PHYS Statistical Mechanics I Assignment 1 Solutions

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Hannah Wilkinson
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1 PHYS Statistical Mechanics I Assignment 1 Solutions 1. That was an easy 5 points, eh? 2. (a) First of all let s assume that the gas is made up of mostly nitrogen (diatomic) molecules, which is correct for air. A bit of searching on the web (actually wikipedia) gives the following values for the van der Waals coefficients: a J m 3 /mol 2, b m 3 /kmol= m 3 /mol. I hate moles, though. We know that Avogadro s constant N A = mol 1 converts to numbers of particles, so a J m 3 /molecule 2 and b m 3 /molecule. Let s check that the units work out. Obviously, b must have the same units as volume, so that s good. We also need a/v 2 to have the same units as pressure. With a in units of J m 3 this gives a/v 2 in units of J/m 3. Pressure is force per unit area or N/m 2 but 1 N=1 J/m so pressure has units of energy per unit volume and we are o.k. By the way, these parameters have important physical meanings. In particular, you can think of b (in units of volume per molecule) as the effective volume taken up by a single molecule, as if it were a spherical infinitely hard ball. So the accessible volume in the ideal gas law is then reduced by the sum of the volumes of all the particles. Assuming that the volume of the molecule is (4/3)πr 3 = , one obtains the hard-core radius r m, or about about 2.5 angstroms. This is consistent with the actual size of a molecule! The parameter a measures the attraction of the particles; for a > 0 the atoms repel each other and thereby increase the effective pressure, and vice versa. In the meantime, atmospheric pressure is P 0 = J/m 3 and room temperature is, say, T 0 = 300 K. Assuming a volume of V 0 = 1 m 3, then neglecting the van der Waals coefficients the ideal gas law becomes N ( J/K ) (300 K) = ( J/m 3 ) (1 m 3 ) which gives N molecules of gas. This was a reasonable assumption since a/v0 2 P 0 and b V 0. We ll assume from now on that this is our value of N and it remains fixed. With the value of N, we can now calculate the van der Waals coefficients for the full system: a = ( ) 2 J m J m 3 and b m m 3. An alternative way of thinking of this is that with N = , there are N/N A 40.7 moles of gas in 1 m 3. So a = J m 3 ; likewise for b. (b) Before we plot anything it is important to render this equation completely unitless, because computers don t know anything about units! Clearly b has units of volume so: ( Nk B T = b P + a ) ( ) ( V V 2 b 1 = b P + a ) b 2 V 2 (V 1),

2 where I have defined the dimensionless volume V V/b. Because V has no units a/b 2 must have units of pressure: Nk B T = ba ( b 2 P b 2 a + 1 ) V 2 (V 1) = a ( P + 1 ) b V 2 (V 1), where P P b 2 /a using the values for a and b above. Finally, a/b must have units of energy because these are the units of the left side. So I can define the dimensionless temperature T = Nk B T b/a to give me the final dimensionless van der Waals equation ( T = P + 1 ) V 2 (V 1). We can now solve for the pressure: P = T V 1 1 V 2. At this point we can start plotting the pressure as a function of volume for various temperatures. This is done on the attached Mathematica notebook. We need to be sure that we are plotting in the requested range, though: 0.1V 0 V V 0. In terms of the rescaled volumes, the range is 0.1V 0 /b V V 0 /b or 62.8 V 628. Likewise, T at room temperature is T (f) Plots of the curves show that the pressure decreases like 1/V, as expected from the ideal gas model and the van der Waals model for a, b 1. So the slope of the P V curve vanishes for V. For sufficiently low temperatures, though, the van der Waals coefficients can change the behaviour. In fact, the curve will be such that the slope could vanish even for finite P. The attached Mathematica notebook does the calculations, but let s summarize them here (I won t usually do this!). Taking the derivative of P with respect to V gives the critical temperature T c = 2(V 1) 2 V 3. Plugging this into the expression for the pressure gives P c = V 2 V 3. For the slope to be zero at finite P, the curve must take the shape of a squashy S. This means that the curvature around P c must also be zero. Setting the second derivative of P with respect to V to zero gives critical values for both temperature and volume: T c = 8/27 and V c = 3. This in turn gives P c = 1/27. Reverting back to the unitful variables gives T c = 8a/27Nk B b, V c = 3b, and P c = a/27b 2. Plugging in the numbers found in part (a) (mutiplying by the number of particles in each case to get the values for the full system) gives T c 128 K, V c m 3, and P c Pa which is approximately 33 atmospheres. So: cold, small, and high pressure! Sounds like a perfect combination of parameters to turn a gas into a fluid or a solid.

3 Assignment 1 Solutions (2b) First let s solve for the pressure in terms of the volume using the unitless version of the van der Waals equation: Clear T, P, V rhs P 1 V^2 V 1 ans FullSimplify Solve rhs T, P P 1 V 2 1 V P T 1 V 1 V 2 P P. ans 1 T 1 V 1 V 2 Now plot the results for room temperature. Note the range in the unitless variables: k ^ 23 Num ^ 25 a ^ 49 Num^2 b ^ 29 Num T Num k 300 b a plot1 Plot P, V, 62.8,

4 2 ass1sol.nb Ok, this looks like what we would expect for an ideal gas. (2c) Now let s obtain the values of the pressure at discrete volume points: For i 1, i 10, V i 62.8; Vval i V; Pval i P; i ; ; tab Table Vval i, Pval i, i, 1, 10 plot2 ListPlot tab 62.8, , 125.6, , 188.4, , 251.2, 75616, 314., 20569, 376.8, , 439.6, , 502.4, , 565.2, , 628., (2d) The plots look the same; let s check by plotting them on the same figure: Show plot1, plot (2e) Now let s lower the temperature until something interesting happens. I find that on this volume scale, we need to lower the unitless temperature to 0.2:

5 ass1sol.nb 3 T 0.2; Plot P, V, 1, Clearly there is a dip in the pressure where there wasn t before! (2f) Now let s solve for the critical value of the temperature. First take the derivative of the pressure with respect to volume, and set it to zero: Clear T, V ; ans1 FullSimplify D P, V ans2 FullSimplify D ans1, V T 2 1 V 2 V 3 2 T 1 V 3 6 V 4 ans1b FullSimplify Solve ans1 0, T T 2 1 V 2 V 3 T T. ans1b V 2 V 3 This gives a critical pressure: FullSimplify P 2 V V 3 Clearly the critical volume ought to be larger than 2! It turns out that the point at which the slope vanishes also must correspond to an inflection point, so we can also set the second derivative of the pressure with respect to volume to zero:

6 4 ass1sol.nb T.; V.; ans2b FullSimplify Solve ans1 0, ans2 0, T, V T, V T, V. ans2b 1 P T 8, V , What is the numerical value of the critical temperature? N This is very close to the empirical value found in part (2e) where the curve started looking different from that of the ideal gas. Now substitute the results for the original (unitful) variables: Tc T a b Num k Vc V b Pc P a b^2 Pc

Quick Review 1. Properties of gases. 2. Methods of measuring pressure of gases. 3. Boyle s Law, Charles Law, Avogadro s Law. 4. Ideal gas law.

Quick Review 1. Properties of gases. 2. Methods of measuring pressure of gases. 3. Boyle s Law, Charles Law, Avogadro s Law. 4. Ideal gas law. 5. Dalton s law of partial pressures. Kinetic Molecular Theory