I was solving problem 5.18 concerning the free expansion of a gas in Fundamentals of Statistical and Thermal Physics by F. Rief. In this process, the internal energy $E$ is constant i.e. $dE=0$. The first part of the problem is to find an expression for $\big(\frac{\partial T}{\partial V}\big)_E$ in terms of the pressure $p$, temperature $T$, $\big(\frac{\partial p}{\partial T}\big)_V$, and the heat capacity at constant volume $C_V$. (The subscripts of the partial derivatives indicate the variable that is held constant). By writing $E=E(T,V)$ and solving for $dE=0$ it can be shown that $$ \label{one} \tag{1} \bigg(\frac{\partial T}{\partial V}\bigg)_E=\frac{1}{C_V}\bigg[p-T\bigg(\frac{\partial p}{\partial T}\bigg)_V\bigg]. $$
The second part asks for an expression for $\big(\frac{\partial S}{\partial V}\big)_E$, $S$ being the entropy, and by examining the first law $dE=TdS-pdV$ and, again, solving for $dE=0$ one can find $$ \tag{2} \bigg(\frac{\partial S}{\partial V}\bigg)_E=\frac{p}{T}. $$
The last part of the problem is where I am struggling. The question asks for the temperature change $\Delta T=T_2-T_1$ resulting from a volume change from $V_1$ to $V_2$ using parts $(1)$ and $(2)$. Here is how I attempted a solution:

From $(2)$: $$ p=T\bigg(\frac{\partial S}{\partial V}\bigg)_E \\ \begin{align} \bigg(\frac{\partial p}{\partial T}\bigg)_V &=\bigg(\frac{\partial}{\partial T}\bigg)_V\bigg(T\bigg(\frac{\partial S}{\partial V}\bigg)_E\bigg) \\ &=\bigg(\frac{\partial T}{\partial T}\bigg)_V\bigg(\frac{\partial S}{\partial V}\bigg)_E+T\bigg(\frac{\partial}{\partial T}\bigg)_V\bigg(\frac{\partial S}{\partial V}\bigg)_E \\ &=\bigg(\frac{\partial S}{\partial V}\bigg)_E+T\bigg(\frac{\partial }{\partial V}\bigg)_E\bigg(\frac{\partial S}{\partial T}\bigg)_V \\ &=\frac{p}{T}+T\bigg(\frac{\partial}{\partial V}\bigg)_E\bigg(\frac{\partial S}{\partial T}\bigg)_V \tag{3} \end{align} $$ where in the last step I swapped the order of differentiation assuming $S$ is a well-behaved function. To find $\big(\frac{\partial S}{\partial T}\big)_V$ I write $S=S(E,V)$. Since the derivative is at constant volume it becomes $$ \bigg(\frac{\partial S}{\partial T}\bigg)_V=\frac{\partial S}{\partial E}\bigg(\frac{\partial E}{\partial T}\bigg)_V=\frac{1}{T}C_V \tag{4}. $$ Therefore $$ \begin{align} \bigg(\frac{\partial}{\partial V}\bigg)_E\bigg(\frac{\partial S}{\partial T}\bigg)_V &=\bigg(\frac{\partial}{\partial V}\bigg)_E\bigg(\frac{C_V}{T}\bigg) \\ &=\frac{1}{T}\bigg(\frac{\partial C_V}{\partial V}\bigg)_E-\frac{C_V}{T^2}\bigg(\frac{\partial T}{\partial V}\bigg)_E \tag{5}. \end{align} $$ Substituting back into $(3)$: $$ \bigg(\frac{\partial p}{\partial T}\bigg)_V=\frac{p}{T}+\bigg(\frac{\partial C_V}{\partial V}\bigg)_E-\frac{C_V}{T}\bigg(\frac{\partial T}{\partial V}\bigg)_E. \tag{6} $$ Rearranging: $$ \bigg(\frac{\partial T}{\partial V}\bigg)_E=\frac{1}{C_V}\bigg[p-T\bigg(\frac{\partial p}{\partial T}\bigg)_V+\bigg(\frac{\partial C_V}{\partial V}\bigg)_E\bigg]. \tag{7} $$ By comparing $(1)$ and $(7)$ one finds that $$ \bigg(\frac{\partial C_V}{\partial V}\bigg)_E=0 \tag{8} $$ or that $C_V$ does not depend on the total volume.

