GREPhysics.NET: GR8677 #67

GR8677 #67

Problem

GREPhysics.NET Official Solution

Statistical Mechanics $\Rightarrow$ }Partition Function

The problem gives three non-degenerate energies, so one can just directly plug this into the canonical(?) partition function to get,

$Z=\sum_i e^{-\frac{\epsilon_i}{kT}}=1+ e^{-\frac{\epsilon}{kT}}+ e^{-3\frac{\epsilon}{kT}},<br />$

where $k$ is the Boltzmann constant, $T$ is the (absolute) temperature.

Since,

$U = Nk T^2/Z \frac{\partial Z}{\partial T} = Nk^2 \left(\frac{\epsilon}{kT^2} e^{-\frac{\epsilon}{kT}}+ \frac{3\epsilon}{kT^2} e^{-3\frac{\epsilon}{kT}} \right).<br />$

For $\epsilon >> kT$ , one can expand $e^x \approx 1+x$ , and thus,

$\frac{U}{Nk^2} \approx \frac{\epsilon}{kT^2}(1-\epsilon/kT) + \frac{3\epsilon}{kT^2}(1-3\epsilon/kT) \approx 4\epsilon, <br />$

where one throws out the higher order $\epsilon$ terms.

( $Z\approx 1$ in denominator) The average energy of each particle is $U/3=\frac{4}{3}\epsilon$ , as in choice (C).

Alternate Solutions

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Comments

Jeremy
2007-11-13 12:21:20

A few things about the long way (official solution)... (1) We don't want the average energy of the system, we want the average energy of each particle. (2) In the first equation for energy $U$ , the $k^{2}$ should be replaced by $k T^{2}$ . (3) $\lim_{t \to \infty} Z=3$ . Also, I think the solution is cleaner using $\beta$ ( $\beta \equiv \frac{1}{k T}$ ).

$Z=1+e^{-\beta \epsilon}+e^{-3\beta \epsilon}$

$E=-\frac{1}{Z} \frac{\partial Z}{\partial \beta}=\frac{\epsilon (e^{-\beta \epsilon}+3 e^{-3\beta \epsilon})}{Z}$

As $t \to \infty$ , $\beta \to 0$ . Therefore the limit causes all exponentials to become $1$ .

$E=\epsilon \frac{1+3}{3}=\frac{4}{3} \epsilon$ .

FortranMan
2008-10-28 11:10:05

So the absolute temperature is when T $\rightarrow \infty$ ? Is that what they mean by absolute temperature?

eliasds
2008-11-03 21:48:34

I think by absolute temperature, they are referring to temperature in degrees kelvin.

physicsisgod
2008-11-05 21:04:47

No, Jeremy is just taking kT >> 1, which makes $\beta \rightarrow$ 0

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hefeweizen
2006-11-30 12:42:40

i think there is a typo -

epsilon << kT

beta*epsilon << 1

and you can just use:

sum(E__i*exp(-beta*epsilon))/Z

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Andresito
2006-03-18 20:31:05

jakevdp, thank you for posting your alternate solution. It seems that that is the shortest and the one ETLS wants you to think of.

Yosun, when you have zeta*(Dzeta/Dtemperature) how come that in the remaining exponential terms (second equation) do not have a factor of 2 in their powers?

I think there should be exp(-2 epsilon/kT) and exp(-6 epsilon/kY

eliasds
2008-11-03 21:47:43

I think by absolute temperature, they are referring to temperature in degrees kelvin.

Reply to this comment

jakevdp
2005-11-01 11:19:24

Alternately, you can realize that at kT>>e, Entropy is near maximum, thus each particle has roughly equal probablility of being in any of the three states. Thus, average energy is simply (0+e+3e)/3 = (4/3)e

nitin
2006-11-16 11:17:24

Yosun

This long and elaborate calculation is truly inappropriate and silly when it comes to answering a GRE MCQ. As jakevdp pointed out, since $kT>>\epsilon$ , all 3 possible nondegenerate energy states are equally likely to be occupied by any of the N particles. Therefore, the average energy of each particle would be $\frac{0+\epsilon+3\epsilon}{3}=\frac{4}{3}\epsilon$ .

grae313
2007-10-07 18:09:55

nitin, if you look at all the answers Yosun gives, one could only conclude that he(?) is presenting the rigorous derivation of each solution where applicable, for those who want to see and study the physics behind the solution. I doubt he thinks it is the best solution or the way to approach the ETS exam. This is for studying only.

Richard
2007-10-31 18:03:39

she...

madfish
2007-11-02 16:24:24

it. j/p

physicsisgod
2008-11-05 21:07:32

My long, elaborate, innappropriate and silly calculation shows that nitin is a douche.

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Post A Comment!

You are replying to:

jakevdp, thank you for posting your alternate solution. It seems that that is the shortest and the one ETLS wants you to think of.
Yosun, when you have zeta*(Dzeta/Dtemperature) how come that in the remaining exponential terms (second equation) do not have a factor of 2 in their powers?
I think there should be exp(-2 epsilon/kT) and exp(-6 epsilon/kY

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