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Quick question, in the lecture notes to a thermodynamic course I'm taking,

$$d\underline{S}=\frac{d\underline{U}}{T}+\Big(\frac{\partial P}{\partial T}\Big)_{\underline{V}} d\underline{V}$$

But everywhere else I've looked I've found (by rearranging the fundamental thermodynamic relation:

$$d\underline{S}=\frac{d\underline{U}}{T}+\frac{P}{T}d\underline{V}$$

The two are obviously extremely similar, however I don't believe they're equivalent in general (I believe you'd have to make further assumptions about the system, particularly $P=kT$ at constant specific volume, to go from one to the other.)

Right now I'm just asking if anyone knows which relation is the more general. I was of the opinion that the fundamental thermodynamic relation was always correct, but now I'm doubting myself. Thanks.

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I've always thought of (hydrostatic) pressure as being defined by the fundamental relation, which implies that your second equation is the correct one. Your second equation is very familiar to me but I've never seen the first one before. I would post this as an answer, but I'd prefer to understand more about where the first equation comes from before I state outright that it's wrong - in your first equation $P$ is a function of $T$ and $V$, whereas in the second it's $U$ and $V$. So I suspect that it might be correct but saying something different from your second equation. – Nathaniel Jun 13 '12 at 8:38
That's a fascinating idea - pressure is defined by this relation (is there a particular text that presents thermodynamics that way?) I haven't actually encountered a derivation for the first equation anywhere, indeed I don't know of a textbook that gives it. I can tell you we need to use the first equation when given an equation of state of the form $P = f(T,\underline{V})$ (the simplest of which is the ideal gas law). So you might be onto something ... – tom Jun 13 '12 at 9:21
BTW - how do you know $P$ is a function of $T$ and $U$ in the second equation? Why not $S$ or another thermodynamic quantity? – tom Jun 13 '12 at 9:21
I'll see if I can come up with a reference for pressure being defined that way. I probably picked it up from one of many papers by Edwin Jaynes. I can tell that $P$ is a function of $T$ and $U$ from the notation $$\left(\frac{\partial P}{\partial T}\right)_V,$$ which means "vary $T$ while keeping $V$ constant". (Personally I find it a somewhat confusing notation and usually prefer to write it as $$\frac{\partial P(T, V)}{\partial T},$$ which is what it really means. But the way you've written it is very standard in thermodynamics.) – Nathaniel Jun 13 '12 at 12:05
The notes are no good, the second equation is right, the first is wrong. – Ron Maimon Jun 14 '12 at 0:45

1 Answer

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Both seem a little strange. So let us see where they are coming from. What we know is from the laws of thermodynamics that the total differential of the internal energy $U$ can be written as $$dU = \delta Q + dW$$ Here $\delta Q$ is an infinitesimal amound of thermal energy and $dW$ an infinitesimal amount of work. The work is the product of a generalized displacement times a generalized force, which are together a conjugated pair of variables. One such pair is pressure and volume, $P\, dV$. For a reversible process $\delta Q = T dS$, so that we get $$dU = T\,dS - P\,dV$$ From here we easily arrive at your second equation: $$ \frac{dU}{T} + \frac{P}{T} dV = dS$$ Now we compare with your first equation, which can only be true if$$\frac{P}{T} = \left(\frac{\partial P}{\partial T}\right)_V$$ which certainly does not hold for all thermodynamic systems. So your first equation is more general and the second one only correct in specific circumstances. We rearrange the condition $$P(T,V) = \left(\frac{\partial P}{\partial T}\right)_V T$$ which can only be satisfied by a general function $f(V)$ and $$P(T,V) = f(V) \,T$$

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Thankyou very much, that's fantastic, and makes 100% sense :) I must admit I'm slightly disturbed by your reply though, considering that in a 3rd year chemical engineering course we have freely been using the first equation with EOS's like $P = \frac{RT}{V-b}-\frac{a}{\sqrt{T}V_m(V_m+b)}$ (a common EOS called the Reidlich-Kwong EOS) which certainly doesn't obey your criterion :( – tom Jun 15 '12 at 5:45
Sorry, did you mean the second equation is more general (considering you derived it)? :) – tom Jun 15 '12 at 6:35
@tom: Sorry for the confusion, I wanted to say: my first equation, which is your second equation. In short $\frac{dU}{T} + \frac{P}{T} dV = dS$ is correct and general. – Alexander Jun 15 '12 at 10:42

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