I obviously failed to find a general expression for $\Delta T$, but I arrived at a puzzling result (at least for me). Is it true that $C_V$ is volume independent for any gas if the internal energy is held fixed? Also, how do I proceed to find an expression for $\Delta T$? Is it possible without foreknowledge of the equation of state?


closed as off-topic by John Rennie, stafusa, Kyle Kanos, Emilio Pisanty, sammy gerbil Aug 25 '18 at 20:11

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  • $\begingroup$ Although I didn't follow your analysis, regarding the last paragraph it is true that, for an ideal gas in a closed system, the change in internal energy for any process is always the specific heat at constant volume time the change in temperature. It may seem counter intuitive since Cv applies to a constant volume process, but its true nonetheless. If I can get a chance I'll demonstrate it. $\endgroup$ – Bob D Aug 17 '18 at 21:44
  • $\begingroup$ Is the problem dealing with an ideal gas? $\endgroup$ – Bob D Aug 18 '18 at 6:16
  • $\begingroup$ No, not necessarily ideal $\endgroup$ – Omar Osama Alsheikh Aug 18 '18 at 12:12
  • $\begingroup$ Please note that this site is not a place to obtain solutions to worked problems. Please see this Meta post on asking homework-like questions and this Meta post for "check my work problems". $\endgroup$ – Kyle Kanos Aug 21 '18 at 10:15

I'm pretty sure that $C_V$ may also depend on the volume and $(8)$ is not generally true. The point in your analysis where it goes wrong is the exchange of derivatives in $(3)$.

You may exchange derivatives like

$$ \left( \frac{\partial}{\partial V} \right)_E \left( \frac{\partial S}{\partial E} \right)_V = \left( \frac{\partial}{\partial E} \right)_V \left( \frac{\partial S}{\partial V} \right)_E, $$

where both partial derivatives are with respect to the same basis, $E$ and $V$. This is assured by Schwarz's theorem as long as the functions are smooth enough.

However, if the derivatives are in different bases, there is no need why they should commute. As a purely mathematical example, think of

$$ S = E = T V. $$


$$ \left( \frac{\partial}{\partial T} \right)_V \left( \frac{\partial S}{\partial V} \right)_E = 0, $$ $$ \left( \frac{\partial}{\partial V} \right)_E \left( \frac{\partial S}{\partial T} \right)_V = 1. $$

In general, you can use the chain rule to transform one of the derivatives into the other's basis, e.g.

$$ \left( \frac{\partial S}{\partial V} \right)_E = \left( \frac{\partial T}{\partial V} \right)_E \left( \frac{\partial S}{\partial T} \right)_V + \left( \frac{\partial S}{\partial V} \right)_T. $$

Then the difference of both orderings follows as

$$ \left( \frac{\partial}{\partial T} \right)_V \left( \frac{\partial S}{\partial V} \right)_E - \left( \frac{\partial}{\partial V} \right)_E \left( \frac{\partial S}{\partial T} \right)_V = \left[ \left( \frac{\partial}{\partial T} \right)_V \left( \frac{\partial T}{\partial V} \right)_E \right] \cdot \left( \frac{\partial S}{\partial T} \right)_V. $$

So, I guess the reason for your strange result is that this difference is missing. Also note that a factor $T$ is missing in front of the last partial derivative in $(7)$.

Regarding how to compute the temperature change in a free expansion, I'm not sure what the intended solution is. Without knowing the first part of the problem, I would have taken the energy $E(T,V)$ as a function of $T$ and $V$ and solved for $T_2-T_1$. One could determine $E(T,V)$ by integrating over its partial derivatives,

$$ \left( \frac{\partial E}{\partial T} \right)_V = C_V \qquad \text{and} \qquad \left( \frac{\partial E}{\partial V} \right)_T = T \left( \frac{\partial p}{\partial T} \right)_V - p, $$


$$ E(T,V) = E(T_0,V_0) + \int_{T_0}^T C_V \mathrm{d} T' + \int_{V_0}^V \left[ T \left( \frac{\partial p}{\partial T} \right)_V - p \right] \mathrm{d} V', $$

but I don't really see much connection to the first parts of the problem. Perhaps there is a simple way to arrive at the temperature difference which I'm overlooking.


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