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?


3 Answers 3


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}

  • $\begingroup$ Not seeing the algebra from step 2 to step 3 $\endgroup$
    – Nova
    Commented Feb 13, 2017 at 1:20
  • $\begingroup$ corrected sign error $\endgroup$
    – GnomeSort
    Commented Feb 13, 2017 at 1:24
  • $\begingroup$ Where does the -1 come from? $\endgroup$
    – Nova
    Commented Feb 13, 2017 at 1:26
  • $\begingroup$ Multiply both sides by step 2 denominator. Subtract del U / del V from both sides. Factor del U / del V from left-hand side. Divide by (del H / del U - 1). $\endgroup$
    – GnomeSort
    Commented Feb 13, 2017 at 1:27
  • 1
    $\begingroup$ Got it. So $$ \left(\frac{\partial H}{\partial U}\right)_{\!P} = \frac{P}{\left(\frac{\partial U}{\partial V}\right)_{\!P}} +1$$ $\endgroup$
    – Nova
    Commented Feb 13, 2017 at 1:37

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. $\endgroup$

  • $\begingroup$ Ok, this is really cool. I guess the one thing I don't understand is steps 5/6. I took multivariable calculus years ago so I guess it must be some rule I forgot... $\endgroup$
    – Nova
    Commented Feb 13, 2017 at 1:46
  • $\begingroup$ Also that same rule probably explains step 4 as well. Cause earlier I was treating them as "fractions" because I remember being told it was ok to manipulate them like that even though it wasn't "technically" right. $\endgroup$
    – Nova
    Commented Feb 13, 2017 at 1:48
  • $\begingroup$ From (3) to (5) is comparing coefficients of dy - on the LHS of eq (3), there's no dy term so the coefficient is zero. From (5) to (6) is just bringing $(\partial x/\partial y)_z$ over to the RHS. It's OK to treat them like fractions up to a certain point. Sometimes it works but sometimes it doesn't as you can see in (8). $\endgroup$ Commented Feb 13, 2017 at 1:48
  • $\begingroup$ No, I wasn't asking about the algebra, but about the calculus reasoning. $\endgroup$
    – Nova
    Commented Feb 13, 2017 at 1:50
  • $\begingroup$ I don't know - there's no real calculus reasoning as far as I can tell, apart from the thing about comparing coefficients $\endgroup$ Commented Feb 13, 2017 at 1:52

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.

  • $\begingroup$ Also, is that always true? For example, is delx/dely always equivalent to (dely/delx)^-1? Can you give an example? Cause I would think taking a derivative of an equation with respect to one variable would yield a different answer than taking the derivative with respect to another... $\endgroup$
    – Nova
    Commented Feb 13, 2017 at 1:00
  • $\begingroup$ The thermodynamic state of a closed system is determined by specifying any two intensive variables. So the molar U can be expressed as a function of pressure and molar V. At constant pressure, U is a unique function of V. This allows you to do what I've done with the reciprocal. $\endgroup$ Commented Feb 13, 2017 at 1:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.