I Change of Variables in Thermodynamics

[cwill53]

I have a question about changing variables in the context of thermodynamics, but I suppose this would extend to any set of variables that have defined and nonzero partial derivatives on a given set of points. First I should define the variables.
$T$ is temperature, $U$ is internal energy, $H$ is enthalpy, $F$ is Helmholtz free energy, $G$ is Gibbs free energy, $P$ is pressure, $V$ is volume, $S$ is entropy, $C_V$ is heat capacity at constant volume, and $C_P$ is heat capacity at constant pressure.

The four functions $U$,$H$,$F$, and $G$ are called thermodynamic potentials or thermodynamic functions. Each one has a set of variables, called natural variables, from which you can derive other thermodynamic variables through partial differentiation in a much cleaner way than using with variables other than these natural variables as the starting point.

For pure substances, the natural variables for $U$ are $S$ and $V$, the natural variables for $H$ are $S$ and $P$, the natural variables of $F$ are $T$ and $V$, and the natural variables of $G$ are $T$ and $P$.

$U=U(S,V);H=H(S,P);F=F(T,V);G=G(T,P)$

So, for a pure substance, the natural variables of the internal energy $U$ are entropy $S$ and volume $V$. The total differential of $U$ is then
$$dU(S,V)=TdS-PdV=\left (\frac{\partial U }{\partial S} \right )_VdS+\left ( \frac{\partial U}{\partial V} \right )_SdV=\frac{\partial U}{\partial S}(S,V)dS+\frac{\partial U}{\partial V}(S,V)dV$$

The question that I have came to mind when I was proving the relations

$$C_V=\left ( \frac{\partial U}{\partial T} \right )_V=T\left ( \frac{\partial S}{\partial T} \right )_V$$

$$C_P=\left ( \frac{\partial H}{\partial T} \right )_P=T\left ( \frac{\partial S}{\partial T} \right )_P$$

When I worked through this, I simply started from the thermodynamic function that was most convenient. For the relation for $C_V$, I started with $U(S,V)$ and did the following:

$$dU(S,V)\rightarrow dU(S(T,V),V)=TdS(T,V)-PdV=T\left [ \left ( \frac{\partial S}{\partial T} \right )_VdT+\left ( \frac{\partial S}{\partial V} \right )_TdV \right ]-PdV$$

$$dV=0\Rightarrow dU(S(T,V),V)=T\left ( \frac{\partial S}{\partial T} \right )_VdT=\left ( \frac{\partial U}{\partial T} \right )_VdT\Rightarrow C_V=T\left ( \frac{\partial S}{\partial T} \right )_V$$

For the relation with $C_P$, I did a similar thing. The natural variables of the enthalpy $H$, for a pure substance, is entropy $S$ and pressure $P$. The total differential for enthalpy is

$$dH(S,P)=TdS+VdP=\left ( \frac{\partial H}{\partial S} \right )_PdS+\left ( \frac{\partial H}{\partial P} \right )_SdP$$

To derive the relation for $C_P$, I went from $dH(S,P)$ to $dH(S(T,P),P)$, where

$$dH(S(T,P),P)=TdS(T,P)+VdP$$

After that I did essentially the same procedure as the one I did for $C_V$.

What I'm wondering now is, how would I do a two-variable change of coordinates starting from the natural variables of a thermodynamic function? In each of the above examples, I sort of used entropy$S$ as a dummy variable and made it a function of a variable that I **wanted** to write the thermodynamic function in terms of, and a variable that the thermodynamic function was already written in terms of, that happened to be one of its natural variables.

But what if I wanted to write the total differential for internal energy, $dU(S,V)$, in terms of another set of variables outside of the natural variables of $U$, like $dU(T,P)$?

Would I have to do something like this?
$$dU(S,V)\rightarrow dU(S(T,P),V(T,P))=TdS(T,P)-PdV(T,P)$$
$$=T\left [ \left ( \frac{\partial S}{\partial T} \right )_PdT+\left ( \frac{\partial S}{\partial P} \right )_TdP \right ]-P\left [ \left ( \frac{\partial V}{\partial T} \right )_PdT+\left ( \frac{\partial V}{\partial P} \right )_TdP \right ]$$

What if I wanted to write the thermodynamic function in terms of *another* thermodynamic function/functions with their own set of natural variables? An example would be going from $U(S,V)$ to $U(H,F)$, where, in terms of natural variables, $U=U(S,V)$, $H=H(S,P)$, and $F=F(T,V)$? How would that work? I know this part might not make physical sense whatsoever, but I want to know just for the sake of the mathematics.

