Derivation: Entropy of Vaporisation using Redlich-Kwong EoS

AI Thread Summary
The discussion focuses on deriving the entropy of vaporization using the Redlich-Kwong equation of state (EoS). The key equation to obtain is ΔS = R[ln((V2-b)/(V1-b))] + (0.5a/(bT^1.5))ln[(V2(V1-b))/(V1(V2-b))]. Participants emphasize the importance of Gibbs free energy and the relationship between entropy and volume changes during vaporization. There's a challenge in applying the EoS, as it is primarily valid for gases, raising concerns about the limits of integration when considering the liquid phase volume. The conversation highlights the need for careful integration and understanding of the EoS in the context of phase changes.
Tom Hardy
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Homework Statement


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:
$$
\Delta S = R\Bigg[ \ln \frac{V_2 -b}{V_1 - b} \Bigg] + \frac{0.5a}{bT^{1.5}}\ln \Bigg[ \frac{V_2(V_1-b)}{V_1(V_2-b)} \Bigg]
$$

Homework Equations


Redlich-Kwong equation:
$$
P = \frac{RT}{V-b} - \frac{a}{T^{0.5}V(V+b)}
$$

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.
 
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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.
 
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Likes Greg Bernhardt and Tom Hardy
I like Serena said:
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.
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?
 
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