In isothermal process $\Delta U =0$. But I am having trouble understanding it.
Say we have an ideal gas, and say my temperature is constant but I move the pressure, volume from $(P, V) \to (P-dP, V+dV) $. So the volume has expanded and system has done some work to the surrounding. So my work is non-zero.
So how come $\Delta U=0$? I am really confused here.
Best Answer
It is not generally true that $\Delta U = 0$ in an isothermal process.
An ideal gas by definition has no interactions between particles, no intermolecular forces, so pressre change at constant temperature does not change internal energy.
Real gases have intermolecular interactions, attractions between molecules at low pressure and repulsion at high pressure. Their internal energy changes with change in pressure, even if temperature is constant.
For an ideal gas, in an isothermal process, $\Delta U = 0 = Q-W$, so $Q=W$.