# Changing Variables in Thermodynamics

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.

• Can you clarify what is the question? Is it about if they can be expressed in other variables? Or is there any meaningful derivative?
– Greg
Oct 15, 2021 at 8:23
• @Greg I'm essentially confused about natural variables somewhat. I'm wondering if you can do a change of variables from the natural variables of a thermodynamic potential; for example $dU(S,V)$ to $dU(T,P)$, by writing $dU(S(T,P),V(T,P))$. Is this allowed? Oct 15, 2021 at 17:29