Expression for $C_p$ using differential expression for Helmholtz free energy

Question

This is a question from Heat and Thermodynamics by Zemansky and Dittman: "From the differential equation for the thermodynamic potential A(T, V), derive expressions for pressure P, entropy S, internal energy U, heat capacity at constant volume $C_v$, heat capacity at constant pressure $C_p$, volume expansivity, and iso- thermal compressibility." I have solved for $C_v$, could someone help how to proceed for $C_p$?

Answer 1

I will get you started:

• Begin with the differential of $A$ $$dA = A_T dT + A_V dV = - S dT - P dV$$ where $A_S=-T$ and $A_V=-P$ are the partial derivatives of $A$ with respect to $T$ and $V$, respectively.
• Express enthalpy as $$ H = A + T S = A - T A_T $$ and the heat capacity as $$C_P = \left(\frac{\partial H}{\partial T}\right)_P = \left(\frac{\partial (A - T A_T)}{\partial T}\right)_P$$
• Finally, calculate the derivatives $$\left(\frac{\partial A}{\partial T}\right)_P\quad\text{and}\quad \left(\frac{T A_T}{\partial T}\right)_P$$ For this last step you will need Jacobians to express the derivatives of $A$ and $A_T$, which view these functions with $T$ and $P$ as independent vatrable, in terms of $A=A(T,V)$, whose independent variables are $T$ and $V$.

Answer 2

So you've got $A(T,V)$ and this induces the natural arguments for $S$ to be $S(T, V)$ when we calculate it as $$S=-\left({\partial A\over\partial T}\right)_V.$$ To take a derivative of it at constant pressure, just do some wishful thinking, the function you want is $$ \bar S(T, P) = S(T, V(T, P)), $$ the bar indicating that these are different functions for computing the same physical property, they just have different arguments involved. (Most physicists don't even make this distinction and that's fine, I'm a bit more mathematical in my inclination.)

This means that you get the famous relation that after $$\left({\partial \bar S\over\partial T}\right)_P = \left(\frac{\partial S}{\partial T}\right)_{V}+ \left(\frac{\partial S}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial T}\right)_{P}$$ and multiplying through by $T$, $$C_P = C_{V} + T\left(\frac{\partial S}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial T}\right)_{P}.$$ Now $\left(\frac{\partial S}{\partial V}\right)_{T}$ is just a second derivative of $A$ in its natural arguments; that fits your purpose great. The last term is the thermal expansion coefficient, it doesn't fit your purposes as is.

Usually at this point I just play. You have a couple of different things you can do.

1. The wishful thinking trick we did above, again. Helps to change from one set of variables to another, but it creates a sum of two terms so you want to make sure that it isn't multiplying your complexities endlessly.

2. The Maxwell relations. Suppose $A$ and $B$ are conjugates, and $X$ and $Y$ are different conjugates, then when you see $$\left(\frac{\partial A}{\partial X}\right)_{B}$$ then there is a way to swap the B and the X at the bottom, by changing the $A$ to $Y$ up-top (but it may introduce a minus sign). This is possible because both $A$ and $Y$ are partial derivatives of some free energy and so the expression you have in front of you is actually a second derivative: and so you can interchange the order of the two derivatives!

3. Inversion. Useful when the “wrong things” are on the bottom, $$\left(\frac{\partial X}{\partial A}\right)_{B} = \left[\left(\frac{\partial A}{\partial X}\right)_{B}\right]^{-1}.$$

Looking at your expansion coefficient, by a Maxwell relation on the Gibbs energy, $\mathrm dG = -\bar S~\mathrm dT + V~\mathrm dP,$ we have $${\partial^2G\over\partial T\partial P} ={\partial^2G\over\partial P\partial T}$$ hence $$\left(\frac{\partial V}{\partial T}\right)_{P}=-\left(\frac{\partial \bar S}{\partial P}\right)_{T}$$ and consulting $\bar S$ above we see that this expands to$$ \left(\frac{\partial V}{\partial T}\right)_{P} =-\left(\frac{\partial S}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial P}\right)_{T} $$ First term is fine, second term just needs an inversion. There might be a better expression out there but that's the one I'd come up with.