Internal energy of an ideal gas independent of any thermodynamic variable at constant temperature

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 ?

• This is not a homework question and is not too broad. What I am asking is pretty clear.
– user170242
May 14 '18 at 14:04
• As an example, an average energy per atom for a monatomic ideal gas $U=\frac 3 2 kT$. Obviously, $dU/dX$ will be zero for variables other than temperature (e.g., P, V).
– V.F.
May 14 '18 at 14:33

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 $U=U(T,N)$.

I don't think a mathematical proof will be helpful here. However, kinetic theory of gases should provide a statisfactory answer.

Since we are considering ideal gases, there are no interactions, neither attractive nor repulsive, between the gas particles. So, the only kind of internal energy the gas particles can only possess is kinetic internal energy.

If you wish to know more about the internal energy, check out this link: What is the meaning of internal Energy?

The kinetic theory of gases defines temperature as the average kinetic energy of the particles in a given system. This implies that as long as temperature is constant the total kinetic energy, or the internal energy, remains constant.

We can come up with our own thermodynamic variable but at constant temperature, internal energy of the system will remain constant regardless of the value of that variable. For instance, if you change volume at constant temperture the internal energy would suffer no change as a consequence of energy conservation (Same temperature, same kinetic energy).