# Can the ideal gas expand adiabatically and irreversely?

Suppose that the ideal gas is expanding adiabatically from the state $(P_1, V_1, T_1)$ to the state $(P_2, V_2, T_2)$

I think this process can be allowed to happen reversibly or irreversibly, but then I face a puzzling contradiction regarding entropy change ($\Delta S$).

If the process occurred reversibly, $\Delta S = 0$ since there are no heat transfer and no entropy generation.

If the process occurred irreversibly, $\Delta S > 0$ since there is entropy generation by the irreversibility of the process.

However, because entropy is a state function it cannot be zero and non zero for the two fixed chosen states $(P_1, V_1, T_1)$ and $(P_2, V_2, T_2)$

• In the irreversible process, there will be entropy generated, so the final entropy will be greater than the initial entropy. So if there is a reversible process between the same two end states, you will have to have $$\Delta S=\int{q_{rev}/T}$$, and $q_{rev}$ will have to be non-zero. This means that the reversible process will have to be non-adiabatic. – Chet Miller May 29 '18 at 2:43
Not really. The fact that you are taking the gas along an adiabatic process already implies that the process is reversible (since it should be at thermal equilibrium with a reservoir during the entire path). If, however, your only requirement is that the initial and final states follow $p_1V_1^\gamma=p_2V_2^\gamma$, then, since entropy is a state function, the entropy of the gas is the same, but the entropy of the whole Universe could (and likely will) be larger.