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Assignment #7 - due Wednesday, Nov 19

Chapter 8: 4c,d(only)
Chapter 9: 2a (a only), 5
Chapter 11: 1, 5, 7, 8, 12

Notes / Answers

Carter Chapter 8
Problem 4cd

(c) Derive: $H=G-T(\del G/\del T)_P$.

$$G=U+PV-TS = H-TS$$

Rearranging to get $H=G+TS$ and using the Maxwell relation that $S=-(\del G/\del T)_P$ gives us the desired relationship.

(d) Derive: $C_P=-T(\del^2 G/\del T^2)_P$. (reversible process)

The 'reversible process' comes in this way: In general, if we heat an object, the ratio of small quantity of heat needed to change an object by small temperature (at constant pressure) is $(\delta Q/\del T)_P$.

If the process is reversible, then this ratio is $(\delta Q_r/\del T)_P =(\del H/\del T)_P = C_P$.

Two ways to do this problem...

  1. $C_P=(\del H/\del T)_P$ Take the expression from (c) and take the partial derivative wrt $P$: $$\begineq\[\frac{\del H}{\del T}\]_P &=& \[\frac{\del}{\del T}\{ G-T\left(\frac{\del G}{\del T}\right)_P\}\]_P\\ C_P &=&\left(\frac{\del G}{\del T}\right)_P - 1*\left(\frac{\del G}{\del T}\right)_P -T\left(\frac{\del^2 G}{\del T^2}\right)_P\endeq$$ \endeq$$
  2. The other way was to start with the $T\,ds$ equation (7.11): $$Tds=c_PdT-T\left(\frac{\del v}{\del T}\right)_PdP.$$ for a constant pressure process, $dP=0$, and we can write (in terms of extensive variables) $[TdS]_P=C_PdT$, which you can re-arrange as $C_P=T\left(\del S/\del T\right)_P$.

    Then, you write $S=-\left( \del G/\del T \right)_P$, and you're within a step of the desired result.

Carter Chapter 9
Problem 2

You could start with the $TdS$ equation in terms of $T$ and $V$ to get an expression for $ds$ that you could integrate to get the expression for $S$. Assume for the time being that $c_V$ and $c_P$ are constants. Remember that the the enthalpy of an ideal gas is a function only of the temperature.

Two typical issues with this problem:

  • Account for $n$ properly. Distinguish carefully with upper and lower case: $c_V=C_v/n$ and $c_P=C_P/n$ $V,S,G$ and $v,s,g$. Then, $G/n = g = \mu$.

  • Justify how you get $H=C_PT+H_0$ (This is not a universal thermodynamic identity): Typically, any exact thermodynamic variable is a function of two parameters, e.g. $H=H(T,P)$, or $H=H(T,V)$. It is only because we know that we're dealing with an ideal gas that we know $H=H(T)$ is only a function of $T$, and thus the Pfaffian of $H$ is $dH= \frac{dH}{dT}dT=C_PdT$ instead of something harder to integrate like $dH=\frac{\del H}{\del T}dT + \frac{\del H}{\del V}dV$.

Carter Chapter 9
Problem 5

If you buy Carter's assertion that there are four phases (and there are two components) I think you can readily show that $f=0$, and therefore this is a eutectic point (fully determined, like the triple point of water). So,I will change this to a more qualitative problem:

Read online a bit about eutectic mixtures. Find online a phase diagram of a binary metal mixture near the eutectic point and sketch that, labelling the different phases. Answer briefly: How do you understand / count the different phases in such a system?

No, not this Eutectic!

Sure, the eutectic point is uniquely determined. But, a "vapor phase"??

Here is the phase diagram for the gold-thallium system around the eutectic point:


Looking around a bit, I find that there are perhaps two ways of counting phases and components:

  1. We've got solid thallium, solid gold, and a liquid mixture of thallium and gold. The eutectic point occurs when any amount of solid gold, any amount of solid thallium, and a liquid mixture of the the two are simultaneously.

    This makes 2 components ($k=2$) and 2 phases of each ($\pi=2$). Using the Gibbs phase rule:

    $f = 2+k-\pi=2+2-\pi=2$.

    The two degrees of freedom are the pressure $P$, and the solid phase "concentration" of gold. Once we've fixed those two parameters, we have no freedom to choose either the temperature $T$, or the gold concentration in the liquid mixture. These are fixed and determine the eutectic point.

  2. Or, some authors talk about solid gold as a single component "phase", and also solid thallium as a "phase". With this way of thinking, we can specify the eutectic point as the point where these 3 "phases" are in equilibrium: solid gold, sold thallium, and a two-component liquid.

    Furthermore, there is a 'reduced phase rule' which comes into play when non-geologic solids are at hand, where the solid phases are pretty insensitive to pressure variations. The pressure doesn't matter, and the effective phase rule becomes...



