3
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Feb 13 '17 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. $\endgroup$ – Chet Miller Feb 13 '17 at 1:30
4
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ Not seeing the algebra from step 2 to step 3 $\endgroup$ – Nova Feb 13 '17 at 1:20
  • $\begingroup$ corrected sign error $\endgroup$ – NewDogOldTricks Feb 13 '17 at 1:24
  • $\begingroup$ Where does the -1 come from? $\endgroup$ – Nova Feb 13 '17 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). $\endgroup$ – NewDogOldTricks Feb 13 '17 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 Feb 13 '17 at 1:37
4
$\begingroup$

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. $\endgroup$

$\endgroup$
  • $\begingroup$ 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 Feb 13 '17 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 Feb 13 '17 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). $\endgroup$ – orthocresol Feb 13 '17 at 1:48
  • $\begingroup$ No, I wasn't asking about the algebra, but about the calculus reasoning. $\endgroup$ – Nova Feb 13 '17 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 $\endgroup$ – orthocresol Feb 13 '17 at 1:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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