clean after five or six washes, and from now on I had to do everything outside!<br />
</span></p>
<p><span style="font-size: 9pt">Like most kids at the time, I was taken over a fair bit by the advent of computer games, and this started a further interest in electronics and particularly computing, but also took me away from what I &#8216;should&#8217; have been doing with dad. As I got older, my interest in computers was also building and I moved up from pure games systems, to the Amiga 1200 which was a cross over between a cheap computer and a games system. I haven&#8217;t mentioned my entrepreneurial streak yet, which is probably due to helping my dad out at boot sales when I was young. This lead to me trying to make money out of my computer, which I did. Once I had gathered together the technical know how, I began producing disk based interactive catalogues of small pieces of freely distributable software (usually games) which I was to sell to my friends. This worked well, and I made a few pounds out of X-Project PD (my &#8216;company&#8217;), but the most useful thing about this in general, was the wealth of knowledge I gained about computing via doing this. After the Amiga I got into building PC&#8217;s for myself and others (making me a few hundred pounds over the years) and one of my other main interests is web site design, although I&#8217;m also a qualified windows expert and budding programmer – but that&#8217;s enough of my CV.<br />
</span></p>
<p><span style="font-size: 9pt">As I reached secondary school age I was becoming more interested in this area of study and while I seemed to excel at most things at school, I particularly liked science. Whilst at school it is safe to say that my interests were well nurtured by the teachers, and I was supported well by mum, dad and others like my granddad. I learned a lot, and this is where my interest in what once was the boring aspects of chemistry and physics, came from.<br />
</span></p>
<p><span style="font-size: 9pt">At school we sometimes used to have pen flicking &#8216;fights&#8217;, where everyone in the class flicked their fountain pen at everyone else, and everyone went home with an inverted pattern of the night sky on their shirts. Now and again I didn&#8217;t really want to be part of the ink &#8216;fights&#8217; but still got covered all the same, sometimes more so for not joining in, but I remember one day when I got my own back quite well.<br />
</span></p>
<p><span style="font-size: 9pt">We had a chemistry lab class on a Wednesday afternoon, just after lunch, and that morning I had got covered in ink, which I wasn&#8217;t pleased about. In the lab class we were doing experiments on very, very weak acids and alkalis, using phenolphthalein indicator, all of which are colourless when separate.<br />
</span></p>
<p><span style="font-size: 9pt">Well, to cut a long story short, whilst using the pipettes, a little alkali got fired across the room and down the backs of some of the people that had covered me in ink during the morning. When they felt it hit it was colourless, and thought that they had been spayed with a little water, which was good from my point of view! Next they were sprayed with a little phenolphthalein, just a bit further up from the alkali spot, and again they thought nothing of it. The class carried on as normal, and over the hour strange things had started to happen to people&#8217;s shirts.<br />
</span></p>
<p><span style="font-size: 9pt">As we had done these experiments before, I knew what to expect. When phenolphthalein is mixed with even the slightest bit of alkali it turns bright purple-red, and as the alkali and the indicator seeped up the shirts, they met and made great big purple blobs – which was great! Luckily I was never found out by the teachers (at least), and I got away with it, although I realise it could have been dangerous now. Anyway, it was good fun at the time, and once people cottoned on, most other classes were also doing it as well!<br />
</span></p>
<p><span style="font-size: 9pt">Well that just about wraps it up for inspirational, life changing events, of course there were others too, but these were a few of those that stick in my mind. My teachers at school, and my college lecturers kept my interest in science ticking over nicely. At the end of my A-levels I still couldn&#8217;t tell what I liked better out of chemistry and physics, the two sciences I really know anything about, and as a result I decided to do a degree that involves both, namely chemical physics, and that is where I am today. I&#8217;m still encouraged by my parents, and they both like the idea of me being nearer to the frontiers of science, and so do I, although I&#8217;ve still a long way to go yet!</span></p>
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		<title>A Description of Past Achievements in Physics, and the Possible Advances to be Made in the Future</title>
		<link>http://www.christopher-dean.co.uk/2001/05/08/a-descrption-of-past-achievements-in-physics-and-the-possible-advances-to-be-made-in-the-future/</link>
		<comments>http://www.christopher-dean.co.uk/2001/05/08/a-descrption-of-past-achievements-in-physics-and-the-possible-advances-to-be-made-in-the-future/#comments</comments>
		<pubDate>Tue, 08 May 2001 14:02:34 +0000</pubDate>
		<dc:creator>Chris</dc:creator>
				<category><![CDATA[Essays]]></category>
		<category><![CDATA[Science]]></category>

		<guid isPermaLink="false">http://www.christopher-dean.co.uk/2007/06/06/a-descrption-of-past-achievements-in-physics-and-the-possible-advances-to-be-made-in-the-future/</guid>
		<description><![CDATA[There have been many technological advances in the fields of chemistry physics in the past, but the most significant of those have been in physics in the last two hundred years or so. As a result of this observation, when considering evaluations of past achievements, I shall be limiting my discussion to approximately the last [...]]]></description>
			<content:encoded><![CDATA[<!-- sphereit start --><p>There have been many technological advances in the fields of chemistry physics in the past, but the most significant of those have been in physics in the last two hundred years or so. As a result of this observation, when considering evaluations of past achievements, I shall be limiting my discussion to approximately the last two hundred years, using mainly (if not exclusively) the physical examples that I find most relevant. I shall also outline these achievements, and their relative merits, in chronological order, so as to give some idea of how science has evolved (to use a more biological term), and hopefully build better foundations for my &#8216;insight&#8217; into the future.</p>
<p><strong><em>Evaluation Of Past Achievements:<br />
</em></strong></p>
<p>As it plays such an important role in our everyday lives it seems sensible to start with the discovery of electricity, without which this essay would not be in such a readable format.</p>
<p>Ever since the ancient Greeks, the electrostatic effect has been known. Back then, and in fact up until the start of the 1800&#8242;s this was just considered as a strange observation &#8211; an attractive effect caused by rubbing a silk cloth on, say, a piece of amber (incidentally call elektron in Greek!). However, with the development of Newton&#8217;s laws in the <em>1660&#8242;s </em>and the &#8216;birth&#8217; of physics as we now know it, there was considerably more interest in describing attractive forces using mathematical equations, than there ever was before. This interest lead to investigations by people such as Gray and Coulomb on the nature of the electrostatic force, and then, once Volta had created the first &#8216;Voltaic Cell&#8217; in <em>1799</em> (what we think of as a &#8216;battery&#8217;), many more similar investigations were launched in the new exciting field of electrical current research. By <em>1827</em> Ampère had produced a summary of the majority of work in a series of mathematical papers and experimental observations, which we still use to describe the macroscopic properties of electrical current today. A little later in the <em>1830&#8242;s</em> Faraday&#8217;s experiments (based around those of Ørsted) lead to the laws of electromagnetic induction (due to the properties of electricity and magnetism being inextricably linked, as he started to prove in <em>1845</em>) which we also use today to create and distribute electricity, and also to make electronic switches in circuits called capacitors &#8211; an integral part of almost every modern electrical appliance.</p>
<p>I don&#8217;t really need to push to hard to make you see the importance of these experiments. They provided, in the short period of about 50 years, the basic laws upon which nearly all modern electronics is based. And as electrical appliances are so influential in our everyday lives, the work of these pioneers of electronics lives on in nearly everything we use. Without these discoveries, we would not have our very important radio and televisual communications networks (later), good refrigerators, life support machines, (practical) vacuum cleaners, washing machines or computers to name just a microscopic sample of uses. And furthermore, without computers, or more generally the laws concerning the behaviour of electrical currents, it&#8217;s probably safe to say that we wouldn&#8217;t know even a small amount of the things we do in all other aspects of science. This is because so many tasks rely on monitoring by electrical apparatus and/or modelling by computer, the structure of DNA and the mapping on the genome being good examples of where both these aspects come into play.</p>
<p>Faraday&#8217;s experiments, along with the work of Ampère, Coulomb (and also Gauss), provided the framework of what we call classical electrodynamics, but it was the work of Maxwell in <em>1864</em>, which really started to close the book on what electricity is. Maxwell was more skilled at maths than Faraday and, using new data came up with four equations that surmise the behaviour of both electricity and magnetism together. He was also the first person to realise that light was (what we now call) an electromagnetic wave and could be treated mathematically using equations that are similar to those used for waves on a string. His work provides the grand unifying theory of electromagnetism, and is really the last word when it comes to understanding how electricity and magnetism interact, and how this can be applied to benefit us.</p>
<p>His work on electromagnetism isn&#8217;t really of much interest to people outside of physics or electrical engineering. All it does (although this shouldn&#8217;t be treated at all half-heartedly) is provide the theoretical basis for the empirical work done by Faraday, and it&#8217;s these equations that we use to model the interactions of electrical systems today. What is of more use to everybody is his description of light as an electromagnetic wave, as it was this that led to the work of Hertz, Marconi and Logie Baird.</p>
<p>Hertz, using Maxwell&#8217;s idea of electromagnetic waves, first showed the existence of radio waves produced by an electric spark, and in <em>1888</em> showed that a tuned circuit up to twenty meters away could detect them. He then went on to show that these waves could be polarised, reflected and showed all other properties of &#8216;standard&#8217; light (the portion of the electromagnetic spectrum that we see), which was an important piece of supporting evidence for Maxwell&#8217;s idea (as it lead onto the investigation for other types of electromagnetic radiation). Hertz died very young, but soon after his death Marconi improved the design of the radio equipment that he used, and showed that radio waves could be transmitted over 1609 meters (approx. 1 mile). Marconi then proceeded to capitalise on &#8216;his&#8217; invention by interesting the British government in the technology.</p>
<p>Ever since electrical current had been discovered, and it had been shown that wires could carry electrical pulses, the telegraph had become the fastest and most important form of communication, apart from letter writing. The telegraph system was adopted widely in Britain, and Morse code was used to send messages all over the UK and the rest of the world through a network of cables. As the British Empire was built the telegraph became an even more important means of communication, however a problem arose when distant points, such as countries or ships at sea had to be connected up to the system. This is where Marconi came in. He convinced the British government that radio transmission was a real alternative to the cable based telegraph, and in <em>1899 </em>he made the first transmission across the English Channel, and radiotelegraphy was born. This transmission caused much interest from the Royal Navy and other parties wishing to exploit the invention, and the system soon became widely adopted as <em>the</em> means of communication in the developed world. Marconi also made the first Trans-Atlantic broadcast in <em>1901</em> via a crude satellite system, which consisted of a kite-born antenna flying above Newfoundland in Canada. In<em> 1915 </em>an American company transmitted the first good quality audio radio signals from Virginia to Paris. By <em>1927</em> he had created a worldwide communication system based upon this technology, which was to change the field of communications forever.</p>
<p>Again, it&#8217;s not hard to see what an impact this has had on the world. We live in an age where the use of communications technology is second nature to us, and provides us with a vast majority of our entertainment. For example, if you live in Britain and want to talk to you family in New Zealand, all you need do is pick up your telephone and dial the code for their telephone, and you can speak to them in a second not weeks, which is what it could have taken (via letter writing) before international radio communications had been developed. In fact nowadays, thanks to John Logie Baird&#8217;s invention of television (based on the same principles of radio communication), you can &#8216;video-chat&#8217; to your family, and actually see an image of them as you talk, or send an instantaneous &#8216;virtual&#8217; (i.e. doesn&#8217;t really exist) letter via e-mail.</p>
<p>The properties that radio waves have also forms the basis for radar detection, a technique that was very influential on the outcome of the Second World War.</p>
<p>Around the time of Marconi&#8217;s first Trans-Atlantic radio transmissions, there was a lot of research beginning into the very nature of matter itself. The beginnings of nuclear physics started with J. J. Thompson in <em>1897</em> when (under improved conditions of vacuum) he discovered that cathode rays could be influenced by a magnetic field. As a result he thought that the rays had a particle like nature, and he therefore tried to measure the charge and mass of this particle. In each experiment he found that the particle has a charge opposite that of a hydrogen nucleus, but was about 1000 times less massive (lighter), no matter what the material was that he used for the cathode. This particle became known as the electron.</p>
<p>Later Rutherford took over from Thompson&#8217;s studies and after many experiments on the radioactive decay found in uranium by Becquerel in <em>1896</em> and in polonium by Curie in <em>1898</em> he deduced (also) in <em>1898</em>, that there were two types of radioactive ray, alpha and beta. Alpha rays were the weakest, and are helium nuclei; beta rays were stronger and were like Thomson&#8217;s electron. Later on in <em>1900</em>, he also discovered gamma rays, which are very high-energy electromagnetic waves (more evidence for Maxwell&#8217;s idea). But it was while working with Geiger that he produced his most important discovery. This was that when alpha particles were fired at a piece of gold foil, they all passed through, except the odd one, which came straight back. In his own words, he described it as &#8216;…quite the most incredible event that has ever happened to me in my life… It was almost as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you&#8217;. This effect led him to deduce in <em>1911</em> that the atom consisted mostly of empty space with a small positive nucleus. From this Bohr created the modern picture of the atom by applying (old) quantum theory to Rutherford&#8217;s model, and coming up with a solar system type arrangement, whereby electrons orbit the nucleus. Rutherford was also involved in the first nuclear fusion reaction in <em>1934</em> when deuterium was bombarded with deuterium nuclei to form tritium (which are all forms of hydrogen, but with different masses).</p>
<p>These discoveries lead to the complete modern picture of atomic structure, which is very important to chemists as it describes how and why reactions may or may not happen, and has lead to many the creation of many new substances that we use everyday. Also the discovery of radioactive decay is very useful in both medicine and industry, where it is used for detection purposes (but in very different ways). Also there is additional evidence for Maxwell&#8217;s idea of electromagnetic waves, which again encouraged others to look for other different waves, leading to x-rays (useful again in medicine and industry), infra-red which is useful for detecting heat sources (such as human bodies) in otherwise opaque conditions (a smoke filled room for example) and microwaves can be used to cook food very quickly in microwave ovens. Most importantly though it lead to experiments on nuclear fusion. These experiments were important as nuclear fusion lead to the creation of the atomic bomb – not in itself a good use for this principle, but it did show that the theory works. The bomb lead onto the idea of &#8216;clean&#8217; atomic energy generation, which is either good or bad depending upon which way you look at the &#8216;problem&#8217; and also lead to the cold war between Russia and America.</p>
<p>The Second World War, which was the next significant period in history, led to the development of many new things such as jet aeroplanes, rockets, atomic bombs, computers and a whole host of other things besides. The development of these technologies continued after the Second World War and, as I stated before, sparked of the arms race for technical superiority between Russia and America. The cold war was again unfavourable but, in my view at least, it was very useful in getting funding into science and advancing technology at a rate of knots. For example, the technology from building missiles has enabled rockets to be propelled into space, and men to visit the moon. This is the very beginning of our exploration of space and has lead onto the idea of space shuttle services, and possible future colonisation of other planets such as Mars. It has also lead to the vast and excellent international communications network we now have via the ability of placing satellites into orbit. There is also the Hubble telescope which has provide the best pictures we have so far of distant galaxies, and has allowed us to say a bit more about what our universe is, and what our place is in it.</p>