Would I write something like

$$dU(S,V)\rightarrow dU(S(H(S,P),F(T,V)),V(H(S,P),F(T,V)))$$

$$=TdS(H(S,P),F(T,V))-PdV(H(S,P),F(T,V))$$

$$=T\left [ \left ( \frac{\partial S}{\partial H} \right )_FdH(S,P)+\left ( \frac{\partial S}{\partial F} \right )_HdF(T,V) \right ]-P\left [ \left ( \frac{\partial V}{\partial H} \right )_FdH(S,P)+\left ( \frac{\partial V}{\partial F} \right )_HdF(T,V) \right ]$$

and continue to expand the $dH(S,P)$ and $dF(T,V)$? Sorry for the long post, but this has been bugging me for a while.

 

[anuttarasammyak]

Would I have to do something like this?

That seems right.

I am not familiar with energy as functions of other free energies. Obviously E=H+F-G

 

[vanhees71]

The math is called "Legendre transformation". Starting from
$$\mathrm{d} U=T \mathrm{d} S -p \mathrm{d} V,$$
which tells you that for $U$ the "natural independent variables" are $S$ and $V$ you can get other thermodynamical potentials with other natural independent variables by a Legendre transformation. From the above differential you get
$$\partial_S U(S,V)=T, \quad \partial_V U(S,V)=-p.$$
Now for the enthalpy, which is the potential with the natural variables $S$ and $P$ you make
$$H=U+p V.$$
Indeed you get
$$\mathrm{d} H = \mathrm{d} U + p \mathrm{d} V +V \mathrm{d} p = T \mathrm{d} S +V \mathrm{d} p$$
and thus
$$\partial_S H(S,p)=T, \quad \partial_p H(S,p)=V.$$
It's analogous for the other potentials.

Another important point is the extensivity and intensivity of the variables. The potentials as well as $S$ and $V$ are extensive while $p$ and $T$ are intensive. Thus you have
$$U(\lambda S,\lambda V)=\lambda U(S,V).$$
Now take the total derivative wrt.\ the scaling factor $\lambda$ and then set $\lambda=1$. This leads to the Gibbs relation
$$S \partial_S U(S,V)+V \partial_V U(S,V)=U(S,V) \; \Rightarrow \; U(S,V)=T S - p V.$$

 

[cwill53]

The math is called "Legendre transformation". Starting from
$$\mathrm{d} U=T \mathrm{d} S -p \mathrm{d} V,$$
which tells you that for $U$ the "natural independent variables" are $S$ and $V$ you can get other thermodynamical potentials with other natural independent variables by a Legendre transformation. From the above differential you get
$$\partial_S U(S,V)=T, \quad \partial_V U(S,V)=-p.$$
Now for the enthalpy, which is the potential with the natural variables $S$ and $P$ you make
$$H=U+p V.$$
Indeed you get
$$\mathrm{d} H = \mathrm{d} U + p \mathrm{d} V +V \mathrm{d} p = T \mathrm{d} S +V \mathrm{d} p$$
and thus
$$\partial_S H(S,p)=T, \quad \partial_p H(S,p)=V.$$
It's analogous for the other potentials.

Another important point is the extensivity and intensivity of the variables. The potentials as well as $S$ and $V$ are extensive while $p$ and $T$ are intensive. Thus you have
$$U(\lambda S,\lambda V)=\lambda U(S,V).$$
Now take the total derivative wrt.\ the scaling factor $\lambda$ and then set $\lambda=1$. This leads to the Gibbs relation
$$S \partial_S U(S,V)+V \partial_V U(S,V)=U(S,V) \; \Rightarrow \; U(S,V)=T S - p V.$$

This makes sense. Can you obtain $dU(T,P)$ as $dU(S(T,P),V(T,P))$ though? Is this the same operation essentially? Do you always have to use a Legendre transform? I thought that the differentials for the thermodynamic potentials just seem implicit from their definitions.

 

[cwill53]

This makes sense. Can you obtain $dU(T,P)$ as $dU(S(T,P),V(T,P))$ though? Is this the same operation essentially? Do you always have to use a Legendre transform? I thought that the differentials for the thermodynamic potentials just seem implicit from their definitions.

I understand now. This is simply an exercise of the chain rule.

However, defining a thermodynamic potential as both the independent and dependent variables of a system violates the state postulate; they're Legendre transformations of each other, so if you tried to write one in terms of two others, those two other dependent variables have additional variables that need to be mentioned to define either. So $U(F,H)$ doesn't make sense.