Entropy changes

What is the entropy change during the expansion of a van der Waals gas for which $C_V$ is a constant?

The equation of state for one mole of a van der Waals gas is

$$P={RT\over V-b}-{a\over V^2};$$

$b$ represents the volume taken up by the finite size of the molecules, and $a/V^2$ is the reduction in pressure due to interactions between the molecules. In this way the two most important corrections neglected in the ideal gas are included. However the heat capacity at constant volume is still independent of temperature and volume, as in an ideal gas.

During an expansion, both the temperature and the volume may change. To calculate the change in entropy, we need

$$\left({\partial S\over\partial T}\right)_{\!\scriptstyle V} ={C_V\over T}$$

and

$$\left({\partial S\over\partial V}\right)_{\!\scriptstyle T} =\left({\partial P\over\partial T}\right)_{\!V}={R\over V-b}$$

so

$$\begin{eqnarray*} {\rm d}S\!\!\!&=&\!\!\!\left({\partial S\over\partial T}\right... ...ion}\\ \!\!\!&=&\!\!\!{C_V\over T}{\rm d}T+{R\over V-b}{\rm d}V \end{eqnarray*}$$

Note: Many problems on this section of the course involve choosing variables, either $(T,V)$ or $(T,P)$, writing ${\rm d}S$ in terms of infinitesimal changes in these variables as in the first line, and then using the definition of the heat capacities, and a Maxwell relation, to obtain something like the second line.

Then we can calculate the total entropy change by integrating, first at constant $T$ and then at constant $V$:

$$\begin{eqnarray*} S(T_1,V_2)\!\!\!&=&\!\!\!S(T_1,V_1)+\int_{V_1}^{V_2}\left({\pa... ...!\!&=&\!\!\!R\ln{V_2-b\over V_1-b}+C_V\ln{T_2\over T_1}\nonumber \end{eqnarray*}$$

where in the second integration we used the fact that $C_V$ is constant. (We can do this integration because the two terms in ${\rm d}S$ are each functions of one variable only. Given that, the bottom line may be obvious to you without all the careful intermediate steps.)

Note that we have not said what kind of process (reversible or non-reversible, isothermal, adiabatic...) has taken place; that will go into the relation between $V_2$ and $T_2$.

We can check our result against those we have already calculated for an ideal gas just by setting $b=0$ and using the correct $C_V$ (eg $3R/2$ for a monatomic gas), as follows: First, for an isothermal expansion ($T_1=T_2$) we get $\Delta S=R\ln(V_2/ V_1)$ as before.
Also, for a reversible adiabatic expansion we can use $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$ (see here) so

$$\begin{eqnarray*} \Delta S\!\!\!&=&\!\!\!R\ln{V_2\over V_1}+C_V\ln\left({V_1\ove... ...)\right)\ln{V_2\over V_1}\nonumber\\ \!\!\!&=&\!\!\!0\nonumber \end{eqnarray*}$$

(using $C_P-C_V=R$ for an ideal gas). But that is as expected: entropy changes for reversible, adiabatic processes are it always zero! (See here if you have forgotten why...)

Conversely, we can use the fact that $\Delta S=0$ for a reversible adiabatic process to see that for a van der Waals gas, $T(V-b)^{R/C_V}$ is constant. But $R\ne C_p-C_V$ so the exponent isn't $\gamma-1$.