# Joule free expansion and differentials in irreversible processes

I am studying thermodynamics (undergraduate level) and I am not sure I understand what happens in irreversible processes.

I thought I understood but then I got confused in the Joule free expansion (not Joule-Thomson). This is my thought process:

I know it's essentially an adiabatic expansion into vacuum. Adiabatic means $$q = 0$$ and free expansion - vacuum means external pressure $$p_{ext} = 0$$, therefore for pV work, $$w = 0$$. So, $$ΔU = q + w = 0$$.

From my lecture notes, I see that: $$dU = 0 = C_V dT_U + \left(\frac{\partial U}{\partial V} \right)_{T} dV_U$$, therefore $$\left(\frac{\partial U}{\partial V} \right)_{T} = -C_V \left(\frac{\partial T}{\partial V} \right)_{U}$$, which makes sense considering for ideal gas it's $$0$$.

But my problem is that since it's a free expansion, it's also an irreversible process. What is the meaning of the differential $$dU$$ in an irreversible process? Isn't the whole problem in irreversible processes that it is not defined since the state is not in equilibrium? Why can we even use data from this experiment in the general expression for $$dU$$?

From my understanding, the only states in equilibrium are the initial and the final, so $$ΔU = 0$$ is correct. But why is $$dU$$ equal to 0?

Also, I am still on the first law and although I know about entropy in general and that $$ΔS > 0$$ in irreversible processes, I would prefer to understand this without involving entropy if it's possible.

• For a tutorial on how to determine the change in entropy for a system that experiences an irreversible process, see the following link:chemistry.stackexchange.com/questions/109654/…. This link also has a worked example for this exact problem. – Chet Miller Feb 19 at 2:01

For the system you describe in which no heat is transferred across the system boundary (insulated) Q=0 and no work is done on the rigid boundary surrounding the contents of the chamber W = 0, the first law of thermodynamics tells us that the change in internal energy of the system is zero: $$\Delta U=0$$For an ideal gas, the internal energy is a function only of temperature (and not volume) U=U(T), so this means that the change in temperature of the gas is zero: $$\Delta T=0$$ For a tutorial on how to determine the change in entropy for a system that experiences an irreversible process, see the following link: https://chemistry.stackexchange.com/questions/109654/entropy-change-of-isothermal-irreversible-expansion-of-ideal-gas