# 10.6: The Transformation of Thermodynamic Variables in General

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Let us suppose that $$M$$, $$Q$$, $$R$$, $$X$$, and $$Y$$ are state functions and that we know the total differentials

$dM={\left(\dfrac{\partial M}{\partial X}\right)}_YdX+{\left(\dfrac{\partial M}{\partial Y}\right)}_XdY \nonumber$

$dQ={\left(\dfrac{\partial Q}{\partial X}\right)}_YdX+{\left(\dfrac{\partial Q}{\partial Y}\right)}_XdY \nonumber$

$dR={\left(\dfrac{\partial R}{\partial X}\right)}_YdX+{\left(\dfrac{\partial R}{\partial Y}\right)}_XdY \nonumber$

To find the total differential of $$M\left(Q,R\right)$$,

$dM={\left(\dfrac{\partial M}{\partial Q}\right)}_RdQ+{\left(\dfrac{\partial M}{\partial R}\right)}_QdR \nonumber$

we solve the total differentials of $$Q\left(X,Y\right)$$ and $$R\left(X,Y\right)$$ to find $$dX$$ and $$dY$$ in terms of $$dQ$$ and $$dR$$. Since $$dQ$$ and $$dR$$ are simultaneous equations in the variables $$dX$$ and $$dY$$, we can apply Cramer’s rule to obtain

$dX=\dfrac{\left| \begin{array}{cc} dQ & {\left({\partial Q}/{\partial Y}\right)}_X \\ dR & {\left({\partial R}/{\partial Y}\right)}_X \end{array} \right|}{J\left(\dfrac{Q,R}{X,Y}\right)}=\dfrac{{\left(\dfrac{\partial R}{\partial Y}\right)}_XdQ-{\left(\dfrac{\partial Q}{\partial Y}\right)}_XdR}{J\left(\dfrac{Q,R}{X,Y}\right)} \nonumber$

and

$dY=\dfrac{\left| \begin{array}{cc} {\left({\partial Q}/{\partial X}\right)}_Y & dQ \\ {\left({\partial R}/{\partial X}\right)}_Y & dR \end{array} \right|}{J\left(\dfrac{Q,R}{X,Y}\right)}=\dfrac{{-\left(\dfrac{\partial R}{\partial X}\right)}_YdQ+{\left(\dfrac{\partial Q}{\partial X}\right)}_YdR}{J\left(\dfrac{Q,R}{X,Y}\right)} \nonumber$

where $$J\left({\left(Q,R\right)}/{\left(X,Y\right)}\right)$$ is the Jacobian of the transformation of variables $$X$$ and $$Y$$ to variables$$\ Q$$ and $$R$$:

\begin{align*} J\left(\dfrac{Q,R}{X,Y}\right) &= \left| \begin{array}{cc} {\left({\partial Q}/{\partial X}\right)}_Y & {\left({\partial Q}/{\partial Y}\right)}_X \\ {\left({\partial R}/{\partial X}\right)}_Y & {\left({\partial R}/{\partial Y}\right)}_X \end{array} \right| \\[4pt] &= {\left(\dfrac{\partial Q}{\partial X}\right)}_Y{\left(\dfrac{\partial R}{\partial Y}\right)}_X + {\left(\dfrac{\partial Q}{\partial Y}\right)}_X{\left(\dfrac{\partial R}{\partial X}\right)}_Y \end{align*}

To find

$dM={\left(\dfrac{\partial M}{\partial Q}\right)}_RdQ+{\left(\dfrac{\partial M}{\partial R}\right)}_QdR \nonumber$

