Expansion of a van der Waals gas

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In part a) we can reason as follows.

If we let a gas expand slowly, then it will perform work. If we know the internal energy as a function of volume and temperature and we keep the temperature the same during the expansion, then we know by how much the internal energy has changed during the expansion. We also know how much work the gas has performed, so we know the heat that the gas has absorbed. If we divide this heat by the temperature we get the entropy increase. In case of an ideal gas the internal energy does not depend on the volume for fixed temperature. So, the heat absorbed by the gas is equal to the work performed by the gas. For the van der Waals gas this is not true and we need to use the internal energy function to compute the entropy change.

In part b) we can use that in case of free expansion no heat is absorbed and no work is performed, so the internal energy stays constant. In case of an ideal gas we then know that means that the temperature stays constant (I'll derive that later). But then you get the same result as in a) for the entropy because temperature and volume are the same in the end state. In case of the van der Waals gas the temperature will change. We can use the internal energy as a function of temperature and volume to find by how much the temperature will change.

So, we see that to solve the problems we need to find the internal energy change in terms of changes in temperature and volume. I derived this for another problem a few days ago, so I have included that derivation here:

The fundamental thermodynamic relation expresses dU in terms of dS and dV

dU = T dS - P dV (1)

If we want to consider U as a function of T and V, then all we need to do is rewrite this in terms of dT and dV. We substitute:

dS = dS/dT dT + dS/dV dV (2)

Here the derivatives are partial derivatives, the derivative ...