<p>LASER technology is another important advancement out of the cold war. It allows us to communicate along phone cables at the speed of light, listen to digitally recorded music in a way that never degrades the recording, and allows us to have tumours removed or other such precise surgery performed with a minimum of physical damage to us.</p>
<p>Modern technology is increasingly being controlled automatically by computers, mainly through necessity than choice. The first digital computer to be designed was by Babbage in <em>1823 </em>after other related inventions, such as a small calculator that could perform certain computations to 8 decimal places. But the beginnings of modern computers didn&#8217;t surface until around <em>1940</em>, and then it wasn&#8217;t until the <em>1960&#8242;s</em> that they became particularly powerful, with the advent of computer languages and the birth of computer science. But again they have become an integral part of our everyday lives, both in terms of communications and entertainment, as well as other practical applications like controlling the washing machine or the microwave oven.</p>
<p>I think that just about wraps it up for advances in physics as I&#8217;m leaving out the whole of quantum mechanics. This in itself is a very important advance for physicists as it allows them to describe very accurately the microscopic world, where the macroscopic laws of physics don&#8217;t always hold true, but it hasn&#8217;t had much practical application yet other than in things such as LASER&#8217;s and also very tiny integrated electrical circuits, such as on a computer chip.</p>
<p><strong><em>So what about the future?<br />
</em></strong></p>
<p>Well the future still has capacity for many new inventions and advances in physics, even though the subject seems to have already passed &#8216;through its golden age&#8217;. The science of modern physics involves quantum mechanics in every aspect, and is basically the science of the atom, and of the microscopic properties of matter. In this vein it tends towards the more physical aspects of chemistry and materials science, but is different enough not to be considered as such.</p>
<p>Recent developments hold speculation of the possibilities of quantum computing, which is a big buzzword in physics at present. The problems involved with creating a quantum computer are vast, but real efforts are being made to create at least preliminary designs. One of the advantages of such a computer is its possible high speeds and its ability for performing great numbers of calculations in a second. One of the problems with modern computers is that the design, and the basic theory behind the way that they work hasn&#8217;t changed really since Babbage. In a quantum computer however, the whole theory behind the way that the computer performs calculations will have to be rediscovered as the mechanics is so different, and once a computer is built it will be a long time before any real programming can be done on it. It&#8217;s also very unlikely (from a present point of view) that this technology will be available as home PC&#8217;s because it is so fragile. However it could form the basis for a network of home terminals in a community, which could be accessible via a TV set like apparatus, similar to that which is already happening with cable TV!</p>
<p>An alternative (or a compliment) to a quantum computer may also be a computer with circuits that have light flowing through them rather than electricity. This technology is also in its infancy, but it is probably more likely to succeed, as the conditions it requires are much less fragile than those required for a quantum computer. Other reassert on light based computing involves the storage and retrieval of information. A weird and very interesting piece of information storage research is based around small &#8216;glass&#8217; cubes that can hold the same amount of information as 1000 DVD&#8217;s or approximately 700000 floppy disks, using lasers to excite the &#8216;glass&#8217; molecules (ref. New Scientist 10th March 2001, page 25). Note that although light based circuitry and storage would be influenced greatly by quantum effects, it is considered to be different to quantum computing as it still uses the classical computational methods, just in a faster and more efficient way.</p>
<p>The applied science of physics is really moving into communications technology and both these examples show this.</p>
<p>One of the other main branches of physics that warrants attention is also based upon atomic physics, and that is the search for a renewable energy source. This is either by utilising the same fusion reaction that the sun undergoes, or by controlling the energy produced when hydrogen and oxygen react explosively to form water (as is used in rocket fuel). Both of these things are also considered as &#8216;Holy Grails&#8217; in physics at the moment, and again, a fair amount of research is being done on this subject, but there is no real evidence that our cars are to be powered by mini-sun&#8217;s or rocket fuel in the immediate future.</p>
<p>Lastly, although I consider it as a bit of a cliché, there&#8217;s the standard glimpse of space travel and of colonisation of the rest of the universe (or at least the solar system). This is not now so far off, as it seems. In the last couple of days, the first commercial space passenger has returned to earth with a great reception, and more and more probes and equipment are being sent to mars in anticipation of the first manned &#8216;flight&#8217; there. It will be many years before the solar system does start to become colonised, and it will probably proceed via &#8216;planet hopping&#8217; if another suitable host planet (or moon) (other than Mars) exists that can be colonised, as it would take far too long to travel directly from the Earth. This is unless we find a way of moving very fast through space using methods other than the standard rockets we use today &#8211; but who can say. Quantum mechanical experiments may yield new fuels and methods of propulsion, but again these are far off as yet, and may never happen.</p>
<p>For now, we will just have to wait and see&#8230;</p>
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		<title>Simon Singh Review As A Writer</title>
		<link>http://www.christopher-dean.co.uk/2001/05/01/simon-singh-review-as-a-writer/</link>
		<comments>http://www.christopher-dean.co.uk/2001/05/01/simon-singh-review-as-a-writer/#comments</comments>
		<pubDate>Tue, 01 May 2001 05:40:27 +0000</pubDate>
		<dc:creator>Chris</dc:creator>
				<category><![CDATA[Essays]]></category>
		<category><![CDATA[Science]]></category>

		<guid isPermaLink="false">http://www.christopher-dean.co.uk/2007/06/06/simon-singh-review-as-a-writer/</guid>
		<description><![CDATA[Simon Singh&#8217;s parents emigrated from the Punjab in India to Britain in 1950. He grew up in Somerset, and then went to Imperial College, London, to study physics, before completing a PhD in particle physics at Cambridge and CERN. In 1990, he joined the Science Department of the BBC as a producer and director on [...]]]></description>
			<content:encoded><![CDATA[<!-- sphereit start --><p>Simon Singh&#8217;s parents emigrated from the Punjab in India to Britain in 1950. He grew up in Somerset, and then went to Imperial College, London, to study physics, before completing a PhD in particle physics at Cambridge and CERN.</p>
<p>In 1990, he joined the Science Department of the BBC as a producer and director on programmes such as Tomorrow&#8217;s World and Horizon. In 1996 he directed Fermat&#8217;s Last Theorem, which won a BAFTA. <span style="color: red"><br />
</span></p>
<p>He has officially written three books but only two of them are really significantly different that they can really be classed as two separate books, and a re-issue.</p>
<p>The first book was based on his award winning documentary Fermat&#8217;s Last Theorem and was the first book about mathematics to become a No.1 bestseller in the UK. It is basically an extended transcript of the TV documentary and re-tells the story of the origins of Fermat&#8217;s Last Theorem and of Andrew Wiles, and his secret struggle to solve the problem.</p>
<p>The second book was The Code Book, The Secret History Of Codes And Code Breaking and was released in 1999. This book uses the background of historical events such as the plot to assassinate Queen Elizabeth, the Second World War, and the invention of the Internet to explain the science of codes and describes the impact of cryptography on history and our everyday lives.</p>
<p>Both of his books have been based around a scientific or mathematic mystery and secrecy, which is eventually solved using a logical approach. As nearly everyone is curious, or just plain nosy, there is something in both of these books that will appeal to most people. And as his roots lie in physics and mathematics, these are the sort of stories he obviously enjoys talking about.</p>
<p>He tackles the subject more from the human-interest side of things (in an essentially romantic way) rather than from the in depth scientific aspect, as it makes the subject more accessible to more people. After all, the science used in cryptography, and the solution of Fermat&#8217;s Last Theorem, are fairly dry subjects to talk about from just a technical point of view. But then again so is the plain history of Elizabethan England. It is clear that this is not a textbook and is not intended for specialists, however it&#8217;s not tailored for technophobes either. It&#8217;s really there for the majority.</p>
<p>Interestingly enough both of his books have strong links with a TV series. Fermat&#8217;s Last Theorem being the product of a documentary on the BBC and the codebook being the inspiration for the channel four series the science of secrecy. As he was producer of Tomorrow&#8217;s World for such a long time, unless it was his idea in the first place, I think that the formula they use in their reporting has rubbed off on him substantially. Just like Tomorrow&#8217;s World his books are very well planned and thought out, and he is good at getting what is potentially a very difficult subject matter across to a very general audience, the majority of which probably only have a fleeting interest in science. This may also explain why his books tend to lend themselves to TV quite well.</p>
<p>I also feel that he uses the TV series as a powerful tool to promote his books and get people interested in them, but the fact that they have been TV shows at all, proves that his chosen subject matter and his way of telling the story has been regarded highly enough by TV bosses to take the &#8216;gamble&#8217; in producing the series in the first place. And it&#8217;s also indicative of the easily understandable style he has in writing his books, as very few TV programmes are produced for niche audiences.</p>
<p>Off the back of the series the science of secrecy (the series based on the code book) came the re-launch of a &#8216;tarted up&#8217; version of the same book under the same name as the TV series. This is not a bad thing from a publishing point of view, but it should be noted (and also by his own admission) these are virtually the same book.</p>
<p>Both books are written in an enthusiastic narrative style with snippets of facts or relevant digressions to technical explanations that help illustrate what he is talking about, and the relative importance of it. However, he knows the difference between writing a book and a scientific paper, and these digressions serve to quench the reader&#8217;s curiosity enough to allow them not to worry further about how a certain process or procedure is undertaken, and to get on with reading the rest of the book, without becoming incomprehensibly technical themselves.</p>
<p>An example of this would be from The Code Book, when he is talking about the future of codes and code breaking in relation to quantum mechanics, and possibly quantum computing. The level of technical detail required to describe this process is vast, and unless you&#8217;re familiar with it, is very difficult to understand. I think that he does well in describing the basics of how the process may work without even attempting any of the basic (but initially tricky) mathematics needed to grasp this subject, and understand it fully.</p>
<p>Although these explanations are good, if you are not at all mathematically minded it&#8217;s also possible to skip parts of the more involved details if you want to. This is probably where his own mathematical ability is played off against his journalistic instincts, but he never really cascades into anything very complex.</p>
<p>This is in itself probably one of the main reasons he is such a popular science writer. He has the knack of being able to intertwine the basics of some very technical concepts with an exciting and inspiring storyline, that treads the fine line between being a technical text book and a full-on mystery novel. Saying that, he is also quite good at testing the reader&#8217;s ability as the book progresses and again, this isn&#8217;t a bad thing as it is a good way of keeping you constantly involved with the book.</p>
<p>Most people know one of the Sherlock Holmes stories, The Hound of the Baskervilles for example, or at least know of Sherlock Holmes&#8217; reputation for solving puzzles logically. That is the best comparison I can make for the style of his books. They are like a good Sherlock Holmes or other detective story for that matter. The only difference is that they arte based on real situations and the characters are actually real people. And he himself plays Doctor Watson (the chronicler of Holmes&#8217; adventures) to a variety of incarnations of Sherlock, and as such he gets his admiration and respect for his characters over, along with the intellectual excitement experienced when actually solving these problems, again without going into details. However Sherlock Holmes is a very idealised character and Singh also tends to idolise his subjects too.</p>
<p>As he is very enthusiastic about this subject matter, he has a tendency to present all of his characters as heroes. This is not a bad thing as such, because the majority of them are in their own right, and it keeps the reader interested, but he can make it seem as though these people had more to do with advancing the quest for the answer, whether it be the solution to Fermat or an unbreakable code than they may have actually done, just for the sake of a good story. However, as the subjects he talks about are all potentially dry and uninteresting, this slightly possibly unrealistic view of the people involved is needed to keep you wanting more and wanting to know what happens next. Without them, the whole story could appear very bland to a general audience!</p>
<p>An example of this may be in the first few lines of Fermat&#8217;s Last Theorem. Here he states that &#8220;The most important mathematics lecture this century look place in Cambridge on June 23<sup>rd</sup> 1993. Two Hundred mathematicians were transfixed, although only a handful of them understood the fully the dense mixture of Greek symbols on the black board. The others were there to observe what they hoped would be a historical occasion. The solution to Fermat&#8217;s Last Theorem, a problem that had plagued the greatest mathematical minds in the world for over three centuries.&#8221;</p>
<p>Firstly it&#8217;s quite reasonable to assume that this wasn&#8217;t the &#8220;most important mathematics lecture this century&#8221;, as it was only important to the mathematicians who were particularly bothered with its solution, and it has no great universal power over us all. After all it&#8217;s not as if it disproves that</p>
<p>1 + 1 = 2, and I&#8217;m sure that in the scheme of things, the solution to Fermat&#8217;s Last Theorem isn&#8217;t as powerful as some of the other advances last century. Of course I could be wrong….</p>
<p>Secondly this paints the picture of Wiles at the blackboard as the genius who has, only now after 300 years, been able to solve this marvellous problem, with his colleagues staring up at him in awe. I realise that this all comes down to personal perception of the words, but this is how I see it.</p>
<p>Lastly though it serves as a brilliant opening line to get you thinking about Wiles, why he was the only one, so far to really solve the problem, and of course, what this centuries old problem is actually all about. After all, human nature is inquisitive and everyone likes a mystery!</p>
<p>Both books have a beginning, middle and an end, as you would expect, but they vary somewhat. Both books are essentially technological history books, and start that way. Fermat&#8217;s last theorem starts (after the brief introduction to what the book is going to be about) with a description of the Pythagorean brotherhood and the birth of modern science. And the code book starts using one of the first important examples (in British history) where code breaking was used to save the life of Queen Elizabeth. These beginnings are good as they set the origins of the stories in context and provide a good hook to get you into the rest of the book.</p>
<p>The middle is obvious in Fermat&#8217;s Last Theorem, as it where he stops talking about the distant past and starts to get on with the story. In the code book however there isn&#8217;t really a middle as each chapter is set up as more or less a self contained story, but it seems to come about when he changes from talking about cracking codes by hand, to having to use computers and more modern technology.</p>
