# Calculating inner energy, understanding maxwell relations

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?

• 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. – matrp Nov 17 '17 at 11:11

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