I have $n$ mole of an ideal gas with pressure $p$, volume $V$, temperature $T$ and constant heat capacity $C_v$. The question is to calculate the inner energy $U$.

Solution: $$ \left( \frac {\partial U}{\partial T} \right)_V=C_v \to U(T,V)=\left( \frac {\partial U}{\partial T} \right)_V T+\left( \frac {\partial U}{\partial V} \right)_T V=C_v T+u(V) $$

where $u(V)$ is a function of the volume. To calculate $u(V)$, we use

$$dU=TdS-pdV \to \left( \frac {\partial U}{\partial V} \right)_T=T \left( \frac {\partial S}{\partial V} \right)_T-p$$ $$ \left( \frac {\partial S}{\partial V} \right)_T = \left( \frac {\partial p}{\partial T} \right)_V$$ $$pV=nRT \to \left( \frac {\partial p}{\partial T} \right)_V=\frac {nR} {V} $$

So finally we get

$$ \left( \frac {\partial U}{\partial V} \right)_T = T \frac {nR} {V}-p=\frac {nRT} {V} - \frac {nRT} {V} = 0  \to u'(V) = 0 \to u(V)=Constant$$

Final answer is: $U=C_V T+constant$

I don't quite understand the concept of putting a variable outside of my paranthesis, I thought that it meant the variable was constant in the given system.

I'm confused by how we can write $dU=TdS-pdV \to \left( \frac {\partial U}{\partial V} \right)_T=T \left( \frac {\partial S}{\partial V} \right)_T-p$, are we not setting the temperature as constant here? Why is that OK? I understand that we do this so it's easier to work with and I can use the maxwell-relation later on, but is that enough of an argument when setting $T$ as constant?

I'm also confused about $ \left( \frac {\partial S}{\partial V} \right)_T = \left( \frac {\partial p}{\partial T} \right)_V $. If $T$ is constant, then the right side should be $0$, and if $V$ is constant, then the left side should be $0$? What am I missing?

  • $\begingroup$ The variable outside the parenthesis is not a constant in the system, it is a constant in the calculated derivative. For example, in $(\partial{S}/\partial{V})_ T$ means the variation of entropy in relation to variation of volume, if $T$ is kept constant. $\endgroup$
    – matrp
    Nov 17, 2017 at 11:11

1 Answer 1


Lets look more closely at expressions like $\left(\frac{\partial P}{\partial T}\right)_{V}$. This quantity describes the following experiment: take a system with constant volume, change its temperature and check how the pressure changes as a result. But wait, what is the volume of the system? It is constant throughout this specific experiment, but you may conduct other experiments that involves changes in $V$. For example, you could do the following one: keep the temperature of the system constant, change the volume and measure the change in its entropy. The relevant quantity this time is $\left(\frac{\partial S}{\partial V}\right)_{T}$.

Thus, the Maxwell relation

$$\left(\frac{\partial P}{\partial T}\right)_{V}=\left(\frac{\partial S}{\partial V}\right)_{T}$$

asserts that the rate of change you measure in each of the experiments will be the same. When you think about those derivative in analogy to experiments, it is easier to understand their meaning. Try to think this way about the other doubts you raised in your question. Is it more clear now?

  • $\begingroup$ This helped a lot, I think I'm starting to understand the concept. I was thinking about it too mathematically, not understanding what the relation actually stood for. $\endgroup$
    – armara
    Nov 17, 2017 at 12:26

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