We substitute these results for $$dX$$ and $$dY$$ into the total differential of $$M=M\left(X,Y\right):$$

\begin{aligned} dM &={\left(\dfrac{\partial M}{\partial X}\right)}_YdX+{\left(\dfrac{\partial M}{\partial Y}\right)}_XdY \\[4pt] &=\dfrac{{\left(\dfrac{\partial M}{\partial X}\right)}_Y\left[{\left(\dfrac{\partial R}{\partial Y}\right)}_XdQ-{\left(\dfrac{\partial Q}{\partial Y}\right)}_XdR\right]}{J\left(\dfrac{Q,R}{X,Y}\right)}+\dfrac{{\left(\dfrac{\partial M}{\partial Y}\right)}_X\left[{-\left(\dfrac{\partial R}{\partial X}\right)}_YdQ+{\left(\dfrac{\partial Q}{\partial X}\right)}_YdR\right]}{J\left(\dfrac{Q,R}{X,Y}\right)} \\[4pt] &=\left[\dfrac{{\left(\dfrac{\partial M}{\partial X}\right)}_Y{\left(\dfrac{\partial R}{\partial Y}\right)}_X-{\left(\dfrac{\partial M}{\partial Y}\right)}_X{\left(\dfrac{\partial R}{\partial X}\right)}_Y}{J\left(\dfrac{Q,R}{X,Y}\right)}\right]dQ +\left[\dfrac{-{\left(\dfrac{\partial M}{\partial X}\right)}_Y{\left(\dfrac{\partial Q}{\partial Y}\right)}_X+{\left(\dfrac{\partial M}{\partial Y}\right)}_X{\left(\dfrac{\partial Q}{\partial X}\right)}_Y}{J\left(\dfrac{Q,R}{X,Y}\right)}\right]dR \end{aligned} \nonumber

where the coefficients of $$dQ$$ and $$dR$$ are $${\left({\partial M}/{\partial Q}\right)}_R$$ and $${\left({\partial M}/{\partial R}\right)}_Q$$, respectively. In §5, we find the other total differentials in terms of $$dP$$ and $$dT$$. If we set $$X=T$$ and $$Y=P$$, we can use these relationships to find the total differential for any state function expressed in terms of any two other state functions.

To illustrate this point, let us use these relationships to find the total differential of $$S$$ expressed as a function of $$P$$ and $$V$$, $$S=S\left(P,V\right)$$. In this case, we are transforming from the variables $$\left(P,T\right)$$ to the variables $$\left(P,V\right)$$. This is a one-variable transformation. To effect it, we make the additional substitutions $$M=S$$, $$Q=V$$, and $$R=P$$. Since we have $$Y=R=P$$, the transformation equations simplify substantially. We have

$\left(\dfrac{\partial R}{\partial Y}\right)_X=\left(\dfrac{\partial P}{\partial P}\right)_X=1 \nonumber$

$\left(\dfrac{\partial R}{\partial X}\right)_Y=\left(\dfrac{\partial P}{\partial T}\right)_P=0 \nonumber$

$\left(\dfrac{\partial Q}{\partial Y}\right)_X=\left(\dfrac{\partial V}{\partial P}\right)_T \nonumber$

$\left(\dfrac{\partial Q}{\partial X}\right)_Y=\left(\dfrac{\partial V}{\partial T}\right)_P \nonumber$

The Jacobian becomes

$J\left(\dfrac{Q,R}{X,Y}\right) = \left(\dfrac{\partial V}{\partial T}\right)_P \nonumber$

and the partial derivatives of $$S$$ become

${\left(\dfrac{\partial S}{\partial V}\right)}_P={{\left(\dfrac{\partial S}{\partial T}\right)}_P}/{{\left(\dfrac{\partial V}{\partial T}\right)}_P={\dfrac{C_P}{T}\left(\dfrac{\partial T}{\partial V}\right)}_P} \nonumber$

and

\begin{align} \left(\dfrac{\partial S}{\partial P}\right)_V &= \dfrac{-{\left(\dfrac{\partial S}{\partial T}\right)}_P{\left(\dfrac{\partial V}{\partial P}\right)}_T+{\left(\dfrac{\partial S}{\partial P}\right)}_T{\left(\dfrac{\partial V}{\partial T}\right)}_P}{{\left(\dfrac{\partial V}{\partial T}\right)}_P} \\[4pt] &={\dfrac{C_P}{T}\left(\dfrac{\partial T}{\partial P}\right)}_V+{\left(\dfrac{\partial S}{\partial P}\right)}_T \end{align} \nonumber

This page titled 10.6: The Transformation of Thermodynamic Variables in General is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Paul Ellgen via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.