# Derivation: Entropy of Vaporisation using Redlich-Kwong EoS

Tags:
1. Apr 10, 2017

### Tom Hardy

1. The problem statement, all variables and given/known data
For some reason it is not letting me add the image here, here is the link to the question:

http://imgur.com/a/3DLWM

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:
$$\Delta S = R\Bigg[ \ln \frac{V_2 -b}{V_1 - b} \Bigg] + \frac{0.5a}{bT^{1.5}}\ln \Bigg[ \frac{V_2(V_1-b)}{V_1(V_2-b)} \Bigg]$$
2. Relevant equations
Redlich-Kwong equation:
$$P = \frac{RT}{V-b} - \frac{a}{T^{0.5}V(V+b)}$$

3. The attempt at a solution
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.

2. Apr 10, 2017

### I like Serena

We can't just take the derivative of P wrt to T since P is also a function of V, unless we would assume V to be constant, which would be unusual for vaporization.

Instead we can write dU=TdS=PdV for a reversible path.
If we assume isothermal vaporization, we get $T\Delta S=\int PdV$, from which the equation follows.

Last edited: Apr 11, 2017
3. Apr 11, 2017

### Tom Hardy

Thank you, my only concern is the limits of integration. Seemingly, the lower is limit is the liquid phase volume, but how can we use this with the EoS that is only valid for gases?