GREPhysics.NET: GR9277 #72

GR9277 #72

Problem

GREPhysics.NET Official Solution

\prob{72}
A system in thermal equilibrium at temperature T consists of a large number $N_0$ of subsystems, each of which can exist only in two states of energy $E_1$ and $E_2$ , where $E_2-E_1=\epsilon \gt 0$ . In the expressions that follow, k is the Boltzmann constant.

The internal energy of the system at any temperature T is given by $E_1N_0+\frac{N_0\epsilon}{1+e^{\epsilon/kT}}$ . The heat capacity of the system is given by which of the following expressions?

$N_0k \left(\frac{\epsilon}{kT}\right)^2\frac{e^{\epsilon/kT}}{\left(1+e^{\epsilon/kT}\right)^2}$
$N_0k \left(\frac{\epsilon}{kT}\right)^2\frac{1}{\left(1+e^{\epsilon/kT}\right)^2}$
$N_0k\left(\epsilon/kT\right)^2e^{-\epsilon/kT}$
$N_0k\left(\epsilon/kT\right)^2$
$3N_0k/2$

Statistical Mechanics $\Rightarrow$ }Heat Capacity

The heat capacity is just $dU/dT$ , where ETS generously supplies U, the internal energy. Since $E_1$ and $N_0$ are constants, the first term is trivial.

The temperature-derivative of the second term is $N_0\epsilon^2/(kT^2)e^{\epsilon/kT} /(1+e^{\epsilon/kT})^2=$ , as in choice (A).

(The temperature derivative is easily done if one applies the chain-rule $\frac{df}{dy}\frac{dy}{du}\frac{du}{dT}$ where $f=1/y$ , $y=1+e^u$ , $u=\epsilon/(kT)$ .)

Alternate Solutions

There are no Alternate Solutions for this problem. Be the first to post one!

Comments

RebeccaJK42
2007-03-23 10:27:19

Where does the k in the numerator come from?

alpha
2007-03-30 22:37:37

k is Boltzmann constant.. The k on the numerator is actually canceled out by the $k^2$ in the denominator

Richard
2007-09-14 00:05:03

The fact of the matter is, you have a factor of
$\epsilon/k$ introduced by the derivative.
To make it look pretty, they multiplied the top and bottom by $k$ giving (with the extra $\epsilon$ ) the factor $k(\frac{\epsilon}{k})^2$ . Then of course, you have the $T^2$ .

Reply to this comment

Andresito
2006-03-29 10:05:29

I could not obtain T^2 in the denominator. Is there an error in the solution provided by the ETS?

radicaltyro
2006-10-31 14:19:25

Hi Andresito,

Review your calculus and try again. The answer is correct.

Blue Quark
2007-11-01 08:00:11

Andresito you made the same mistake I did. The T is in the denominator, not the numerator. Thus when you differentiate you get a (-T^(-2)) in front of the exponential in addition to the normal e/k

Reply to this comment

Post A Comment!

You are replying to:

Where does the k in the numerator come from?

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...

Chat Archives | Delete left banner ad | Donate

(Click to view chat.)

Hate being Anonymous? Login or Register

Upgrade to Poser 7 Now

Huge Textbook Selection, Low Prices – Phat Campus.

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