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GR9277 #72
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\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?

1. $N_0k \left(\frac{\epsilon}{kT}\right)^2\frac{e^{\epsilon/kT}}{\left(1+e^{\epsilon/kT}\right)^2}$
2. $N_0k \left(\frac{\epsilon}{kT}\right)^2\frac{1}{\left(1+e^{\epsilon/kT}\right)^2}$
3. $N_0k\left(\epsilon/kT\right)^2e^{-\epsilon/kT}$
4. $N_0k\left(\epsilon/kT\right)^2$
5. $3N_0k/2$

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
 jmason862009-09-03 19:19:22 I did this one by limits and general test taking strategy. Limits: Q --> finite value as T --> $\infty$. Eliminates (D) and (E) ETS will generally force you to choose between very similar answers. Eliminates (C) With only (A) and (B) left, it is a good idea to guess. But if you can get a hunch, that is even better. It's probably totally flawed thinking, but I saw the lack of an exponential in the numerator for (B), which made it look similar to a Bose-Einstein distribution. The full expression in (A) reminded me of the Maxwell-Boltzmann distribution that comes from the problem statement. Choose (A) Reply to this comment
ernest21
2019-09-30 05:00:43
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BerkeleyEric
2010-09-17 17:04:59
If you remember that $C=\frac{dU}{dT}$, then immediately you can see that there will have to be the squared sum in the denominator, so only A and B remain. And there still has to be the exponential in the numerator (a derivative won't get rid of that), so the answer must be A.
 flyboy6212010-10-22 16:11:48 +1
jmason86
2009-09-03 19:19:22
I did this one by limits and general test taking strategy.

Limits: Q --> finite value as T --> $\infty$. Eliminates (D) and (E)

ETS will generally force you to choose between very similar answers. Eliminates (C)

With only (A) and (B) left, it is a good idea to guess. But if you can get a hunch, that is even better. It's probably totally flawed thinking, but I saw the lack of an exponential in the numerator for (B), which made it look similar to a Bose-Einstein distribution. The full expression in (A) reminded me of the Maxwell-Boltzmann distribution that comes from the problem statement. Choose (A)
RebeccaJK42
2007-03-23 10:27:19
Where does the k in the numerator come from?
 alpha2007-03-30 22:37:37 k is Boltzmann constant.. The k on the numerator is actually canceled out by the $k^2$ in the denominator
 Richard2007-09-14 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 $T^2$.
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?
 Blue Quark2007-11-01 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
 drizzo012012-11-08 07:40:44 am I the only one who doesn't see a T in the denominator of the second term? I appreciate that T^2 would be in the denominator if there was a T to being with, but as far as I can see, the only T's are within the expression for the exponential, which don't come out upon taking a derivative.
 drizzo012012-11-08 07:46:58 wait nvm, got it.

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}\$ $\frac{a}{b}$