Carter Chapter 11
Problem 1

In section (d) - The notation, may be a bit obscure: You should show that $$\sigma=\[\bar{\,v^2\,}-(\bar v)^2\]^{1/2}$$ "The square root of [ the average value of ($v^2$) minus (the average value of $v$)${}^2$]."

Carter Chapter 11
Problem 5

When integrating exponentials, you can tell Mathematica to make assumptions about some of your constants, e.g. that one is positive.

Carter Chapter 11
Problem 7

In part (b) - Use integration by parts to show this result. Hint: One of the functions could be $f(x)=x$.

Carter, Chapter 11
Problem 12

Assignment #8 - due Wednesday, Nov 26

Read Chapter 12, Section 4: Boltzmann's argument about entropy
Chapter 12: 1, 4, 6*, 8, 10
Chapter 13: 1

Notes / Answers

Carter Chapter 12
Problem 1

a) 50 honest coins. # of microstates? We need to count the 50-symbol-long sequences like "HHTTHTTTTHTTHHTTHHHT...". There are 2 possibilities for each symbol in the sequence. So the total number of microstates is $$2^{50} = 1.13\times 10^{15}$$

b) The most probable macrostate is the one with equal numbers of heads and tails (25). In this case the number of configurations (equation 12.4) involves large factorials, but Mathematica can deal with it... $$w_\text{max}=\frac{50!}{25! 25!}\approx 1.26 \times 10^{14}.$$

c) The true probability is $w_\text{max}/\Omega$, where $\Omega$ is the total number of microstates (from part a). [Mathematica] so the probability of the most probable state is: $$w_\text{max}/\Omega=\frac{1.26 \times 10^{14}}{1.13\times 10^{15}}=0.11= 11%.$$

Carter Chapter 12
Problem 4
- For 1000 coins, we are to compare $w_{500}= \frac{1000!}{500! 500!}$ with $w_{600}=\frac{1000!}{600! 400!}$. Let's just compare the denominators of these two expressions, since the numerators are the same.

Using Stirling's approximation: $$\begineq \ln\left(\frac{400!600!}{500!500!}\right)&=& \ln 400!+\ln 600! - 2\ln 500!\\ &=&400\ln 400 -400 +600\ln 600 -600 - 2(500\ln 500 -500)\\ &=&400\ln 400 +600\ln 600 - 2(500\ln 500)\\ &=&20.14\endeq $$ And $e^{20.14}=0.56\times 10^9$ so the ratios work out to just about a billion to one.

The easy way is to use Mathematica, which tells us that $$\frac{400!600!}{500!500!} = \frac{8.1\times 10^{2276}}{1.5\times 10^{2268}}=5.4\times 10^8\approx 0.5\,\text{billion} $$ So, if the denominators have this ratio, the corresponding probabilities must have the ratio of about 2 billionths.

Carter Chapter 12
Problem 6
- In the statement of the problem, it looks like there are only 4 energy levels (they list 0, 1$\epsilon$, 2$\epsilon$, 3$\epsilon$.) But I could not find 9 "distributions" of the particles in so few energy levels. Instead, assume that the energy levels just keep going: 0, 1$\epsilon$, 2$\epsilon$, 3$\epsilon$, 4$\epsilon$, 5$\epsilon$, 6$\epsilon$, 7$\epsilon$, 8$\epsilon$,.... Now, I was able to find exactly 9 different distributions of 4 particles with total energy $U=6\epsilon$.

c) $\overline{N_0}=(3*4+2*12+2*12+1*12+2*6+1*24+1*4)/84=112/84=1.33$,
$\overline{N_2}=60/84=0.71$, $\overline{N_3}=40/84=0.48$, $\overline{N_4}=24/84=0.29$,
$\overline{N_5}=12/84=0.14$, $\overline{N_6}=4/84=0.05$.

Carter, Chapter 12
Problem 8

The possible configurations of two particles that have $U=\sum N_i E_i=2\epsilon$ are (0,2), (1,1) and (2,0). There are 3 ways of organizing the particles, so the entropy is $$S=k_B\ln w=k_B\ln 3.$$ [The notation: (0,2) means the first particle has energy $0\epsilon$ and the second particle has energy $2\epsilon$.]

Now, if another particle is added, the possible configurations of the 3 particles are (0,0,2), (0,2,0), (2,0,0), (1,1,0), (1,0,1), and (0,1,1). 6 different configurations, so $$S=k_B\ln 6$$ The entropy has increased by the ratio $$\frac{\ln 6}{\ln 3}=1.63.$$

Carter, Chapter 12
Problem 10

Here, a spreadsheet is perhaps the easiest way to do this. You can put in separate columns the quantum numbers $N_x$, $N_y$, and $N_z$, and make sure you get all combinations (up to about the ones with 4s in them...) Then you can calculate $N_j^2$ in another column, and sort on the $N_j^2$ column to get the energy levels in order:

Notice that another interesting thing starts to happen at the 14th level, where combinations with different numbers have the same $N_j^2$. The degeneracy of this level is 4.