<p>The end in Fermat&#8217;s last theorem is the main climax of the book also with the actual presentation of wiles proof to the world, but in the code book the ending is more obscure still. The code book is ended by a glimpse into the future and speculation on what may happen, and although this is fairly inspiring to the reader as they have just learned how to crack codes for themselves, it&#8217;s also a pretty standard and unimaginative way to end a book. What was clever though was to include a huge puzzle to be decrypted for a cash prize at the end. This really does pull the end of the book up well because a t the end of the code book the read feel s that he has learned a lot and has a lot of new knowledge that he wants to try out for himself, and this puzzle provides a good vent for the enthusiasm for the subject produced. Without it I think the reader may have felt a little cheated as he has had an introduction to the subject of code breaking, and has learned some techniques on how to solve codes, but doesn&#8217;t really have the technical knowledge (without doing a massive amount of further reading) to attempt real world examples. So instead as I&#8217;ve just said, he provides a very useful &#8216;vent&#8217; for this potentially frustrated enthusiasm. Also the solution to the final puzzle could yield a £10,000 prize, which is an excellent reason to buy the book in the first place, even if you previously had no real interest in the subject matter.</p>
<p>Both his books also have a knack of making you feel that you want to go out and learn more about the subject in hand. The way that he writes is very inspiring and encourages you to do this, via references to other relevant material at the end.</p>
<p>In both of his books it&#8217;s easy to see that he has a passion for his subject matter, and is very good at writing about it. Really, number theory, ciphers, coding, and aspects of history can be very dry subjects, but he adds enthusiasm to them. His characters are real people, his stories of them are real, his highs and lows are real, and it&#8217;s very hard to get away from that.</p>
<p>He tells the story in hand using many shorter sub-stories each with their own small beginning, middle and end, this causes the reader to never looses interest in what is being said as it&#8217;s something slightly different each time. For example Fermat&#8217;s Last Theorem could really have been a 10 page pamphlet on how the problem was eventually solved, but instead he builds up to the climax using the stories of the more interesting people who attempted the problem.</p>
<p>Money or reward seems to be a constant theme in his books as well. In Fermat he uses the fact that the unlocking of the theorem yields a cash prize and other fame and fortune, if it&#8217;s solved. In code breaker he puts up £10,000 in prize money himself, and further more for the science of secrecy he puts up a holiday to Egypt.</p>
<p>Lastly I&#8217;ll just say that the books are very entertaining (probably due to the reader participation), but there is little or no humour in any of them, this is not to say that he is humourless, merely that his stories and the way in which they are presented don&#8217;t require the use of humour to keep the reader interested, instead he makes you feel good about the characters he is writing about when they feel elated and sad when they are sad, and this is enough.</p>
<p><strong><em>So what can we learn from Simon Singh?</em></strong></p>
<p>He is good at matching an exciting and interesting story line with just the right amount of technical detail, to get the importance and difficulty of the subject matter across without actually going in depth into any abstract maths or other concepts that may confuse and annoy the reader.</p>
<p>He presents both heroes and villains in his work just as a good novelist would and treats them accordingly.</p>
<p>He chooses good descriptions in order to make you feel the same way he obviously does about his subject matter.</p>
<p>He basically works from the human-interest side of science in order to produce a generally popular book.</p>
<p>And Lastly hew never really leaves the audience along to get bored as he is always testing them or starting a new sub-story within the context of the main book.</p>
<p><strong><em>His Books:<br />
</em></strong></p>
<p><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2121-simonsinghr1.png" title="Simon Singh Review As A Writer" alt="060607 2121 simonsinghr1 Simon Singh Review As A Writer" /><br />
<img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2121-simonsinghr2.png" title="Simon Singh Review As A Writer" alt="060607 2121 simonsinghr2 Simon Singh Review As A Writer" /><br />
<img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2121-simonsinghr3.png" title="Simon Singh Review As A Writer" alt="060607 2121 simonsinghr3 Simon Singh Review As A Writer" /></p>
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		<title>The Pumping Speed of a Rotary Pump</title>
		<link>http://www.christopher-dean.co.uk/2001/04/27/the-pumping-speed-of-a-rotary-pump/</link>
		<comments>http://www.christopher-dean.co.uk/2001/04/27/the-pumping-speed-of-a-rotary-pump/#comments</comments>
		<pubDate>Fri, 27 Apr 2001 09:48:42 +0000</pubDate>
		<dc:creator>Chris</dc:creator>
				<category><![CDATA[Essays]]></category>
		<category><![CDATA[Science]]></category>

		<guid isPermaLink="false">http://www.christopher-dean.co.uk/2007/06/06/the-pumping-speed-of-a-rotary-pump/</guid>
		<description><![CDATA[Aim: The aim of this experiment was to observe the different factors that influence the effective pumping speed of a rotary pump. Background: The very first pumps were used for raising water and mining, as they still are today, but nowadays pumps have many more applications, other than just moving fluids around (although that&#8217;s still [...]]]></description>
			<content:encoded><![CDATA[<!-- sphereit start --><p style="margin-left: 21pt"><strong>Aim:<br />
</strong></p>
<p>The aim of this experiment was to observe the different factors that influence the effective pumping speed of a rotary pump.</p>
<p><strong>Background:<br />
</strong></p>
<p>The very first pumps were used for raising water and mining, as they still are today, but nowadays pumps have many more applications, other than just moving fluids around (although that&#8217;s still essentially all they do).</p>
<p>In this experiment the pump is used to create a vacuum in a chamber, and is based upon one of the first useful designs for such a piece of apparatus. Vacuum pumps have a great deal of practical applications in today&#8217;s world and not only for physics experiments. They are used for example, to create the vacuums in light bulbs or televisions tubes, for producing an atmosphere devoid of any gasses that may react (either chemically or physically) with sensitive apparatus as in the making of complex electrical circuits, and sometimes in other industrial processes where a vacuum is necessary, such as freeze-drying or the melting of reactive metals such as titanium. In general though, pumps are classified into one of three groups depending on the way in which energy is imparted to the fluid in order to move it from one place to another:</p>
<ol>
<li>Volumetric displacement</li>
<li>Addition of kinetic energy</li>
<li>Use of electromagnetic force.</li>
</ol>
<p>Volumetric displacement is where the fluid is transported either mechanically (by the sweeping action of a rotating vane – as in this experiment) or by the use of another fluid (using air to pump water from a coal mine, for example). A pump that mechanically displaces the fluid is generally known as a positive displacement pump<span style="color: red"><br />
</span>and generally moves relatively low volumes of fluid at high pressure.<span style="color: red"><br />
</span></p>
<p>The addition of kinetic energy to a fluid can be achieved by either using an impeller rotating at high speed (cf the rotating vane &#8211; above) or by providing an impulse in the direction of flow (in a bicycle pump for example). And generally kinetic pumps<strong><br />
</strong>impel high volumes at low pressure.</p>
<p>And lastly if we were to use an electromagnetic force to provide the pumping action, the fluid has to be a good electrical conductor in order to be moved at all.</p>
<p>The pump that is being considered in this experiment is the oil sealed rotary vane pump, and can be considered as both a volumetric displacement and a kinetic pump, all rolled into one. It is very useful as a &#8216;roughing&#8217; pump and can be used to evacuate the majority of gas (at around atmospheric pressure) from a closed vessel before a more specialised pump, such as diffusion pump, is used to increase the vacuum further for use in applications such as electrical micro-circuit production.</p>
<p>Gaede first developed the rotary vane pump in 1905, and it has three main parts, these are: The stator (pump housing), the rotor and vanes, and lastly the oil seal.</p>
<p>The stator is really a large sealed tube that houses the rotor and vanes, and has both an inlet and exhaust hole. The rotor is a bar that runs along the length of the stator and has the (spring loaded) vanes (solid plates, similar to the blades of a propeller) attached to it. The vanes are kept in contact with the stator wall via an oil coating, which is used as a seal between each &#8216;compartment&#8217; formed by any two neighbouring vanes and the stator wall. The whole pump set up is finally surrounded by a jacket of oil to minimise gas leaking to and from the atmosphere.</p>
<p>The pump basically works according to the following method:</p>
<ol>
<li><span style="font-size: 10pt; font-family: Times New Roman">The inlet valve between the pump and the system to be evacuated is opened to increase the system&#8217;s volume. Gas expands to fill the volume and therefore enters the pump. The valve is then closed.<br />
</span></li>
</ol>
<ol>
<li><span style="font-size: 10pt; font-family: Times New Roman">The rotor (and vanes) inside the stator rotate, and effectively move the gas from one greater volume &#8216;side&#8217; of the pump to the other smaller volume side. This process causes the gas to be compressed in the smaller volume and also therefore leaves a vacuum in the &#8216;compartment&#8217; where the gas once was.<br />
</span></li>
</ol>
<ol>
<li><span style="font-size: 10pt; font-family: Times New Roman">Finally, when the gas is compressed so much that its pressure is greater than that of the atmosphere (which holds a valve to the outside shut), it pushes open the exhaust valve and is expelled into the atmosphere.<br />
</span></li>
</ol>
<p>The pumping speed (the quantity of gas moved per second) is greatest from atmospheric pressure to about one torr, then as the ultimate pressure (the pressure to which the pump is capable of pumping) is reached the speed decreases to zero. The ultimate pressure for this pump is about 10<sup>-1</sup> torr or 10<sup>-3</sup> torr if two such pumps are connected in series. For this pump the ultimate pressure is reached when the mean free path of the gas molecules is greater than the dimensions of the vessel being evacuated, and therefore very few molecules actually pass into the mouth of the pump. If the mouth of the pump (the size of the inlet valve) was variable them we may have been able to achieve slightly lower pressures by increasing its size. However at this stage a second pump (such as a diffusion pump) really needs to be employed to significantly lower the pressure any more. This is why the rotary vane pump is a good roughing pump for other more specialised devices such as the diffusion pump.</p>
<p>Lastly the ultimate pressures possible are limited by leakage between the high and low-pressure &#8216;sides&#8217; of the pump<strong>. </strong>This<strong><br />
</strong>is<strong><br />
</strong>mainly due to non-exhausted gas being carried back to the low-pressure side, or vapours that were once dissolved in the sealing oil evaporating off when exposed again, to the low-pressure side. The decomposition of the oil when exposed to &#8216;hot spots&#8217; caused by friction can also affect the ultimate pressure as can bad pitting on the stator or vane (caused by pumping a corrosive gas, for example) which leads to the oil seal being partially or even completely broken down.</p>
<p><span style="font-size: 10pt; font-family: Times New Roman">Using some sort of cooling apparatus along side the pump may reduce the decomposition of oil and the vapours in the sealing oil can be removed by using either a vapour trap or by gas ballasting.<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman">Gas ballasting is the deliberate input of air (via a one way gas ballast valve) into the high-pressure side of the pump, so that any vapours being pumped do not condense under the high pressure there and pollute the oil. In other words the total pressure on the high-pressure side is increased (so that the outlet valve is opened &#8216;early&#8217;) without increasing the partial pressure of the vapour, so that any vapours can be released before they have the chance to liquefy.<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman">The vapour trap is a more costly and more difficult arrangement to set up as it involves passing the gas being pumped through a chemical that will absorb any vapours before pumping. This would be fairly useful if the only contaminant was, say water vapour, but as any number of different contaminant vapours could be present, gas ballasting is an all-round better way of doing things.<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman">The overall efficiency of the pump mainly depends upon how well the pump is sealed and the quality of the surface finishes throughout the pump.<br />
</span></p>
<p>Lastly, in this experiment we will be using a Pirani Thermal conductivity gauge to measure the low-pressure situations around 10<sup>-1</sup> torr. It determines the pressure based upon the rate at which heat is dissipated from a hot filament. It is basically a Wheatstone<strong><br />
</strong>bridge, with one arm being the heated filament in the vacuum system.<span style="color: #3366ff"><br />
</span>The filament&#8217;s resistance depends upon its temperature, which, in turn, depends on the rate of dissipation of thermal energy through the gas. As the wire is struck by passing gas molecules, it gives up some if its heat to them, thus causing the temperature of the wire and therefore the resistance to drop slightly. The bridge is powered from a constant voltage source, and any unbalanced current (due to the temperature changes) is converted directly to torr.<span style="color: #3366ff"><br />
</span></p>
<h2>A diagram of the pump taken from the lab script is shown on the next page</h2>
<p><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings1.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings1 The Pumping Speed of a Rotary Pump" /><strong>Assumptions:<br />
</strong></p>
<ol>
<li>When working out the volume of the chamber and its pipes and connectors that were to be pumped out we assumed that the valves had the same volume as the pipes they were connecting, the chamber could be considered as a composite of straight sections and domes only, and lastly that all pipe work was connected at sharp right angles and not &#8216;curved around&#8217;.</li>
</ol>
<ol>
<li>That the Pirani gauges held no volume of gas to be pumped what so ever.</li>
</ol>
<ol>
<li>The time taken to open the valves was instantaneous and had no effect on the pumping speed</li>
</ol>
<ol>
<li>For the first experiment we could have closed V1 and opened V2 to let the air from the chamber fill the pipes as much as possible before entering the pump, and therefore give us a better idea of the effective pumping speed. This would be difficult in the second experiment however, as the gas would have taken some time to fill the pipe work through the aperture, and then subsequent pumping would not have give an accurate representation of the effect of the restrictive aperture. And as we later wanted to compare both values for the effective pumping speed, we decided that we could make a better comparison if this was not done in either experiment. The result being that we assume that the pump is evacuating the chamber directly and not the associated pipe work past V2.</li>
</ol>
<ol>
<li>That the dimensions of the space to be evacuated are as stated in the results analysis.</li>
</ol>
<ol>
<li>The temperature of the room is constant</li>
</ol>
<ol>
<li>The quantity of gas entering the pump is the same as that expelled from the atmosphere, i.e. that the pump is working perfectly</li>
</ol>
<p><strong>Method:<br />
</strong></p>
<ol>
<li>The apparatus was carefully assembled as in the following diagram (from the lab script) except the blank off, which was already permanently sealed by glass, and not a bung or other kind of removable seal:</li>
</ol>
<p><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings2.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings2 The Pumping Speed of a Rotary Pump" /></p>
<ol>
<li>AA1 (the Air Admittance Valve) was closed off and the rotary vane pump was started (from now on I will refer to the rotary vane pump just as &#8216;the pump&#8217; as it is the only pump being used in this experiment) with the gas ballast off. As the glugging of the pump stopped after a few moments, we knew that we didn&#8217;t have any large leaks.</li>
</ol>
<ol>
<li>Next with the ¼ swing valve closed we attached P1 to the control box and switched it on. We then slowly opened V1 and V3 and saw the gauge reading fall down to just below 10<sup>-1</sup> torr so we knew that there were no leaks in our system of pipes.</li>
</ol>
<ol>
<li>We then mounted the glass chamber (after checking it for cracks and other obvious imperfections) on top of its &#8216;connector&#8217; and mounted the blanking tube on top of the chamber. We then put the metal safety cage around the chamber and connected P2.</li>
</ol>
<ol>
<li>Then we closed both AA2 and V3 and slowly opened V2 to pump the chamber out as a test run. The chamber pressure reached just below 10<sup>-1</sup> torr, so we knew we were all right to start the experiment for real.</li>
</ol>
<ol>
<li>We closed V1 to cut the pump off from the chamber and then very slowly opened the air inlet so that the chamber was returned to atmospheric pressure. Once the chamber was returned to normal pressure the air inlet was closed again, V2 was closed and we opened V1 to evacuate the tubing up to V2. We then opened the gas ballast valve for 10 minutes to attempt to remove any unwanted vapours dissolved in the oil, and achieve the best results from the pump.</li>
</ol>
<ol>
<li>Next we took the measurements we required to determine the effective pumping speed of the pump. We both drew a copy of <em>table 1</em> (see results), and using a stopwatch timed how long the pump took to evacuate the chamber down to the specified pressures in the table. To give some idea of our reaction time errors we repeated the experiment once each.</li>
</ol>
<ol>
<li>After taking this data we restored the chamber pressure (as before) and isolated the pump by closing V1.</li>
</ol>
<ol>
