Differentiation of H = U + pV

I was asked to use direct differentiation of "H = U + pV" to find a relationship between:


$$ \left(\frac{\partial H}{\partial U}\right)\_{\!P}$$ and $$\left(\frac{\partial U}{\partial V}\right)\_{\!P} $$


So my first step was to take a differential of the equation twice, once in terms of U and once in terms of V:


$$ \left(\frac{\partial}{\partial U}H\right)\_{\!P}=\left(\frac{\partial}{\partial U}(U +PV)\right)\_{\!P}\rightarrow\left(\frac{\partial H}{\partial U}\right)\_{\!P}= 1 + P\left(\frac{\partial V}{\partial U}\right)\_{\!P}$$
and


$$ \left(\frac{\partial}{\partial V}H\right)\_{\!P}=\left(\frac{\partial}{\partial V}(U +PV)\right)\_{\!P}\rightarrow\left(\frac{\partial H}{\partial V}\right)\_{\!P}= \left(\frac{\partial U}{\partial V}\right)\_{\!P} + P$$


When I tried to substitute one equation into the other, they would simply end up all canceling out. Is there another way?




A way to do it would be:


\begin{align}
\left( \frac{\partial H}{\partial U} \right)\_P \left( \frac{\partial U}{\partial V} \right)\_P &= \left( \frac{\partial H}{\partial V} \right)\_P = \left( \frac{\partial U}{\partial V} \right)\_P + P\\
\left( \frac{\partial H}{\partial U} \right)\_P &= \frac{\left( \frac{\partial U}{\partial V} \right)\_P + P}{\left(\frac{\partial U}{\partial V} \right)\_P}\\
\left(\frac{\partial U}{\partial V} \right)\_P &= \frac{P}{\left( \frac{\partial H}{\partial U} \right)\_P - 1}\\
\end{align}


 4

Not really a full answer per se, but since you are curious about why the partial derivatives can be manipulated in the way they are:


Consider $z = z(x,y)$. We have


$$\require{begingroup} \begingroup \newcommand{\md}[0]{\mathrm{d}} \newcommand{\pdiff}[3]{\left( \frac{\partial #1}{\partial #2} \right)\_{\!#3}} \md z = \pdiff{z}{x}{y}\md x + \pdiff{z}{y}{x}\md y \tag{1}$$


but also $x = x(y,z)$, so


$$\md x = \pdiff{x}{z}{y}\md z + \pdiff{x}{y}{z}\md y \tag{2}$$


Substitute $(1)$ into $(2)$:


$$\md x = \pdiff{x}{z}{y}\pdiff{z}{x}{y}\md x + \pdiff{x}{z}{y}\pdiff{z}{y}{x}\md y + \pdiff{x}{y}{z}\md y \tag{3}$$


Comparing coefficients of $\md x$ gets you to


$$1 = \pdiff{x}{z}{y}\pdiff{z}{x}{y} \tag{4}$$


This is what you need to relate the partial derivatives $(\partial V/\partial U)\_P$ and $(\partial U/\partial V)\_P$; as Chester Miller already noted they are reciprocals of one another.


[Incidentally: not relevant, but if you compare coefficients of $\md y$, you will get:]


$$\begin{align}
\pdiff{x}{z}{y}\pdiff{z}{y}{x} + \pdiff{x}{y}{z} &= 0 \tag{5} \\
\pdiff{x}{z}{y}\pdiff{z}{y}{x} &= -\pdiff{x}{y}{z} \tag{6} \\
&= -\pdiff{y}{x}{z}^{-1} \tag{7} \\
\pdiff{x}{z}{y}\pdiff{z}{y}{x}\pdiff{y}{x}{z} &= -1 \tag{8}
\end{align}$$


where in going from $(6)$ to $(7)$ we have used equation $(4)$. And yes, the negative sign is supposed to be there! The partial derivatives don't "cancel out" like fractions, so don't think of them as fractions. $\endgroup$


 4

The partial of V with respect to U at constant P is the reciprocal of the partial of U with respect to V at constant P.


 3