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The part I'm stuck on is the last part. Basically, the question is to obtain the following equation for the entropy of vaporisation using the Redlich-Kwong equation: $$ \Delta S = R\Bigg[ \ln \frac{V_2 -b}{V_1 - b} \Bigg] + \frac{0.5a}{bT^{1.5}}\ln \Bigg[ \frac{V_2(V_1-b)}{V_1(V_2-b)} \Bigg] $$

Solution Attempt:

I think I should start by using the known fact that at equilibrium, the free energy of both phases must be the same. Using Gibbs free energy: $$ dG = -SdT + VdP \implies -S_1dT + V_1dP = -S_2dT + V_2dP $$ Therefore, I can write an equation for change in entropy due to vaporisation: $$ S_1 - S_2 = (V_1 - V_2)\frac{dP}{dT} $$ However when I simply differentiate the given EoS and multiply by $V_1 - V_2$ I don't get the same result. Clearly, in the answer they have integrated wrt v at some point but I just don't get why or how. Any help will be appreciated, thank you.


1 Answer 1


I arrived to an answer very similar to what it says it should end up being except for a difference in the sign of a term.

Here is how I proceeded:

The total differential of a two-variable function can be written as $$ dS(T,V) = \left( \frac{\partial S}{\partial T} \right) _V dT + \left( \frac{\partial S}{\partial V} \right) _T dV $$

Since a phase change occurs at constant temperature, $ dT = 0 $. Hence, the equation becomes simply

$$ dS = \left( \frac{\partial S}{\partial V} \right) _T dV $$

Replacing the right hand side of the above equation using other thermodynamic relations for the partial derivative in terms of $ P $ and $ T $, we get

$$ dS = \left( \frac{\partial P}{\partial T} \right) _V dV $$

Now we differentiate $ P $ with respect to $T$ at constant $V$: $$ \left( \frac{\partial P}{\partial T} \right) _V = \frac{R}{V-b} + \frac{a}{2T^{3/2}} \frac{1}{V(V+b)} $$

Noting that $ \Delta S = \int dS $,

$$ \Delta S = S_2 - S_1 = \int_{V_1}^{V_2} \left( \frac{\partial P}{\partial T} \right) _V dV \\ = \left[ R \ln (V-b) + \frac{a}{2b T^{3/2}} \ln \left( \frac{V}{V+b} \right) \right]_{V_1}^{V_2} \\ = R \ln \frac{(V_2-b)}{(V_1 - b)} + \frac{a}{2b T^{3/2}} \ln \left[ \frac{V_2}{V_1} \left( \frac{V_1+b}{V_2+b} \right) \right] $$

Either the question is wrong in their signs or I made a mistake with the signs in my solution. Either way, this should hopefully guide you in the right direction.


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