<li>For our second experiment we fitted a small disk with a 1mm hole in the centre between the pipe work and the chamber base so that we could measure the effect of such a restrictive aperture on the pumping speed. We then evacuated the pipe work up to V2 as before and opened the gas ballast valve for 10 minutes.</li>
</ol>
<ol>
<li>Once the aperture was in place the experiment was completed as the previous one was, but only one of us took readings due to time restrictions (see <em>table 2</em>).</li>
</ol>
<p><strong>Results:<br />
</strong></p>
<p><span style="font-size: 9pt; font-family: Arial"><strong><em>Table 1:    <br />
</em></strong></span></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings3.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings3 The Pumping Speed of a Rotary Pump" /><span style="font-size: 9pt; font-family: Arial"><br />
</span></p>
<p><span style="font-size: 9pt; font-family: Arial"><strong><em>Graph 1:<br />
</em></strong></span></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings4.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings4 The Pumping Speed of a Rotary Pump" /><span style="font-size: 9pt; font-family: Arial"><br />
</span></p>
<p><span style="font-size: 9pt; font-family: Arial"><strong><em>Table 2:    <br />
</em></strong></span></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings5.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings5 The Pumping Speed of a Rotary Pump" /><span style="font-size: 9pt; font-family: Arial"><br />
</span></p>
<p><span style="font-size: 9pt; font-family: Arial"><strong><em>Graph 2 (a):<br />
</em></strong></span></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings6.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings6 The Pumping Speed of a Rotary Pump" /><span style="font-size: 9pt; font-family: Arial"><br />
</span></p>
<p><strong>Sources of Errors:<br />
</strong></p>
<p>A big error in our results is obviously due to our own personal reaction times, we have attempted to minimise this by taking two lots of results and averaging them, although this is still not perfect. Ideally we would have used an electronic data logger to record our results, or we would have had to repeat the experiment many times to obtain a representative distribution of results.</p>
<p>Another error comes from the accuracy of our meters and measurements as are shown in the tables.</p>
<p>The other errors could be due to not opening valves quick enough and small undetected leaks in the apparatus.</p>
<p>In the results analysis below the major errors come from the slopes of the graphs (via <span style="text-decoration: underline">choosing</span> the best fit) and even more so, the assumptions involved in approximating the volume to be pumped.</p>
<p><strong>Results Analysis:<br />
</strong></p>
<p><strong><em>Experiment 1:<br />
</em></strong></p>
<p>We want measure the effective pumping speed of the pump by using it to evacuate the chamber and measuring how long it takes to get from one pressure p<sub>a</sub> (760 torr, atmospheric pressure) to a lower pressure p<sub>b</sub>.</p>
<p>If a volume dV of air from the chamber enters the pump in a time dt its speed is given by: <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings7.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings7 The Pumping Speed of a Rotary Pump" /></p>
<p>From Boyle&#8217;s Law we know that pV is constant at constant T, so: <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings8.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings8 The Pumping Speed of a Rotary Pump" /></p>
<p>This can be re-written as: <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings9.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings9 The Pumping Speed of a Rotary Pump" /></p>
<p>It follows that: <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings10.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings10 The Pumping Speed of a Rotary Pump" /></p>
<p>And: <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings11.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings11 The Pumping Speed of a Rotary Pump" /></p>
<p>And when we plot a graph of p against t on semi-log axes we can observe 2 linear regions on the graph. And the corresponding plot of ln p against t has a gradient of <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings12.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings12 The Pumping Speed of a Rotary Pump" /></p>
<p>The first slope has a gradient of <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings13.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings13 The Pumping Speed of a Rotary Pump" /></p>
<p>The second slope has a gradient of <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings14.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings14 The Pumping Speed of a Rotary Pump" /></p>
<p>The volume to be evacuated is the same in both scenarios, and this is calculated (approximately, with reference to diagram) as follows:</p>
<p><span style="font-size: 10pt"><span style="font-family: Times New Roman"><em>Note</em>: All measurements are </span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 0.5 mm<br />
</span></span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman; text-decoration: underline">Measurements:<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman"><em>Section A:</em>     Dimensions:    Length = 115 mm;    Internal Diameter = 25 mm<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman"><em>Section B:</em>     Dimensions:    Length = 101 mm;    Internal Diameter = 24 mm<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman"><em>Section C (i):</em>     Dimensions:    Length = 53 mm;    Internal Diameter = 27 mm<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman"><em>Section C (ii):</em>     Dimensions:    Length = 155 mm;    Internal Diameter = 15 mm<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman"><em>Section D (i):</em>     Dimensions:    Length = 43 mm;    Internal Diameter = 82 mm<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman"><em>Section D (ii):</em>     Dimensions:    Length = 18 mm;    Internal Diameter = 59 mm<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman"><em>Section E:</em>     Dimensions:    Length = 140 mm;    Internal Diameter = 26 mm<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman"><em>Section F:</em>     Dimensions:    Length = 20 mm;    Internal Diameter = 7 mm<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman"><em>Section G:</em>     Dimensions:    Length = 70 mm;    Internal Diameter = 15 mm<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman"><em>Section H:</em>     Dimensions:    Length = 150 mm;    Internal Diameter = 140 mm<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman">Note: I am assuming further that the shaded parts of the diagram on the next page do not contribute to the overall volume being pumped<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman; text-decoration: underline">Volumes:<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman"><em>Section A:    <br />
</em></span></p>
<p><span style="font-size: 10pt"><span style="font-family: Times New Roman">Cylinder = </span><span style="font-family: Symbol">p</span><span style="font-family: Times New Roman">(0.5 </span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 0.025)<sup>2</sup><br />
</span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-family: Times New Roman"><span style="font-size: 10pt"> 0.115 + Dome = 0.5(<sup>4</sup></span><span style="font-size: 9pt">/</span><span style="font-size: 10pt"><sub>3</sub></span></span><span style="font-size: 10pt"><span style="font-family: Symbol">p</span><span style="font-family: Times New Roman"> (0.5 </span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 0.025)<sup>3</sup>) = 6.05 </span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt"><span style="font-family: Times New Roman"> 10<sup>-5</sup>m<sup>3</sup> (</span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 0.02</span></span><span style="font-size: 8pt; font-family: Arial"> X</span><span style="font-size: 10pt; font-family: Times New Roman"> 10<sup>-5</sup> m<sup>3</sup>)<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman"><em>Section B:</em>     <br />
</span></p>
<p><span style="font-size: 10pt"><span style="font-family: Times New Roman">Cylinder = </span><span style="font-family: Symbol">p</span><span style="font-family: Times New Roman">(0.5 </span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 0.024)<sup>2</sup><br />
</span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 0.101 = 4.57 </span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt"><span style="font-family: Times New Roman"> 10<sup>-5</sup> m<sup>3 </sup>(</span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 0.02</span></span><span style="font-size: 8pt; font-family: Arial"> X</span><span style="font-size: 10pt; font-family: Times New Roman"> 10<sup>-5</sup> m<sup>3</sup>)<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman"><em>Section C:</em>     <br />
</span></p>
<p><span style="font-size: 10pt"><span style="font-family: Times New Roman">Cylinder = </span><span style="font-family: Symbol">p</span><span style="font-family: Times New Roman">(0.5 </span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 0.027)<sup>2</sup></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt"><span style="font-family: Times New Roman">0.053 + Cylinder = </span><span style="font-family: Symbol">p</span><span style="font-family: Times New Roman">(0.5 </span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 0.015)<sup>2</sup></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman">0.155 = 5.07</span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt"><span style="font-family: Times New Roman">10<sup>-5</sup> m<sup>3 </sup>(</span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 0.02</span></span><span style="font-size: 8pt; font-family: Arial"> X</span><span style="font-size: 10pt; font-family: Times New Roman"> 10<sup>-5</sup> m<sup>3</sup>)<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman"><em>Section D:</em>     <br />
</span></p>
<p><span style="font-size: 10pt"><span style="font-family: Times New Roman">Cylinder = </span><span style="font-family: Symbol">p</span><span style="font-family: Times New Roman">(0.5</span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman">0.082)<sup>2</sup><br />
</span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt"><span style="font-family: Times New Roman"> 0.043 + Cylinder = </span><span style="font-family: Symbol">p</span><span style="font-family: Times New Roman">(0.5</span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman">0.059)<sup>2</sup><br />
</span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 0.018 = 2.76</span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt"><span style="font-family: Times New Roman">10<sup>-4</sup> m<sup>3 </sup>(</span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 0.02</span></span><span style="font-size: 8pt; font-family: Arial"> X</span><span style="font-size: 10pt; font-family: Times New Roman"> 10<sup>-5</sup> m<sup>3</sup>)<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman"><em>Section E:</em>     <br />
</span></p>
<p><span style="font-size: 10pt"><span style="font-family: Times New Roman">-Cylinder = </span><span style="font-family: Symbol">p</span><span style="font-family: Times New Roman">(0.5</span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman">0.026)<sup>2</sup><br />
</span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-family: Times New Roman"><span style="font-size: 10pt"> 0.140 + Dome = 0.5(<sup>4</sup></span><span style="font-size: 9pt">/</span><span style="font-size: 10pt"><sub>3</sub></span></span><span style="font-size: 10pt"><span style="font-family: Symbol">p</span><span style="font-family: Times New Roman"> (0.5</span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman">0.026)<sup>3</sup>) = -7.89 </span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt"><span style="font-family: Times New Roman"> 10<sup>-5 </sup>m<sup>3 </sup>(</span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 0.02</span></span><span style="font-size: 8pt; font-family: Arial"> X</span><span style="font-size: 10pt; font-family: Times New Roman"> 10<sup>-5</sup> m<sup>3</sup>)<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman"><em>Section F:</em>     <br />
</span></p>
<p><span style="font-size: 10pt"><span style="font-family: Times New Roman">Cylinder = </span><span style="font-family: Symbol">p</span><span style="font-family: Times New Roman">(0.5 </span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 0.007)<sup>2</sup><br />
</span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 0.020 = 7.70 </span><span style="font-size: 8pt; font-family: Arial">X </span><span style="font-size: 10pt"><span style="font-family: Times New Roman">10<sup>-7</sup> m<sup>3 </sup>(</span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 0.02</span></span><span style="font-size: 8pt; font-family: Arial"> X</span><span style="font-size: 10pt; font-family: Times New Roman"> 10<sup>-5</sup> m<sup>3</sup>)<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman"><em>Section G:</em>     <br />
</span></p>
<p><span style="font-size: 10pt"><span style="font-family: Times New Roman">Cylinder = </span><span style="font-family: Symbol">p</span><span style="font-family: Times New Roman">(0.5 </span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 0.015)<sup>2</sup><br />
</span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 0.070 = 1.24 </span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt"><span style="font-family: Times New Roman"> 10<sup>-5</sup> m<sup>3 </sup>(</span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 0.02</span></span><span style="font-size: 8pt; font-family: Arial"> X</span><span style="font-size: 10pt; font-family: Times New Roman"> 10<sup>-5</sup> m<sup>3</sup>)<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman"><em>Section H:</em>     <br />
</span></p>
<p><span style="font-size: 10pt"><span style="font-family: Times New Roman">Cylinder = </span><span style="font-family: Symbol">p</span><span style="font-family: Times New Roman">(0.5 </span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 0.140)<sup>2</sup><br />
</span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-family: Times New Roman"><span style="font-size: 10pt"> 150 + Dome = 0.5(<sup>4</sup></span><span style="font-size: 9pt">/</span><span style="font-size: 10pt"><sub>3</sub></span></span><span style="font-size: 10pt"><span style="font-family: Symbol">p</span><span style="font-family: Times New Roman"> (0.5 </span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 0.140)<sup>3</sup>) = 3.03 </span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt"><span style="font-family: Times New Roman"> 10<sup>-3</sup> m<sup>3 </sup>(</span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 0.02</span></span><span style="font-size: 8pt; font-family: Arial"> X</span><span style="font-size: 10pt; font-family: Times New Roman"> 10<sup>-5</sup> m<sup>3</sup>)<br />
</span></p>
<p><span style="font-size: 10pt"><span style="font-family: Times New Roman">Total Volume Evacuated </span><span style="font-family: Symbol">»</span><span style="font-family: Times New Roman"> 3.40 </span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt"><span style="font-family: Times New Roman"> 10<sup>-3</sup> m<sup>3</sup> (</span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 0.02</span></span><span style="font-size: 8pt; font-family: Arial"> X</span><span style="font-size: 10pt"><span style="font-family: Times New Roman"> 10<sup>-5</sup> m<sup>3</sup>) = 3.4 litres (</span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 0.2</span></span><span style="font-size: 8pt; font-family: Arial"> X</span><span style="font-size: 10pt; font-family: Times New Roman"> 10<sup>-3</sup> litres)<br />
</span></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings15.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings15 The Pumping Speed of a Rotary Pump" /><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings16.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings16 The Pumping Speed of a Rotary Pump" /></p>
<p>Gradient 1 = &#8211; 0.34 s<sup>-1</sup> = <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings17.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings17 The Pumping Speed of a Rotary Pump" /><span style="font-family: Symbol">Þ</span> S = 0.34V = 0.34 s<sup>-1</sup><br />
<span style="font-size: 8pt; font-family: Arial">X</span> 3.40 <span style="font-size: 8pt; font-family: Arial">X</span> 10<sup>-3</sup> m<sup>3</sup> = 1.156 <span style="font-size: 8pt; font-family: Arial">X</span> 10<sup>-3</sup> m<sup>3</sup> s<sup>-1</sup></p>
<p>= 69.36 l min<sup>-1</sup></p>
<p><span style="font-size: 10pt"><span style="font-family: Times New Roman">Gradient 2 = &#8211; 0.01 s<sup>-1</sup> = <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings18.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings18 The Pumping Speed of a Rotary Pump" /></span><span style="font-family: Symbol">Þ</span><span style="font-family: Times New Roman"> S = 0.01V = 0.01 s<sup>-1</sup><br />
</span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 3.40 </span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 10<sup>-3</sup> m<sup>3</sup> = 3.4 </span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 10<sup>-5</sup> m<sup>3</sup> s<sup>-1<br />
</sup></span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman">= 2.04 l min<sup>-1</sup><br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman; text-decoration: underline">Deducing C:<br />
</span></p>
<p><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings19.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings19 The Pumping Speed of a Rotary Pump" />    <span style="font-family: Symbol">Þ</span><br />
<img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings20.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings20 The Pumping Speed of a Rotary Pump" />l min<sup>-1<br />
</sup></p>
<p><span style="font-size: 10pt; font-family: Times New Roman">C = 3.77 </span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt"><span style="font-family: Times New Roman"> 10<sup>-3</sup> m<sup>3</sup> s<sup>-1</sup> (</span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 5 </span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 10<sup>-3</sup> m<sup>3</sup> s<sup>-1</sup>)<br />
</span></p>
<p><strong><em>Table of Effective Pumping Speeds for Experiment 1:<br />
</em></strong></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings21.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings21 The Pumping Speed of a Rotary Pump" /><strong><em><br />
