https://physics.stackexchange.com/a/199976/170242
In this answer (in the link), it is mentioned that "the internal energy of an ideal gas is independent of volume when considered as a function of volume and temperature."
Later, it is also mentioned that $$\left(\dfrac{\partial U}{\partial X}\right)_T=0$$ for any variable $X$.
From this, it is easy to conclude that internal energy of an ideal gas is a function of temperature only.
But, can one derive this expression mathematically ?
Best Answer
Yes, this can be proved with only thermodynamics (i.e. without a microscopic theory).
We will take for granted the following: the fundamental equation of thermodynamics in the form $dU = TdS - pdV$ for a gas, and that any thermodynamic quantity may be expressed in terms of only two variables. We shall take those to be $V,T$ and prove in this case $U=U(V,T)=U(T)$ only. That suffices to show that $U$ depends only on $T$ no matter what we pick as our second variable.
The proof is basically that $\frac{\partial U}{\partial V}_T =0$. To show this we use the fundamental relation and the chain rule:
$\frac{\partial U}{\partial V}_T=\frac{\partial U}{\partial V}_S+\frac{\partial U}{\partial S}_V\frac{\partial S}{\partial V}_T = -p + T\frac{\partial S}{\partial V}_T $
Then we use a trick called a Maxwell relation. This is done by considering the quantity $F = U - TS$ such that $dF = -SdT-pdV$. Considering this expression tells us that $\frac{\partial S}{\partial V}_T=\frac{\partial p}{\partial T}_V$
This all adds up to saying that $\frac{\partial U}{\partial V}_T = T\frac{\partial p}{\partial T}_V-p$. But an ideal gas by definition satisfies $pV=nRT$ and so the RHS of this is zero (it's linear so $\frac{\partial p}{\partial T} =\frac{p}{T}$).
This means that $U$ depends only on $T$ for any ideal gas depending on only two variables. If you let it also depend on $N$ then of course $U$ now depends on $U=U(T,N)$.