In university physics textbook he says :
The internal energy of an ideal gas depends only on its temperature, not on its pressure or volume.
I know that the only contribution to the internal energy comes from the translational kinetic energy (for monatomic ideal gas) according to $$U=K_{trans}=\frac{3}{2} nKT$$
So, obviously, the internal energy $(U)$ depends only on the temperature $(T)$ and the number of moles $(n)$ of the gas. But if someone did work on the gas leading to an increase in pressure and decrease in volume, will this affect the temperature accordingly?
If no, is it because the increase in pressure cancels out with the decrease in volume, and the temperature remains costant according to $$T=\frac{1}{nR}PV$$
If yes, why did he say that the internal energy depends only on the temperature, not on the pressure or the volume ?
Best Answer
For an ideal gas, you have that $U = \frac{3}{2}nRT$ and also $PV = nRT$, which means that you can write
$$U = \frac{3PV}{2}$$
if you'd like.
It doesn't make sense to say that $U$ is a function of $T$ in no way affected by $P$ and $V$, because (via the ideal gas law) $P,V$, and $T$ are all related to one another. Instead, think of it as the fact that $U$ is determined completely by $T$. If you know $T$, then you know $U$, full stop.
In particular, knowing how $T$ changes tells you immediately how $U$ changes. What happens to $U$ during an isothermal process? Well, if $T$ doesn't change, then $U$ doesn't change. That's it.