</em></strong></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings22.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings22 The Pumping Speed of a Rotary Pump" /><strong><em><br />
</em></strong></p>
<p><strong><em>Experiment 2:<br />
</em></strong></p>
<p>Plotting the graph on semi-log axes allows us to again see 2 linear regions on the graph and therefore the plot of ln p against t has a gradient of <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings23.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings23 The Pumping Speed of a Rotary Pump" /></p>
<p>The first slope has a gradient of <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings24.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings24 The Pumping Speed of a Rotary Pump" /></p>
<p>The second slope has a gradient of <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings25.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings25 The Pumping Speed of a Rotary Pump" /></p>
<p>Gradient 1 = -0.04 s<sup>-1</sup> = <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings26.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings26 The Pumping Speed of a Rotary Pump" /><span style="font-family: Symbol">Þ</span> S = 0.04V = 0.04 s<sup>-1</sup><br />
<span style="font-size: 8pt; font-family: Arial">X</span> 3.40 <span style="font-size: 8pt; font-family: Arial">X</span> 10<sup>-3</sup> m<sup>3</sup> = 1.36 <span style="font-size: 8pt; font-family: Arial">X</span> 10<sup>-4</sup> m<sup>3</sup> s<sup>-1</sup></p>
<p>= 8.16 l min<sup>-1</sup></p>
<p><span style="font-size: 9pt"><span style="font-family: Times New Roman">Gradient 2 = -7.76 </span><span style="font-family: Arial">X</span><span style="font-family: Times New Roman"> 10<sup>-3</sup> s<sup>-1</sup> = <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings27.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings27 The Pumping Speed of a Rotary Pump" /></span><span style="font-family: Symbol">Þ</span><span style="font-family: Times New Roman"> S = 7.76 </span><span style="font-family: Arial">X</span><span style="font-family: Times New Roman"> 10<sup>-3</sup>V = 7.76 </span><span style="font-family: Arial">X</span><span style="font-family: Times New Roman"> 10<sup>-3</sup> s<sup>-1</sup><br />
</span><span style="font-family: Arial">X</span><span style="font-family: Times New Roman"> 3.40 </span><span style="font-family: Arial">X</span><span style="font-family: Times New Roman"> 10<sup>-3</sup> m<sup>3</sup> = 2.64 </span><span style="font-family: Arial">X</span><span style="font-family: Times New Roman"> 10<sup>-5</sup> m<sup>3</sup> s<sup>-1<br />
</sup></span></span></p>
<p><span style="font-size: 9pt; font-family: Times New Roman">= 1.58 l min<sup>-1</sup><br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman; text-decoration: underline">Deducing C:<br />
</span></p>
<p><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings28.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings28 The Pumping Speed of a Rotary Pump" />    <span style="font-family: Symbol">Þ</span><br />
<img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings29.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings29 The Pumping Speed of a Rotary Pump" />l min<sup>-1<br />
</sup></p>
<p><span style="font-size: 10pt"><span style="font-family: Times New Roman">C = 0.15 m<sup>3</sup> s<sup>-1</sup> (</span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 5 </span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 10<sup>-3</sup> m<sup>3</sup> s<sup>-1</sup>)<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman; text-decoration: underline">Deducing C (for the aperture):<br />
</span></p>
<p><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings30.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings30 The Pumping Speed of a Rotary Pump" />     NB: C<sub>1</sub> is the value of C from experiment 1</p>
<p><span style="font-family: Symbol">Þ<img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings31.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings31 The Pumping Speed of a Rotary Pump" /></span>l min<sup>-1<br />
</sup></p>
<p><span style="font-size: 10pt"><span style="font-family: Times New Roman">C = 0.15 m<sup>3</sup> s<sup>-1</sup> (</span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 5 </span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 10<sup>-3</sup> m<sup>3</sup> s<sup>-1</sup>)<br />
</span></p>
<p><strong><em>Table of Effective Pumping Speeds for Experiment 2:<br />
</em></strong></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings32.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings32 The Pumping Speed of a Rotary Pump" /><strong><em><br />
</em></strong></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2107-thepumpings33.png" title="The Pumping Speed of a Rotary Pump" alt="060607 2107 thepumpings33 The Pumping Speed of a Rotary Pump" /><strong><br />
</strong></p>
<p>On graph 1 we can see that there are effectively two linear regions (roughly) shown by the two dotted lines, with a curved &#8216;cross-over&#8217; section in between them. These correspond to the change over from macroscopic to microscopic gas laws. For the &#8216;first&#8217; linear region (that with the steepest gradient) the assumptions that we make in modelling a macroscopic gas hold true, as there is a great deal of gas in the system. Therefore the pump works normally and achieves its maximum pumping speed. As more and more gas is removed however, these assumptions break down causing the effective pump speed to decrease rapidly, until it virtually stops altogether (when the mean free path of the gas molecules becomes comparable to the dimensions of the chamber).</p>
<p><span style="font-size: 10pt; font-family: Times New Roman">The same kind of effect is shown in the second experiment but is much less pronounced due to the aperture restricting the gas flow from the start.<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman">The conductance of the pipes was 3.77 </span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt"><span style="font-family: Times New Roman"> 10<sup>-3</sup> m<sup>3</sup> s<sup>-1</sup> (</span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 5 </span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt"><span style="font-family: Times New Roman"> 10<sup>-3</sup> m<sup>3</sup> s<sup>-1</sup>). With the aperture the total conductance was 0.15 m<sup>3</sup> s<sup>-1</sup> (</span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 5 </span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 10<sup>-3</sup> m<sup>3</sup> s<sup>-1</sup>) and the conductance of the aperture alone was also<br />
</span></p>
<p><span style="font-size: 10pt"><span style="font-family: Times New Roman">0.15 m<sup>3</sup> s<sup>-1</sup> (</span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 5 </span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 10<sup>-3</sup> m<sup>3</sup> s<sup>-1</sup>). This shows the large effect an aperture has on the conductance of the pipes, and when in place it&#8217;s conductance is effectively the conductance of the pipes. So if we wanted to regulate the conductance of some pipe work for use in pumping apparatus, we could do so in the limit of the diameter of the pipe, by using an aperture of known conductance in the same way as we did here.<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman">On the first graph of pumping speed against pressure you can see more dramatically how the pumping speed decreases with respect to the pressure. But the second graph appears to show us that pumping speed actually increases with the decrease in pressure. However if we examine the data we find that both at high pressure and at low pressures the pumping speed increases with pressure as expected, but there appears to be two different curves showing this effect. I don&#8217;t know the reason for this, it may that our data is wrong, but if not there must, at some point be a cross over between one curve and the other. This change over may be an effect of changing over from, again, the macroscopic to the microscopic gas regimes, but I can&#8217;t be sure.<br />
</span></p>
<p><strong>Conclusion:<br />
</strong></p>
<p><span style="font-size: 10pt; font-family: Times New Roman">All in all the experiment was a success.<br />
</span></p>
<p><span style="font-size: 10pt; font-family: Times New Roman">The conductance of the pipes was 3.77 </span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt"><span style="font-family: Times New Roman"> 10<sup>-3</sup> m<sup>3</sup> s<sup>-1</sup> (</span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 5 </span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 10<sup>-3</sup> m<sup>3</sup> s<sup>-1</sup>)<br />
</span></p>
<p><span style="font-size: 10pt"><span style="font-family: Times New Roman">The conductance of the aperture was 0.15 m<sup>3</sup> s<sup>-1</sup> (</span><span style="font-family: Symbol">±</span><span style="font-family: Times New Roman"> 5 </span></span><span style="font-size: 8pt; font-family: Arial">X</span><span style="font-size: 10pt; font-family: Times New Roman"> 10<sup>-3</sup> m<sup>3</sup> s<sup>-1</sup>).<br />
</span></p>
<p>The effective pumping speeds (without obstruction) were: 69.36 l min<sup>-1</sup> for high pressure</p>
<p>2.04 l min<sup>-1</sup> for low pressure</p>
<p>The effective pumping speeds (with obstruction) were: 8.16 l min<sup>-1 </sup>for high pressure</p>
<p><span style="font-family: Times New Roman"><span style="font-size: 9pt">1.58 l min<sup>-1</sup><br />
</span><span style="font-size: 10pt">for low pressure<br />
</span></span></p>
<p><span style="font-family: Times New Roman"><span style="font-size: 10pt">To improve the experiment an exact volume reading of the chamber could have been taken, the results could have been taken using a data logger. The length of the pipe work could have been decreases and the valves replace with others of a more suitable design (that open quicker). Also the best-fit lines could have been calculated rather than &#8216;guessed&#8217;.</span><span style="font-size: 9pt"><br />
</span></span></p>
<p><strong>Bibliography:<br />
</strong></p>
<p>Encyclopaedia Britannica 1999, CDROM Version</p>
<p><span style="font-size: 10pt; font-family: Times New Roman">Grolier Year 2000 CDROM Encyclopaedia<br />
</span></p>
<p>Physics For A Level, 2<sup>nd</sup> Edition, F. Azzopardi.</p>
<p>Longmans general physics, 1962, 6<sup>th</sup> Edition.</p>
<p><strong><br />
</strong> </p>
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		<title>The Periodic Table in Terms of the Electronic Structure of Atoms</title>
		<link>http://www.christopher-dean.co.uk/2001/04/24/the-periodic-table-in-terms-of-the-electronic-structure-of-atoms/</link>
		<comments>http://www.christopher-dean.co.uk/2001/04/24/the-periodic-table-in-terms-of-the-electronic-structure-of-atoms/#comments</comments>
		<pubDate>Tue, 24 Apr 2001 15:43:03 +0000</pubDate>
		<dc:creator>Chris</dc:creator>
				<category><![CDATA[Essays]]></category>
		<category><![CDATA[Science]]></category>

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		<description><![CDATA[The modern picture of the electronic configuration of atoms takes the form of the &#8216;atomic shell model&#8217; whereby the configurations of all atoms are considered as modified versions of the configuration of the hydrogen atom, the simplest atom in the periodic table. The atomic shell model states that the electrons orbit the nucleus of the [...]]]></description>
			<content:encoded><![CDATA[<!-- sphereit start --><p>The modern picture of the electronic configuration of atoms takes the form of the &#8216;atomic shell model&#8217; whereby the configurations of all atoms are considered as modified versions of the configuration of the hydrogen atom, the simplest atom in the periodic table. The atomic shell model states that the electrons orbit the nucleus of the atom in shells of well-defined energy that increase in radius from the nucleus outward, just like the shelled layers of an onion. Then, on closer examination, each of these larger shells contain &#8216;sub-shells&#8217; which are basically, smaller intervals of energy difference in the range defined by the particular shell being considered.</p>
<p>Each shell is labelled by both a letter and a number. The number is the principle quantum number &#8216;n&#8217;, and has values from 0 to infinity in integer steps. The letters range from J (corresponding to n = 0) &#8216;upward&#8217; in relation to the values of n; K represents n = 1, L represents n = 2 and so on.</p>
<p>A letter and a number also define each sub-shell. The number here is the orbital angular momentum quantum number &#8216;l&#8217; and has values that range from 0 to n-1 again in integer steps. The letter that corresponds to the number in this case is slightly different<span style="color: red">.</span> For the numbers l = 0 to l = 3 the corresponding letters are s, p, d, f, and for the numbers where l = 4+, the letters just follow the pattern of the alphabet (continuing on from f) so l = 4 has the letter g, l = 5 the letter h, and so on.</p>
<p>The periodic table is an ordered chart of the known elements based upon this model of the atoms electrons, and also more importantly, their chemical and physical properties. In the table, only the elements ground state (lowest energy) electron configurations are represented, partly because this is the usual state of most elements atoms in their pure form under standard conditions, but mostly because a table that included any excited states wouldn&#8217;t really have much significance, and would be impractical.</p>
<p>The table can basically be considered as an ordered list of increasing atomic number, which is read from left to right and has a new line (or period) for every shell that is being filled with electrons. The table is also arranged in columns. In the most general sense each column is of varying width and causes the table to be divided into 4 blocks, each of which corresponds to a particular sub-shell being filled. The blocks are labelled as s, p, d, and f, as there are no stable atoms that have electrons in the g and beyond sub-shells in their ground states.</p>
<p>Each sub-shell can hold a different number of electrons (which is why the very general column widths vary), which is given by the number of orbitals it has. Each orbital can hold 2 electrons whose spins are anti-parallel, and the number of orbitals in each sub-shell is given by the magnetic quantum number, m<sub>l</sub>. The values that the magnetic quantum number can take (and hence the number of orbitals that can be filled) is given by the range 0 to <span style="font-family: Symbol">±</span> l, so for example the d sub-shell has l = 2, so m<sub>l</sub> ranges from –2 to +2 in integer steps, and therefore there are five orbitals that can be filled.<span style="color: red"><br />
</span></p>
<p>Generally the table takes the following form, note that the f block is removed from the main part of the table, this is mainly for convenience, because the f block elements are generally rare, radioactive, not really in common usage and as they are so different, they disturb the regularity of the rest of the table<span style="color: red">.</span></p>
<p><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2117-theperiodic1.png" title="The Periodic Table in Terms of the Electronic Structure of Atoms" alt="060607 2117 theperiodic1 The Periodic Table in Terms of the Electronic Structure of Atoms" />Also you can see that H and He are included in the s and p blocks respectively, this isn&#8217;t a strictly true representation, but for alternative reasons. H is included at the head of the first column of the s-block as it is filling the s sub-shell, however it has very few similar properties when compared to the other s-block elements. He is included at the head of the last column of the p-block, this is because its properties are similar to the rest of the elements in that column. Although He isn&#8217;t actually filling its p sub-shell, it does have a full s sub-shell and consequently a full outer shell of electrons. Overall these elements should be detached from the bulk of the table, as they only fit some of the trends that the others do.</p>
<p>Each of these four blocks has its separate columns and each column corresponds to the position of the last electron to orbit the nucleus. For example the very first column in the period table is the first column of the s-block, this means that the last electron to orbit the nucleus is an s-electron and we can also deduce that there is one possible orbital for it to be in and that it&#8217;s the first electron in that orbital. The same applies for the second column in the s-block, but here we can further say that the spin of this electron is opposite to whatever that of the first electron in this orbital was due to the Pauli Principle.</p>
<p>When looking at the columns across the table we generally only consider the columns that contain what are known as the main group elements. These are the elements contained within the s and p-blocks only and as a result we label each column as groups 1 – 8 from left to right across the table. In some cases we also consider the &#8216;sub-groups&#8217; formed from the columns of the d-block elements, this adds another 10 groups to the table, making a total of 18.<span style="color: red"><br />
</span></p>
<p>So, each group relates atoms with a similar final electron configuration. As the final configuration also designates what chemical and physical properties the various elements have, the groups also relate elements with similar characteristics, and this results in a number of readily observable trends within the periodic table, with emphasis on the main group elements.</p>
<p>For example, group eight contains the nobel gasses. They all have full shells and are therefore in their most stable (lowest energy) state. The result of this is that they are all monatomic gasses (as either ionic or covalent &#8216;attraction&#8217; between the atoms has to occur for them to form liquids or solids, and although there may be some very momentary &#8216;van der Waals&#8217; attraction between the atoms at room temperature, this is not long lived enough to cause liquefaction or solidification!) that barely react with anything (as for a &#8216;normal&#8217; reaction to take place there has to be a net loss of energy between the two reacting atoms, via the transfer or &#8216;sharing&#8217; of an electron(s) between them, as the shells are full there is very little opportunity for these atoms to &#8216;release&#8217; or &#8216;accept&#8217; electrons).</p>
