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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?

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3 Answers 3

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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}

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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. $\endgroup$

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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