I am having trouble understanding why in an isothermal process, the change in internal energy is zero. I know that $\Delta U$ or $\Delta E=q+w$, and so in isothermal process $q=0$. But how does one show that $w =0$? Or is it necessary that if we are talking of isothermal process, we are not doing work on the system; why or why not? Or is it necessary that if $w$ is not equal to 0, then the process can't be isothermal; why? Please explain this. I found a similar question here but I was not able to understand anything from it.
[Physics] Change in internal energy is 0 in isothermal process
energythermodynamicswork
Best Answer
The quick answer is $\Delta U \neq 0$.
Let's look at some details. In the special case where you are dealing with ideal gas. $$U = \frac{3}{2} nRT$$ Thus $$\Delta U = \frac{3}{2}nR\Delta T $$ Since the process is isothermal, $\Delta T$ is zero. Therefore $\Delta U = 0$. So it is not true that $q = 0$(that would be called adiabatic). Rather, $q = -w$.
The above analysis fails if the gas is NOT ideal. Since $U = \frac{3}{2}nRT$ is generally not true. But usually the ideal gas approximation works fine.