<p>All other atoms (aside from those in group eight) &#8216;want&#8217; to be in their lowest possible energy configurations and as a result they will try to &#8216;accept&#8217;, &#8216;loose&#8217; or &#8216;share&#8217; each other&#8217;s electrons to achieve this. For example the reaction between Cs and F gives the most stable &#8216;ionic&#8217; compound known:</p>
<p>Group one is the alkali metals, they all have one &#8216;outer&#8217; electron and as a result they would &#8216;prefer&#8217; to loose this to form a singly charged positive ion, which has a full outer shell of electrons, just like the noble gasses. As you descend group 1 the proton number Z increases, as does the corresponding number of electrons (as single atoms are electrically neutral). This means that the distance of the final electron from the nucleus at the bottom of the group is as large as possible and, as the electrostatic influence of the nucleus decreases exponentially with distance, the last electron on the atom at the bottom of group 1 is very &#8216;loosely bound&#8217; (as shielding effects tend to cause each electron to see only one proton, the only force acting on that electron is the electrostatic attraction between the one proton and the said electron. This is not the full picture as the electron also &#8216;momentarily&#8217; feels the pull of the whole nucleus, but overall one electron is bound by one proton). Cs is the last element at the bottom of group one (that isn&#8217;t radioactive), and as a result can loose its outer electron very easily.</p>
<p>Conversely the atoms in group seven all have a single &#8216;gap&#8217; for an electron to go into to form a full outer shell, and therefore a negative ion. The effect of size increase down the group implies that the effect of the nucleus on the electrons is greatest at the top (where the atoms are smaller), therefore the atoms with the strongest pull for an electron (high electronegativity) are at the top of group seven. Fluorine is at the top of group seven and therefore has the strongest pull for an electron.</p>
<p>It should be noted that any ion is actually at a higher energy than the neutral atom even if it does have a full shell of electrons, but this energy is recouped when a positive ion(s) meets a negative ion(s) and forms an ionic lattice. The formation of the lattice releases a vast amount of energy that compensates for the ionisations of the component atoms and actually leads to a net decrease in the energy of the system. The quantity of energy released also depends upon how well the ions can pack together in the lattice, which is also why a very large Cs<sup>+</sup> ion along with a very small F<sup>-</sup> ion makes for a very favourable combination overall.</p>
<p>All s-block atoms can react in the same way with p-block atoms in groups six and seven (and partially group 5) to form ionic compounds with different ratios between the numbers of individual ions involved (depending on where the atoms come from). For example Mg has two electrons to loose to form a full shell, and again F can only accept one, but two F atoms can react with one Mg atom to form MgF<sub>2</sub>.</p>
<p>Reactions with the atoms of groups three and four and some of five are fairly different as they form the borderline cases whereby they don&#8217;t &#8216;want&#8217; to loose electrons, but they don&#8217;t want to gain them either. When these atoms react they have to share electrons between themselves to lead to a net energy decrease. This is covalency and an example would be Mg<sub>3</sub>Al<sub>2</sub> for s-block atoms reacting with p-block, but as these atoms are ambiguous they can also react with any of the other atoms in the p-block, CO<sub>2</sub> being another example.</p>
<p>The majority of elements are metallic, and a number of physical properties depend upon the strength of the metallic bonding between the atoms, which also depends upon the electronic structure. As you move from left to right across the d-block valences increase. This means that the number of electrons available to form a (delocalised) metallic bond over the whole structure increases, and therefore the &#8216;strength&#8217; of the bond increases. The greater the metallic bond, the higher the electrical and thermal conductivity, and the melting and boiling points. A similar effect happens across the main block elements, until they change from being metals to non-metals. So overall these properties increase as you move from left to right.</p>
<p>As you move across a period (up until you reach non-metals) the density of the elements increases, this is due to the fact that no new shells are started across a period, but atomic number (and therefore atomic mass) increases with every element. So there is a vast mass increase with only a tiny increase in the atomic volume, leading to a greater density.</p>
<p>And lastly, the d-block elements are also the most coloured of all the elements, and patterns in the coloration which correspond to their electronic configurations can be noted as you move across the block.</p>
<p><strong>Conclusion:<br />
</strong></p>
<p>There are many more trends in chemical and physical properties throughout the periodic table than have been mention here, and the majority of them are due to the electronic structure of the elements. The ordering of the table, based upon the ground state electronic structure of the atoms, helps us to see these trends easily and makes the periodic table one of the most useful tools available when studying the properties of elements, whether it be from a chemical or a physical point of view.</p>
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		<title>The Physics of the Cyclotron Particle Accelerator</title>
		<link>http://www.christopher-dean.co.uk/2001/04/08/the-physics-of-the-cyclotron-particle-accelerator/</link>
		<comments>http://www.christopher-dean.co.uk/2001/04/08/the-physics-of-the-cyclotron-particle-accelerator/#comments</comments>
		<pubDate>Sun, 08 Apr 2001 19:26:38 +0000</pubDate>
		<dc:creator>Chris</dc:creator>
				<category><![CDATA[Essays]]></category>
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		<description><![CDATA[When a charged particle is accelerated in a uniform magnetic field it will move along a circular path in a plane perpendicular to the direction of the magnetic field. This is due to the fact that the magnetic force acting on a charged particle in a uniform magnetic field is always perpendicular to the velocity [...]]]></description>
			<content:encoded><![CDATA[<!-- sphereit start --><p style="text-align: justify"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2103-thephysicso1.png" title="The Physics of the Cyclotron Particle Accelerator" alt="060607 2103 thephysicso1 The Physics of the Cyclotron Particle Accelerator" /><span style="font-size: 10pt; font-family: Times New Roman">When a charged particle is accelerated in a uniform magnetic field it will move along a circular path in a plane perpendicular to the direction of the magnetic field. This is due to the fact that the magnetic force acting on a charged particle in a uniform magnetic field is always perpendicular to the velocity of the particle. So as the corresponding acceleration of the particle is always perpendicular to the magnetic field, the field doesn&#8217;t do any actual work on the particle, it just changes the direction of the velocity, as is shown in the diagram on the right.<br />
</span></p>
<p style="text-align: justify">The crosses are the tails of the magnetic field vectors, the circle is the path mapped out by the positive charge, the black arrows are the velocity vectors (v), and the green arrows are the magnetic force vectors (F). The red arrow represents the radius of the circle (r).</p>
<p style="text-align: justify">As the discoveries of the workings of the atom developed, physists wanted a way to accelerate protons to do experients on atomic nucli. One way of accelerating a proton would be to place it at one end of a long channel, and then apply a large negative voltage to the other, and use this to acclerate the positive proton towards it, and thus into the target nucleus. However, this way of doing things didn&#8217;t acclerate the protons fast enough to be of any real use, and dealing with the high voltages would have been both dangerous, expensive and complicated to say the least!</p>
<p style="text-align: justify"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2103-thephysicso2.png" title="The Physics of the Cyclotron Particle Accelerator" alt="060607 2103 thephysicso2 The Physics of the Cyclotron Particle Accelerator" />Then in 1932 Ernest Lawrence and Stanley Livingston developed the first 11-inch (27.5 cm) (classical) cyclotron magnetic resonance particle accelerator, based upon the principle above. It was capable of accelerating protons to energies of about 10 MeV <em>without</em> using a high voltage.</p>
<p style="text-align: justify">Nowadays though, modern particle accelerators can accelerate particles to speeds very close to the speed of light and take many forms and sizes, but are usually categorised into two main types, synchrotron (circular) and linac (linear), cyclotrons are the basis for the former, and the latter could be the cathode ray tube of a TV set, for a simple example.</p>
<p style="text-align: justify">The biggest and most complex cyclotron particle accelerator in the world is the TRIUMF accelerator in Canada, and accelerates protons to three quarters of the speed of light. The particles at TRIUMF make 1500 turns in 1/3000 of a second, with a total path length of 45 km. The accelerated protons can then also be used to form other sub-atomic particles such as<span style="color: red"><strong><br />
</strong></span>neutrons or other charged particles such as muons and pions (pions are very short lived mesons which are mass-produced in their billions every second at TRIUMF, using the very intense proton beams they have there) all travelling at high velocities.</p>
<p style="text-align: justify"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2103-thephysicso3.png" title="The Physics of the Cyclotron Particle Accelerator" alt="060607 2103 thephysicso3 The Physics of the Cyclotron Particle Accelerator" />The classical cyclotron basically consists of two hollow (copper) metal semi-circular electrodes called dees (due to their D shape), inside a vacuum chamber (to prevent collisions with air molecules), across which a high frequency oscillating voltage is put. The dees are open along their straight edge so that they effectively form a &#8216;broken cylinder&#8217;, and this is shown schematically in the diagrams on the right and on the next page.</p>
<p style="text-align: justify">Above and below the &#8216;broken cylinder&#8217; there is a circular electromagnet, which provides a uniform magnetic field of fixed intensity across the dees and &#8216;acceleration cavity&#8217;. This is used to guide the travelling particles through the accelerator.</p>
<p style="text-align: justify"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2103-thephysicso4.png" title="The Physics of the Cyclotron Particle Accelerator" alt="060607 2103 thephysicso4 The Physics of the Cyclotron Particle Accelerator" />And in the very centre of the apparatus, between the dees there is an ion source, which releases the charged particles (protons) that are to be accelerated into the machine.</p>
<p style="text-align: justify"><span style="font-size: 10pt; font-family: Times New Roman">The source of the changing voltage is an oscillator (similar to that used for transmitting radio waves) that corresponds to the frequency of revolution of the particles in the magnetic field. This (accelerating) oscillating potential difference across the dees causes an electric field to be produced perpendicular to their plane, which is concentrated in the gap between them. The oscillation of electronic field in the cavity is then timed so that when the charged particle leaves one of the dees it is attracted towards the other.<br />
</span></p>
<p style="text-align: justify">It is only when the ions are in the gap between the dees that they are accelerated by the electric field, as when they enter the dees they feel no electric field (as no electric field exists within a conductor – <em>Gauss&#8217;s law</em>) but the &#8216;external&#8217; magnetic field isn&#8217;t affected (shielded) by the metal of the dees and causes the linear path of the proton to curve around into a semi-circle and &#8216;shoot&#8217; back out of the other side of the dee, and into the gap again, where it is accelerated further.</p>
<p style="text-align: justify">The reason that the cyclotron actually works at all is the fact that the orbits of ions in a uniform magnetic field are isochronous; i.e. &#8220;the time taken by a particle of a given mass to make one complete circuit is the same at any speed or energy as long as the speed is much less than that of light&#8221;. This means that it is possible to accelerate the particles many times using a high voltage, which reverses its polarity at a constant frequency. This reversing of polarity at constant frequency has to fulfil the cyclotron&#8217;s &#8216;resonance condition&#8217;. In other words the &#8216;extra&#8217; energy from the accelerating voltage must be &#8216;given&#8217; to the proton at a frequency equal to the natural frequency of the proton circulating in the magnetic field for the proton to accelerate in a well-defined and predictable way. The equation for the resonance condition is qB = 2<span style="font-family: Symbol">p</span>mf<sub>osc</sub>, and as m and q are fixed for the proton (and f<sub>osc</sub> is assumed to be constant), we can see that we only need to alter the magnetic field, B until this condition is satisfied and a beam of energetic protons emerges.</p>
<p style="text-align: justify">As the particles move faster the radius of the path increases until they are &#8216;ejected&#8217; from the accelerator at the required velocity, which can be found using the expression:</p>
<p style="text-align: right"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2103-thephysicso5.png" title="The Physics of the Cyclotron Particle Accelerator" alt="060607 2103 thephysicso5 The Physics of the Cyclotron Particle Accelerator" />                        <strong>(1)</strong></p>
<p style="text-align: justify">This equation can be adapted as follows to produce an expression fore the cyclotron frequency:</p>
<p style="text-align: right"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2103-thephysicso6.png" title="The Physics of the Cyclotron Particle Accelerator" alt="060607 2103 thephysicso6 The Physics of the Cyclotron Particle Accelerator" />                        <strong>(2)<br />
</strong></p>
<p style="text-align: right"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2103-thephysicso7.png" title="The Physics of the Cyclotron Particle Accelerator" alt="060607 2103 thephysicso7 The Physics of the Cyclotron Particle Accelerator" />                        <strong>(3)<br />
</strong></p>
<p style="text-align: justify">Equation (3) is characteristic of a particular particle moving in a particular magnetic field. It also tells us that the particles take exactly the same amount of time to travel along the larger semi-circular paths as they do for the initial shorter paths, which (again) is the critical characteristic in the operation of the cyclotron.</p>
<p style="text-align: justify">The frequency of the oscillator must match that of the cyclotron frequency, which, as is shown by equation (3) can be determined by the mass and charge of the particle, along with the value of the magnetic field. Again the final velocity of the particle can be found via equation (1) and for speeds much less than the speed of light the final kinetic energy can be given by K = ½ mv<sup>2</sup></p>
<p style="text-align: justify">Classical cyclotrons have maximum energies of around 10 MeV. However, for particles of the same mass, but of higher charge the energies increase as the square of the charge.</p>
<p style="text-align: justify">Note that in the case of the cyclotron the final energy actually depends upon the magnetic field, and not the potential difference. The magnetic field can&#8217;t accelerate the particles, but it does control the number of revolutions that they make in the accelerator before they are ejected, and as the particles pick up energy on each revolution (i.e. on each crossing of the gap between the two dees) the more revolutions the particles make the greater their final energy.</p>
<p style="text-align: justify">If we increase the potential difference across the gap, the energy of the particle isn&#8217;t really affected as although the &#8216;thrust&#8217; from the potential difference causes the particle to initially have more energy, it consequently moves in a path with a greater radius, the greater the radius the sooner the particle reaches the maximum radius where it leaves the accelerator. Therefore the particle crosses the gap fewer times and the net result is not that different from what would be achieved at lower potential difference, with the same magnetic field. And conversely, if the potential difference is lowered across the gap the particle gets less of a thrust on each revolution, but it makes more revolutions in total before it leaves, which again leads to the same resultant energy as a particle at higher potential difference for the same magnetic field.</p>
<p style="text-align: justify">As the speed of a particle approaches that of light, the effect of relativity causes its mass to increase. So the principles of the classical cyclotron have to be modified to allow for this relativistic increase. That is what happens in modern synchrocyclotrons, where speeds very close to that of light are achieved, with immense energies in the region of 1000 GeV, as at the Fermi-lab particle accelerator near Chicago – but that is something fairly different!<span style="color: red"><br />
</span></p>
<p style="text-align: justify"><strong><em>Conclusion:<br />
</em></strong></p>
<p style="text-align: justify">Cyclotrons are essential for performing modern atomic and nuclear physics experiments where the building blocks of nature are constantly being probed and new particles are being found out. Without them, there would be great difficulties and expenses involved in accelerating the particles to high enough energies to perform even the most basic of experiments and essentially, if they didn&#8217;t exist we may not yet know about such things as quarks and mesons which are the foundations of what were previously thought to be some of the smallest particles imaginable (i.e. protons and neutrons). And really the best thing about the cyclotron in general, is that it&#8217;s a ready application of some very basic physical principles (neglecting the effect of relativity in synchrocyclotrons of course! – which in-itself is added evidence for relativity being real and not just a possibility).</p>
