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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.

  • $\begingroup$ You correctly wrote "at equilibrium, the free energy of both phases must be the same" but your corresponding equation says the the change in the Gibbs free energy must be the same, which is not equivalent. $\endgroup$ – Chemomechanics Apr 10 '17 at 19:05
  • $\begingroup$ @Chemomechanics hmm, should it be dG_1 = -dG_2 ? $\endgroup$ – MathsIsHard Apr 10 '17 at 19:26
  • $\begingroup$ $ΔG=G_1−G_2=0$ for the individual phases and $dG=0$ for the entire system, from which you can derive that $\mu_1=\mu_2$. But anyway, to solve for the entropy of each phase at the constant-temperature phase transition, try writing $dS=\left(\frac{\partial S}{\partial T}\right)_V\,dT+\left(\frac{\partial S}{\partial V}\right)_T\,dV=\left(\frac{\partial S}{\partial V}\right)_T\,dV$ and using a Maxwell relation. If this is agreeable, I'll write it out as an answer for your approval. $\endgroup$ – Chemomechanics Apr 11 '17 at 4:00
  • $\begingroup$ @Chemomechanics Yes please! My only concern is the limits of integration. Seemingly, they will be $V_L$ for the lower limit and $V_V$ for the upper. However, the EoS doesn't work for liquids, it can only predict gas behaviour, so how can we put down $V_L$ as a limit in the integration? $\endgroup$ – MathsIsHard Apr 11 '17 at 7:25
  • $\begingroup$ My understanding is that the Redlich-Kwong equation (and its modifications) is a special case of an equation of state that reasonably describes both the gaseous and liquid states, which is probably why it was selected for your problem. $\endgroup$ – Chemomechanics Apr 11 '17 at 7:51

The reason that the equations don't match is that the free energy of both phases being equal (which is an accurate statement) is not equivalent to their $dG$ values being equal. One way to see this is to note that although the gas and liquid free energy curves coincide at a phase transition, their slopes are not equal.

A better approach is to express the entropy $S$ as a function of $T$ and $V$ (i.e., $dS=\left(\frac{\partial S}{\partial T}\right)_V\,dT+\left(\frac{\partial S}{\partial V}\right)_T\,dV$, which is simply $\left(\frac{\partial S}{\partial V}\right)_T\,dV$ at the constant-temperature condition of a first-order phase transition. Using the Maxwell relation $\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$, it's relatively straightforward to obtain the intended expression for $S$ and then for $\Delta S$.

More expressions for systems following the Redlich-Kwong equation of state are given here.

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