# Work done by an ideal gas in an adiabatic free expansion

A system consists of the volume ($$V_1+V_2$$) within a container with a partition separating it into two volumes $$(𝑉_1, 𝑉_2)$$. In $$V_1$$, $$𝑛$$ moles of an ideal gas are present, and the surrounding is the rigid adiabatic wall of the container. Upon removal of the partition, the gas expands to fill both volumes.

The expansion of the gas is an irreversible process between the initial and final equilibrium states. Therefore, we may replace the process with a reversible, isothermal expansion of the (ideal) gas between the same two equilibrium states. The work done by the gas on the surroundings in this reversible process is $$nRT\log[(V_1+V_2)/V_1]$$.

However, the work done by the ideal gas in the irreversible process is zero, since the container is rigid. Is there a contradiction here ?

The following query does not address the issue: Internal Energy Change for a free adiabatic expansion

## 2 Answers

There is no contradiction, the reversible process you imagine may have nothing to do with the real irreversible process. All you ask is that the reversible process takes the system from an initial equilibrium ($$V_{i}$$,$$P_{i}$$,$$T_{i}$$) to the same final state ($$V_{f}$$,$$P_{f}$$,$$T_{f}$$) as your real process to yield the same final energy and entropy. So in this particular case, you may for instance deduce the entropy variation of your real process with that of your reversible process. Since you know $$T_{i}$$=$$T_{f}$$ (for an ideal gas) it has to be a reversible isothermal process.

So, real process: irreversible, adiabatic, no work done, $$\delta Q=0$$, $$\delta W=0$$, $$dU=0$$

Reversible (imaginary process) : isothermal $$\delta Q_{rev}>0$$, $$\delta {W}_{rev}<0$$ $$dU=0$$, (and so $$dS=\delta Q_{rev}/T_{i}= -\delta W_{rev}/T_{i}=pdV/T_{i})$$

• The work done in an adiabatic free expansion is unique - either it is zero or it is non-zero. Which ? – Frost Apr 13 '20 at 10:25
• The work done is 0, the work you computed is that of the ideal reversible process. Work is process dependent. – user8736288 Apr 13 '20 at 10:55
• Yes. The OP needs to become aware that in determining the entropy change for a irreversible process, one needs to determine the integral of dq/T for an alternate reversible process between the same two ends states. That is the only reason that the alternate reversible process is introduced. So you are entirely correct; there is no contradiction. – Chet Miller Apr 13 '20 at 11:56
• Thank you for the clarification – Frost Apr 13 '20 at 13:36
• Is there a rigorous proof for the validity of this procedure viz. introducing an alternate reversible process for computing the entropy of irreversible processes, when the end-states are equilibrium states ? – Frost Apr 13 '20 at 13:45

You described a system with an adiabatic wall and then you consider an isothermal expansion. There it's the problem!

The process will be an adiabatic expansion. Hence, since there is no external pressure for the gas to expand against, the work done by the system is also $$0$$.