<p style="text-align: justify">Cyclotrons also have applications outside of academic research labs, for example, some of the applied research at TRIUMF involves, the use of a proton beam to destroy cancerous eye tumours, the development of a new method for detecting illegal drugs or plastic explosives hidden in mail, luggage and cargo containers, and they are also readily used in hospitals to create some of the short lived</p>
<p>radioactive particles used in medicine (as tracers for example).</p>
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		<title>An EXPLOSIVE Chemical Kinetics Project</title>
		<link>http://www.christopher-dean.co.uk/2001/03/19/an-explosive-chemical-kinetics-project/</link>
		<comments>http://www.christopher-dean.co.uk/2001/03/19/an-explosive-chemical-kinetics-project/#comments</comments>
		<pubDate>Mon, 19 Mar 2001 16:33:51 +0000</pubDate>
		<dc:creator>Chris</dc:creator>
				<category><![CDATA[Essays]]></category>
		<category><![CDATA[Science]]></category>

		<guid isPermaLink="false">http://www.christopher-dean.co.uk/2007/06/06/an-explosive-chemical-kinetics-project/</guid>
		<description><![CDATA[     My project is going to concern the chemistry of the gas phase 2H2 + O2 à 2H2O reaction. This reaction is usually thought of as being solely explosive (it&#8217;s used in rocket fuel, for example), but this isn&#8217;t necessarily always the case. The reaction itself has many individual &#8216;sub-reactions&#8217;, is very complex and as [...]]]></description>
			<content:encoded><![CDATA[<!-- sphereit start --><p><span style="font-size: 10pt; font-family: Times New Roman">    <br />
</span></p>
<p>My project is going to concern the chemistry of the gas phase 2H<sub>2</sub> + O<sub>2</sub><br />
<span style="font-size: 8pt; font-family: Wingdings">à</span> 2H<sub>2</sub>O reaction.</p>
<p>This reaction is usually thought of as being solely explosive (it&#8217;s used in rocket fuel, for example), but this isn&#8217;t necessarily always the case. The reaction itself has many individual &#8216;sub-reactions&#8217;, is very complex and as a result it&#8217;s quite hard to get any accurate data for any specific part of the reaction mechanism, aside from the general view of what actually happens to the system overall.</p>
<p>Also there are really two main definitions of a gas phase explosion either or both of which must be true for an explosion to occur:</p>
<ol>
<li>Thermal Explosion:</li>
</ol>
<p>The distribution of heat in a reacting mixture is stable when the heat produced by a reaction in a volume element is equal to the heat lost by the same volume element. When the heat loss can&#8217;t compensate for the heat being produced by the reaction, the internal heat of the reaction increases exponentially, thus driving the rate of reaction faster and faster until a point of explosion is reached.</p>
<ol>
<li>Branched Chain (Radical) Explosion:</li>
</ol>
<p>For a reaction mixture of fixed composition and fixed initial temperature we can see that there will be a corresponding critical pressure, above which an explosion will occur. Below this the steady state approximation applies (see later).</p>
<p>The main reactions involved within the overall reaction are as follows:</p>
<p><em>(NB: A <span style="font-size: 9pt"><span style="font-family: Symbol">·</span><br />
</span>implies species is a radical)<br />
</em></p>
<table border="0" style="border-collapse: collapse">
<tr>
<td style="padding-right: 7px; padding-left: 7px; border: 0.5pt solid"><span style="font-size: 9pt"><strong><em>Initiation:</em>    </strong>H<sub>2</sub><br />
</span><span style="font-size: 8pt; font-family: Wingdings">à</span><span style="font-size: 9pt"> H<span style="font-family: Symbol">·</span> + H<span style="font-family: Symbol">·</span> (1)<br />
</span></p>
<p><span style="font-size: 9pt"><strong><em>Branch 1:</em>    </strong>H<span style="font-family: Symbol">·</span> + O<sub>2</sub><br />
</span><span style="font-size: 8pt; font-family: Wingdings">à</span><span style="font-size: 9pt"><br />
<span style="font-family: Symbol">·</span>OH + O<span style="font-family: Symbol">·</span> (2)<strong><br />
</strong></span></p>
<p><span style="font-size: 9pt"><strong><em>Branch 2:</em>    </strong>O<span style="font-family: Symbol">·</span> + H<sub>2</sub><br />
</span><span style="font-size: 8pt; font-family: Wingdings">à</span><span style="font-size: 9pt"><br />
<span style="font-family: Symbol">·</span>OH + H<span style="font-family: Symbol">·</span> (3)</span><strong><br />
</strong></td>
<td style="border-right: 0.5pt solid; padding-right: 7px; border-top: 0.5pt solid; padding-left: 7px; border-left: medium none; border-bottom: 0.5pt solid"><span style="font-size: 9pt"><strong><em>Propagation:</em>    </strong><span style="font-family: Symbol">·</span>OH + H<sub>2</sub><br />
</span><span style="font-size: 8pt; font-family: Wingdings">à</span><span style="font-size: 9pt"> H<sub>2</sub>O + H<span style="font-family: Symbol">·</span> (4)<strong><br />
</strong></span></p>
<p><span style="font-size: 9pt"><strong>        </strong>H<span style="font-family: Symbol">·</span> + O<sub>2 </sub></span><span style="font-size: 8pt; font-family: Wingdings">à</span><span style="font-size: 9pt"> HO<sub>2</sub><span style="font-family: Symbol">·</span> (5)<br />
</span></p>
<p><span style="font-size: 9pt">        HO<sub>2</sub><span style="font-family: Symbol">·</span> + H<sub>2</sub><br />
</span><span style="font-size: 8pt; font-family: Wingdings">à</span><span style="font-size: 9pt"> H<sub>2</sub>O + <span style="font-family: Symbol">·</span>OH (6)</span><strong><br />
</strong></td>
<td style="border-right: 0.5pt solid; padding-right: 7px; border-top: 0.5pt solid; padding-left: 7px; border-left: medium none; border-bottom: 0.5pt solid"><span style="font-size: 9pt"><strong><em>Termination:<br />
</em></strong></span></p>
<p><span style="font-size: 9pt">Combination of radicals:<br />
</span></p>
<p><span style="font-size: 9pt">E.g. H<span style="font-family: Symbol">·</span> + H<span style="font-family: Symbol">·</span><br />
</span><span style="font-size: 8pt; font-family: Wingdings">à</span><span style="font-size: 9pt"> H<sub>2 </sub>(7)<br />
</span></p>
<p><span style="font-size: 9pt">Reaction with walls of vessel:<br />
</span></p>
<p><span style="font-size: 9pt">E.g. Absorbance of HO<sub>2</sub></span><span style="font-size: 8pt; font-family: Symbol">·</span><span style="font-size: 9pt"> on wall (8)</span></td>
</tr>
</table>
<p>As the external conditions, temperature (T) and pressure (P), are changed, the importance of various individual &#8216;sub-reactions&#8217; rises and falls dramatically. This in part exemplified by reactions (2) and (3).</p>
<p><span style="font-size: 10pt; font-family: Times New Roman">Reactions (2) and (3) are the &#8216;key-stones&#8217; in terms of whether the gaseous mixture will explode or not, as all you need for an explosion to happen is a way of generating many radicals in a very short space of time. Which is exactly what equations (2) and (3) do under the right conditions.<br />
</span></p>
<p>Many chemical reactions involve the formation of (very reactive) radicals as intermediary species, and usually this is fine as they tend to be used up as soon as they are produced (i.e. they are straight chain reactions). However, in some reactions under the correct conditions, these intermediate reactions involving the radicals are able to branch out along other reaction pathways, which in some cases form more and more radicals which go on branching out through the whole reaction mixture until the concentration of radicals is so great that explosive effects occur.</p>
<p>The effect of branching means that in the overall scheme of things these intermediary products can no longer be ignored (as their concentrations are so high), and therefore the steady state (SS) approximation (i.e. the approximation that the overall concentration of intermediary species is zero) breaks down quite readily.</p>
<p>To initiate the reaction the H—H bond has to be broken in some way, either by raising the temperature of the reaction to such an extent that it occurs &#8216;naturally&#8217; or by forcing it at lower (room) temperature via a (high energy) electrical spark for example.</p>
<p>This is a typical example of a thermodynamically driven reaction with a kinetic barrier (the barrier being the splitting of the H—H bond, whilst not forgetting the loss in entropy of the system, which is a lesser problem). I.e. If there is no activation via sparking, or other such high energy phenomena, at room temperature, then a mixture of H<sub>2</sub> and O<sub>2</sub> is quite safe, but as soon as the first few H—H bonds are broken the reaction will be over in a fraction of a second.</p>
<p>At low T i.e. approx T &lt; 730 K the reaction is heterogeneous (doesn&#8217;t occur solely in the gas phase), and the gases react via the walls of the reaction vessel. This in turn means that at room temperature reactions (2) and (3) are really very slow and the gas mixture is quite safe. Also any radicals that may be produced in the bulk of the gas itself are able to diffuse through to the walls of the vessel (as the environment of the gas at RTP [Room temperature and Pressure] isn&#8217;t particularly harsh), where they are &#8216;destroyed&#8217; by adsorption.<strong><br />
</strong>If however, whilst within this temperature range, we set up a situation where P is the only variable, then only below the critical pressure (the point where the pressure is so great explosion will occur) will H<span style="font-size: 9pt; font-family: Symbol">·</span> diffuse through the gas to the walls. So, overall, below 730 K and the critical pressure a steady state will occur, but the reaction is so slow in this region that it&#8217;s hard to measure the rate of any reactions happening at all.</p>
<p>The point at which the conditions are right for an explosion to occur is called <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2043-anexplosive1.png" title="An EXPLOSIVE Chemical Kinetics Project" alt="060607 2043 anexplosive1 An EXPLOSIVE Chemical Kinetics Project" />an explosion limit. The limit depends on both temperature and pressure as the diagram on the right for the H<sub>2</sub>/O<sub>2</sub> reaction illustrates:</p>
<p>In the range 730 K &lt; T &lt; 890 K, the situation is quite complex, as there are regions where then change in pressure has a quite marked effect on the likelihood of explosion.</p>
<p>And for the range T &gt; 890 K the reaction is almost always explosive. This is because the formation of radicals becomes more frequent than their removal; the overall concentration of radicals increases exponentially and leads to a branched-chain explosion. The explosion again occurs at a critical pressure, which is usually very low.</p>
<p>From this diagram we can also see that as P is increased above the 1<sup>st</sup> explosion limit, the reaction remains explosive until a second critical pressure is reached, above which a second steady state is observed until we reach the third (and final) explosion limit.</p>
<p><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2043-anexplosive2.png" title="An EXPLOSIVE Chemical Kinetics Project" alt="060607 2043 anexplosive2 An EXPLOSIVE Chemical Kinetics Project" />A graph, which corresponds to this diagram, is the plot of rate vs. log P as in the following:</p>
<p>From this graph it&#8217;s easy to see how the rate depends upon the pressure. As the pressure approaches the 1<sup>st</sup> explosion limit the reaction rate rises slowly and steadily to begin with, and then exponentially as it approaches the limit. Then, as the pressure increases further the rate remains very fast, until the effect of reaction (5) comes into play and the rate decreases exponentially, but this time the rate of the steady state reaction is faster overall than the rate at the start. Lastly the rate rises slowly and steadily as before until we approach the 3<sup>rd</sup> explosion limit where the rate increases exponentially again but this time it stays there!</p>
<p>There are three main factors that can influence the formation of radicals in the H<sub>2</sub>/O<sub>2</sub> reaction (and for that matter most other explosions) and therefore the likelihood of the onset of explosive conditions, these are:</p>
<ol>
<li>The rate of initiation of the reaction – How many individual reactions are able to start within a
<p style="margin-left: 126pt">certain initial time period.</p>
</li>
<li>The rate of chain branching      – The rate at which the reaction is able to branch throughout</li>
</ol>
<p>                 the reaction mixture.</p>
<ol>
<li>The rate of radical termination    – How quickly the radicals are used up or &#8216;removed&#8217; from the</li>
</ol>
<p>                 reacting mixture.</p>
<p>These factors can be related and modelled by the following differential equation:</p>
<p><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2043-anexplosive3.png" title="An EXPLOSIVE Chemical Kinetics Project" alt="060607 2043 anexplosive3 An EXPLOSIVE Chemical Kinetics Project" />        [r] = Concentration of radicals        R<sub>i </sub>= rate of initiation</p>
<p>             <span style="font-family: Symbol">e</span> = rate of branching            k = rate of termination</p>
<p>Which has the solution:</p>
<p style="text-align: center">[r](t) = <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2043-anexplosive4.png" title="An EXPLOSIVE Chemical Kinetics Project" alt="060607 2043 anexplosive4 An EXPLOSIVE Chemical Kinetics Project" />         <span style="font-family: Symbol">f</span> = <span style="font-family: Symbol">e</span> &#8211; k</p>
<p><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2043-anexplosive5.png" title="An EXPLOSIVE Chemical Kinetics Project" alt="060607 2043 anexplosive5 An EXPLOSIVE Chemical Kinetics Project" /></p>
<p>This plot shows the two possible regimes:</p>
<p style="margin-left: 18pt">(a): <span style="font-family: Symbol">f</span> is negative:</p>
<p style="margin-left: 18pt">Termination is the dominant factor, the rate is slow and the reaction comes to a steady state.</p>
<p>(b): <span style="font-family: Symbol">f</span> is positive:</p>
<p>Branching occurs much faster than termination, the concentration of radicals increases exponentially and the mixture explodes.</p>
<p>Which is what should be expected for a reaction of this type.</p>
<p><em>Figure Adapted From Atkins, Physical Chemistry 6<sup>th</sup> Edition<br />
</em></p>
<p>When the H<sub>2</sub>/O<sub>2</sub> reaction is studied it is usually examined in a spherical Pyrex &#8216;bulb&#8217; whose surface is coated with KCl. The KCl acts as a good adsorber for the radicals that make it to the walls and thus &#8216;removes&#8217; them from the reaction mixture, which helps to control the explosion. An important radical to be controlled is HO<sub>2</sub><span style="font-size: 9pt; font-family: Symbol">·</span> as this is mainly responsible for the position of the second explosion limit.</p>
<p>Reaction (5) producing the HO<sub>2</sub><span style="font-size: 9pt; font-family: Symbol">·</span> radical can be considered as a &#8216;chain braking&#8217; reaction as this radical can diffuse quite easily to the wall and it basically serves to &#8216;mop up&#8217; H<span style="font-size: 9pt"><span style="font-family: Symbol">·</span><br />
</span>in this respect. In this regime as pressure increases (5) becomes more frequent, the rate of removal of radicals is higher than the rate</p>
<p>of formation, and as a result it takes an even higher T to cause he explosion to occur. This is exemplified in the diagrams above in the region between the 2<sup>nd</sup> and 3<sup>rd</sup> explosion limits, where a steady state is attained for the second time.</p>
<p>Under the &#8216;wrong&#8217; conditions though, the HO<sub>2</sub><span style="font-size: 9pt"><span style="font-family: Symbol">·</span><br />
</span>can be regarded as a propagator as in the reaction:</p>
<p style="text-align: right">HO<sub>2</sub><span style="font-size: 9pt; font-family: Symbol">·</span> + H<sub>2</sub><br />
<span style="font-size: 8pt; font-family: Wingdings">à</span> H<sub>2</sub>O<sub>2</sub> + H<span style="font-size: 9pt"><span style="font-family: Symbol">·</span>                    (9)</span></p>
<p>This is the reaction of HO<sub>2</sub><span style="font-size: 9pt; font-family: Symbol">·</span> in the gas phase, and it counteracts the chain braking effect in (5), to such an extent that it has to be included when forming a rate equation for the H<sub>2</sub>/O<sub>2</sub> reaction.</p>
<p>This reaction is favoured rather than (5) as a true picture of what is going on, as large amounts of H<sub>2</sub>O<sub>2</sub> are found to exist in the reaction mixture, between the 2<sup>nd</sup> and 3<sup>rd</sup> limits. But also the traces of H<sub>2</sub>O<sub>2</sub> disappear quite fast under these harsh conditions via the reaction:</p>
<p style="text-align: right">HO<sub>2</sub><span style="font-size: 9pt; font-family: Symbol">·</span> + H<sub>2</sub>O<sub>2</sub><br />
<span style="font-size: 8pt; font-family: Wingdings">à</span> H<sub>2</sub>O + O<sub>2</sub> + HO<span style="font-size: 9pt"><span style="font-family: Symbol">·</span>                </span>(10)</p>
<p>Absorption spectroscopy has to be used to give any good indication of the quantities present, and in fact that this is really happening at all.</p>
<p>Note that the reactions that produce H<sub>2</sub>O<sub>2</sub> with or via the walls are also possible, but are so slow that they can be neglected. A piece of evidence to support this is that the high reflectance walls (Pyrex, etc.) actually cause the reaction rate for reaction (10) to increase a lot. Therefore the reaction must be occurring mostly in the gas phase, as there are few reactions happening at the walls, but the rate of production of H<sub>2</sub>O<sub>2</sub> still increases.</p>
<p>If the vessel is coated with something other than KCl the &#8216;reflectance&#8217; of the HO<sub>2</sub><span style="font-size: 9pt; font-family: Symbol">·</span> at the walls can be affected greatly. For just clean silica or Pyrex (which cause more radicals to be reflected back into the gas mixture, and therefore a vast reduction in the chain breaking processes) the 2<sup>nd</sup> explosion limit is much higher and the tip of the explosion peninsula (the region between the 1<sup>st</sup> and 2<sup>nd</sup> explosion limits) moves to such a low temperature and pressure that it is hard to observe the 1<sup>st</sup> explosion limit at all. I.e. The explosion region becomes greater at lower temperatures, if the walls are coated with a chemical that isn&#8217;t a very good adsorber of radicals.</p>
