# Proving a thermodynamic relation between $(\partial H/\partial T)_p$ and $(\partial U/\partial T)_V$

(This question is taken from Problem 1.1(b) of the book Chemical Thermodynamics: Principles and Applications.)$$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$$

Prove $$\pd HTp = \pd UTV + \left[ V - \pd HpT \right] \pd pTV. \tag{1}$$

First, I expressed $$H$$ as $$U + PV$$, and took the partial derivative against $$T$$, keeping pressure constant:

$$\pd HTp = \pd UTp + p\pd VTp \tag{2}$$

Expressing $$U$$ as a function of $$(T, V)$$:

$$\pd UTp = \pd UTV + \pd UVT \pd VTp \tag{3}$$

Using the well-known identity that $$\pd pVT \pd TpV \pd VTp = -1, \tag{4}$$

$$\pd UVT \pd VTp = -\pd UpT \pd pTV = \left[ V - \pd UpT - V\pd ppT \right]\pd pTV \tag{5}$$

This, of course, means that

$$\pd HTp = \pd UTV + \left[ V - \pd HpT \right]\pd pTV + p\pd VTp \tag{6}$$

Based on my working, there is an additional $$p\pd VTp$$ term. Is my working incorrect? I can’t find my mistake.$$\endgroup$$

• You might want to make a new command for a partial derivative to save yourself some pain. I can give you an example in a while... Jan 24, 2020 at 14:12
• put at the start of your post: $\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$ and at the end of your post $\endgroup$... and between those lines you can now use $\pd{V}{T}{p}$ (or even \pd VTp although I don't recommend making the latter a habit, unless you understand why it works without braces). Jan 24, 2020 at 14:16
• I (very cautiously) think the question might be incorrect. My working out of the RHS (albeit by a different approach) is yielding an inconsistency too. Can someone check and confirm? Jan 24, 2020 at 15:18
• @theorist Truthfully, I don't fully understand it myself... but it has something to do with how TeX scans for arguments. Usually it just reads in one thing at a time (either a letter or a control sequence like \alpha) and treats that as an argument. Grouping several things together with curly braces allows you to pass more than one character at a time. That also means that if you only want a single-character argument, it's unnecessary to enclose it in braces (although having braces is a lot more familiar for most people) [...] Jan 25, 2020 at 3:11
• [...] the \! is just cosmetic. You probably know it is a "negative space". I find that without it, the subscripted value is too far away from the brackets; but that's just personal preference, really. If you're interested in the braces, there's always Knuth's good old TeXBook which I think explains this behaviour very carefully. Jan 25, 2020 at 3:13

$$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$$

I would start from

$$dH = \pd HTp dT + \pd HpT dp$$

This gives rise to

$$\pd HTV = \pd HTp + \pd HpT \pd pTV$$

in which you can recognize some components of the solution. Since

$$dH = dU +PdV + VdP$$

so that

$$\pd HTV = \pd UTV + V \pd pTV$$

it follows that

$$\pd UTV + V \pd pTV = \pd HTp + \pd HpT \pd pTV$$

which is readily rearranged to obtain the desired equation. $$\endgroup$$

• Thanks for your answer! Could I inquire where I went wrong in my incorrect proof? I’m not too sure why my first expression is wrong. Jan 25, 2020 at 8:08
• @ANZGCFlyingFalcon It looks like you flubbed the substitution of Eq 4 into Eq 3 (see the middle expression in Eq 5). Jan 25, 2020 at 10:32
• If I were to take the RHS of (5) and multiply it by (4), wouldn’t I get the LHS? (Obviously the sign would become +ve but you get what I mean.) Jan 25, 2020 at 15:29
• To get $(\frac{\partial U}{\partial p})_T$ I recommend starting from the differential form of U=H-PV and taking the derivative wrt p at constant T. I'm not sure how you arrive at your result, it is hard for me to follow. Jan 25, 2020 at 19:22