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

kroner
2009-10-08 09:47:38

If $\epsilon$ is the difference in energy between two states, then as $\epsilon/kT \to 0$ , the ratio of the probabilities of being in each state $e^{\epsilon/kT}$ goes to 1, so the states become equally likely (as jakevdp and others point out).

Then the average energy of each particle is $(0 + \epsilon +3\epsilon)/3 = 4\epsilon/3$ .

Reply to this comment

Comments

kroner
2009-10-08 09:47:38

chemicalsoul
2009-11-03 21:23:51

This is it ! the average value of a three sided dice. No need for lengthy derivation.

Reply to this comment

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

Reply to this comment

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

Reply to this comment

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.

Herminso
2009-09-01 12:35:18

kT>>e means that Entropy is near maximum, and remember that the maximum for the entropy is achieved when the system reach the equilibrium thermodynamic. The Postulate of Equal Priori Probability: When a macroscopic system is in thermodynamic equilibrium, its state is equally likely to be any state satisfying the macroscopic conditions of the system (k. Huang 2ed pag 129).

That justify why each particle has roughly equal probability of being in any of the three states, just as jakevdp did.

Reply to this comment

Post A Comment!

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...

<a href="http://www6.shoutmix.com/?grephysics">View shoutbox</a>
ShoutMix chat widget

1GHz Smartphone for 1 cent - best deal on a Samsung Galaxy Phone

Free 3D Virtual World. Largest User-Created Fantasy World. Don't Play, Experience. Join Now!

Bored, got time to kill, and want to earn micro-cash by mindlessly clicking on links? Join CrownGPT. Click Offers. Click the Paid to Click button... then click to your heart's delight.

FREE 3D Virtual World Chat - Join Second Life Today!

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.

Bare Basic LaTeX Rosetta Stone