<p>Lastly, if we raise P until a third limit is reached we find that above this limit the conditions are so harsh that the reaction is always explosive. This 3<sup>rd</sup> limit is probably a combination of both thermal and branched chain limits. In other words, as at high P the effect of T is more important than the effect of chemical termination i.e. reaction (8) is reduced as this depends upon HO<sub>2</sub><span style="font-size: 9pt; font-family: Symbol">·</span> being long lived, then under these conditions HO<sub>2</sub><span style="font-size: 9pt; font-family: Symbol">·</span> reacts as in reaction (6) and the gas phase reaction chain begins again, thus causing the 3<sup>rd</sup> explosion limit.</p>
<p><span style="text-decoration: underline"><strong>Conclusion:<br />
</strong></span></p>
<p>Studying the H<sub>2</sub>/O<sub>2</sub> reaction is very useful when it comes to trying to learn about the way gas phase explosions occur and what reactions may be happening with the bulk of the reacting gas.</p>
<p>It&#8217;s good because it contains the elements of both thermal and radical explosions and the reaction overall can be quite carefully controlled, using different coatings on the reaction vessel wall for example.</p>
<p>Also the products are just radicals and water – no corrosive acids or poisonous gasses that may cause harm or injury to the experimenter, or in the case of corrosive acids as in the H<sub>2</sub>/Cl<sub>2</sub> reaction react with the reaction vessel itself or the coatings on it, causing all kinds of problems.</p>
<p>It&#8217;s also good as it allows us to predict the properties and the ways of dealing with explosive or potentially explosive reactions. A good example of this may be in predicting the properties of the reactions where hydrocarbons are burned in a combustion engine. The explosions in this case have to be controlled very specifically and this is can be achieved by controlling the number of radicals produced. In the H<sub>2</sub>/O<sub>2</sub> reaction we use the coated walls of the vessel to terminate the reaction, but as</p>
<p>this is fairly impractical in the case of a car engine for example, and additive is put into the petrol mixture to help regulate the explosions c.f. leaded petrol.</p>
<p>Lastly the outcome of this reaction could be used to predict the results, or the characteristics (explosion limits, etc.) for any high pressure/temperature reaction, which is potentially very useful c.f. the production of ammonia.</p>
<p><span style="text-decoration: underline"><strong>References:<br />
</strong></span></p>
<p><strong>The Foundations Of Chemical Kinetics, Benson, 1982<br />
</strong></p>
<p><strong>Physical Chemistry, Atkins, 6<sup>th</sup> Edition<br />
</strong></p>
<p><strong>Combustion, Flames And Explosions Of Gasses, Lewis &amp; von Elbe, 2<sup>nd</sup> Edition<br />
</strong></p>
<p><strong>Elementary Reaction Kinetics, Latham &amp; Burgess, 3<sup>rd</sup> Edition<br />
</strong></p>
<p><strong>Reaction Kinetics Volume One, Laidler, 1970<br />
</strong></p>
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		<title>The Oscillatory Phenomena Of Mass-Spring Systems</title>
		<link>http://www.christopher-dean.co.uk/2001/01/07/the-oscillatory-phenomena-of-mass-spring-systems/</link>
		<comments>http://www.christopher-dean.co.uk/2001/01/07/the-oscillatory-phenomena-of-mass-spring-systems/#comments</comments>
		<pubDate>Sun, 07 Jan 2001 21:09:55 +0000</pubDate>
		<dc:creator>Chris</dc:creator>
				<category><![CDATA[Essays]]></category>
		<category><![CDATA[Science]]></category>

		<guid isPermaLink="false">http://www.christopher-dean.co.uk/2007/06/06/the-oscillatory-phenomena-of-mass-spring-systems/</guid>
		<description><![CDATA[1. The System In Equilibrium: When the system is in equilibrium the total force F on the system is 0 as there is no driving force F(t). There is no y diplacement, and therefore no velocity term and hence no drag term either. The force of gravity and the restoring force of the spring are [...]]]></description>
			<content:encoded><![CDATA[<!-- sphereit start --><p><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat1.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat1 The Oscillatory Phenomena Of Mass Spring Systems" /><span style="font-size: 20pt">1.<br />
</span></p>
<p><strong>The System In Equilibrium:<br />
</strong></p>
<p>When the system is in equilibrium the total force F on the system is 0 as there is no driving force F(t).</p>
<p>There is no y diplacement, and therefore no velocity term and hence no drag term either.</p>
<p>The force of gravity and the restoring force of the spring are the only forces able to act on the static system, and by vertue of the fact that it is static, the gravity force term mg equals the restoring force F<sub>s</sub>.</p>
<p>The equation describing the system is initially:<img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat2.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat2 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p>As force is the effect of an accelerated mass, and acceleration is the second derivitve of the diplacement with respect to time, this equation becomes: <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat3.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat3 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p><strong>The Displaced System:<br />
</strong></p>
<p>When the mass is dispaced from the equilibrium position y<sub>e</sub> (&#8216;by pulling down&#8217;) to position y and is the released the total force on the system is not 0.</p>
<p>F<sub>s</sub> depends on y in all cases; when the mass is displaced and the spring is stretched, F<sub>s</sub> &gt; 0.</p>
<p>mg opposes the total force F which is in the positive y direction.</p>
<p><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat4.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat4 The Oscillatory Phenomena Of Mass Spring Systems" />is the drag term which also is always in the opposite direction to the total force on the system, therefore it is initially in the negative y direction in this case.</p>
<p>F(t) is the driving force of the sytem which is due to the release of the displaced mass, and is directed in the positive y direction due to the fact that the mass was initially &#8216;pulled down&#8217; i.e. directly away from the fixed end of the spring.</p>
<p>The equation for this system can be based upon that derived for a suystem in equlibrium above: <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat5.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat5 The Oscillatory Phenomena Of Mass Spring Systems" /> but now we have to consider the facts that the total force on the system now has as driving force F(t) as its main component in the positive y direction and a drag term opposing the total force which is initially in the negative y direction, making the overall equation for the system:</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat6.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat6 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p>If we rearrange this equation, assuming a Hooke&#8217;s Law spring: <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat7.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat7 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p>We see that it is a linear, non-homogenous ordinary differential equation, with constant co-efficients of order 2 and degree1.</p>
<p>Appropriate initial conditions would be:</p>
<p>y(0) = y<sub>e</sub>, y(t<sub>1</sub>) = y<sub>1, </sub></p>
<p><span style="font-size: 20pt"><strong>2.<br />
</strong></span></p>
<p>This equation can be expressed as a system of first order differential equations as follows:</p>
<p>The displacement of the mass with respect to time is the velocity given by <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat8.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat8 The Oscillatory Phenomena Of Mass Spring Systems" /> the second derivative of the displacement of the mass with respect to time is the acceleration of t he mass and can be expressed as the derivative of the velocity <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat9.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat9 The Oscillatory Phenomena Of Mass Spring Systems" />. Therfore the equation overall is as follows:</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat10.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat10 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p>this second way of writing the equation is better from a computational point of view as you only have to write the code to solve a first order differential equation and apply it however many times you need to solve the problem, where as the first equation requires a pice of code to perform a single differentiation and also a second differentiation which takes more time as is shown in the programmes Q2.a and Q2.b on the disk porovided.</p>
<p><span style="font-size: 20pt"><strong>3.<br />
</strong></span></p>
<p>Solving <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat11.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat11 The Oscillatory Phenomena Of Mass Spring Systems" /> :</p>
<p>Use the characteristic equation y=e<sup>rt</sup> where r is a constant:</p>
<p><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat12.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat12 The Oscillatory Phenomena Of Mass Spring Systems" />    <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat13.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat13 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p>Assume the differential equation has the following form where a<sub>1</sub>, a<sub>2</sub> a<sub>3</sub>, are all arbitary constants:</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat14.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat14 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p>It then follows that:</p>
<p style="text-align: center">a<sub>2</sub>r<sup>2</sup> e<sup>rt</sup> + a<sub>1</sub>re<sup>rt</sup> + a<sub>0</sub>e<sup>rt</sup> = 0</p>
<p>Which gives:</p>
<p style="text-align: center">e<sup>rt</sup>(a<sub>2</sub>r<sup>2</sup> + a<sub>1</sub>r + a<sub>0</sub>)</p>
<p>So the solution for r is given by:</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat15.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat15 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat16.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat16 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat17.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat17 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p>The general solution is initially:</p>
<p style="text-align: center">y = C<sub>1</sub>e<sup>2it</sup> + C<sub>2</sub>e<sup>-2it</sup></p>
<p>But:</p>
<p style="text-align: center">e<sup>i<span style="font-family: Symbol">q</span><br />
</sup>= cos<span style="font-family: Symbol">q</span> + isin<span style="font-family: Symbol">q</span></p>
<p>So from the principle of superposition:</p>
<p style="text-align: center">y = y<sub>1</sub> + y<sub>2</sub> = C<sub>1</sub>(cos(2t) + isin(2t)) + C<sub>2</sub>(cos(-2t) + isin(-2t))</p>
<p><span style="color: red">And:         y = y<sub>1</sub> &#8211; y<sub>2</sub> = C<sub>1</sub>(cos(2t) + isin(2t)) &#8211; C<sub>2</sub>(cos(-2t) + isin(-2t))        ??????<br />
</span></p>
<p style="text-align: center">As y(0) = 0</p>
<p style="text-align: center">0 = C<sub>1</sub>(1) + C<sub>2</sub>(1)</p>
<p>i.e.</p>
<p style="text-align: center">-C<sub>1 </sub>= C<sub>2<br />
</sub></p>
<p>Unfortunately I am lost with how to rearrange all this to get the required result, which i know is that:</p>
<p><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat18.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat18 The Oscillatory Phenomena Of Mass Spring Systems" /><span style="color: red">where C is a constant who&#8217;s value is obtained thus:        ??????????<br />
</span></p>
<p><span style="color: red">y(0) = 0 <span style="font-family: Symbol">Þ</span> C = 0                                ??????????<br />
</span></p>
<p><span style="font-size: 20pt"><strong>4.<br />
</strong></span></p>
<p><span style="font-size: 20pt"><strong>5.<br />
</strong></span></p>
<p><span style="font-size: 20pt"><strong>6.<br />
</strong></span></p>
<p><span style="font-size: 20pt"><strong>7.<br />
</strong></span></p>
<p><span style="font-size: 20pt"><strong>8.<br />
</strong></span></p>
<p><span style="font-size: 20pt"><strong>9.<br />
</strong></span></p>
<p>If moved to the moon the spring would feel a significant difference on it extension due to the change in the force of gravity. On Earth the force of gravity acting on the mass would extend the spring to an equilibrium position y<sub>e</sub> as in the two diagrams at the start of this project.But on the moon the force of gravity is less and as a result the spring would have a different equilibrium position that would be vertically higher than that which it has on the Earth(i.e. if y<sub>e</sub> = 0 on Earth, then on the moon y<sub>e</sub> &gt; 0).</p>
<p>On the Earth the force of gravity acting on the mass along with the drag term cause a resistance to the motion of the oscillating spring, the spring on the moon would almost be a perfect simple harmonic oscillator and go on oscillation for ever once started. However therte is gravity on the moon so the it will eventually stop oscillating but not as quickly as would be expected on the Earth, which may cause problems for apperatus that needs to be critically damped.</p>
<p>Also due the lack of resistance forces the motion of the mass on the sping will seem faster than if it were on the Earth as there is less to slow it down.</p>
<p><span style="font-size: 20pt"><strong>10.<br />
</strong></span></p>
<p><span style="font-size: 20pt"><strong>11.<br />
</strong></span></p>
<p>Solving: <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat19.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat19 The Oscillatory Phenomena Of Mass Spring Systems" />    <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat20.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat20 The Oscillatory Phenomena Of Mass Spring Systems" />    <span style="font-family: Symbol">w</span> = 4    y(0) = 0        <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat21.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat21 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat22.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat22 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p>Assume a homogenious solution i.e.</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat23.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat23 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p>Assume a particular solution as f(t) = 3sin(4t) (it&#8217;s differneial family is cos, sin).</p>
<p style="text-align: center">y<sub>p</sub> = C<sub>1</sub>sin(4t) + C<sub>2</sub>cos(4t)</p>
<p style="text-align: center">y<sub>p</sub>&#8216; = 4C<sub>1</sub>cos(4t) &#8211; 4C<sub>2</sub>sin(4t)</p>
<p style="text-align: center">y<sub>p</sub>&#8221; = -16C<sub>1</sub>sin(4t) &#8211; 16C<sub>2</sub>cos(4t)</p>
<p style="text-align: center">-16C<sub>1</sub>sin(4t) &#8211; 16C<sub>2</sub>cos(4t) +16(C<sub>1</sub>sin(4t) + C<sub>2</sub>cos(4t)) = 0</p>
<p>Therefore the general solution to the homogenious version of the equation is:</p>
<p style="text-align: center">y<sub>p</sub> = C<sub>1</sub>sin(4t) + C<sub>2</sub>cos(4t)</p>
<p>As y(0) = 0,<span style="font-family: Symbol">Þ</span> C<sub>2</sub> = 0, so the solution becomes:</p>
<p style="text-align: center">y<sub>p</sub> = C<sub>1</sub>sin(4t)</p>
<p>Now we need to find the complimentary solution to the homogenious solution i.e.</p>
<p><span style="font-size: 20pt"><strong>12.<br />
</strong></span></p>
<p><span style="font-size: 20pt"><strong>13.<br />
</strong></span></p>
<p><span style="font-size: 20pt"><strong>14.<br />
</strong></span></p>
<p><span style="font-size: 20pt"><strong>15.</strong></span></p>
<p style="text-align: center">Solving: <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat24.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat24 The Oscillatory Phenomena Of Mass Spring Systems" />    y(0) = 0, 4, 5, 6     <img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat25.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat25 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat26.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat26 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p style="text-align: center">y = e<sup>rt</sup> r = constatnt</p>
<p style="text-align: center">y&#8217; = re<sup>rt</sup></p>
<p style="text-align: center">y&#8221; = r<sup>2</sup>e<sup>rt</sup></p>
<p>As in question 3.</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat27.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat27 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p>It then follows that:</p>
<p style="text-align: center">a<sub>2</sub>r<sup>2</sup> e<sup>rt</sup> + a<sub>1</sub>re<sup>rt</sup> + a<sub>0</sub>e<sup>rt</sup> = 0</p>
<p>Which gives:</p>
<p style="text-align: center">e<sup>rt</sup>(a<sub>2</sub>r<sup>2</sup> + a<sub>1</sub>r + a<sub>0</sub>)</p>
<p>So the solution for r is given by:</p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat28.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat28 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat29.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat29 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p style="text-align: center"><img src="http://www.christopher-dean.co.uk/wp-content/uploads/2007/06/060607-2113-theoscillat30.png" title="The Oscillatory Phenomena Of Mass Spring Systems" alt="060607 2113 theoscillat30 The Oscillatory Phenomena Of Mass Spring Systems" /></p>
<p>The general solution is initially:</p>
<p style="text-align: center">y = C<sub>1</sub>e<sup>(¼+4.987i)t</sup> + C<sub>2</sub>e<sup>-(¼+4.987i)t</sup></p>
<p>But:</p>
<p style="text-align: center">e<sup>i<span style="font-family: Symbol">q</span><br />
</sup>= cos<span style="font-family: Symbol">q</span> + isin<span style="font-family: Symbol">q</span></p>
<p>So:</p>
<p style="text-align: center">y<sub>1</sub> = C<sub>1</sub>(cos(2t) + isin(2t)</p>
<p style="text-align: center">y<sub>2</sub> = C<sub>2</sub>(cos(-2t) + isin(-2t)</p>
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