I have seen a discussion of what would happen for an ideal gas expands irreversibly and adiabatically until absolute zero degree K. The entropy change is like that:

$\Delta S=C_v \ln(T_2/T_1)+R \ln(V_2/V_1)$

It is impossible for T2 to be zero K in the equation and so it becomes one justification of the third law. I wonder whether it is valid and whether an irreversible process can be represented by the above equation. For irreversible process the heat exchange is not equal to Tds so can the Gibbs equation be applied?

The video is in https://www.youtube.com/watch?v=r4fGG_7NQr8 , 46:12


The heat is removed from the gas because it does work as it expands, so the amount of cooling depends on the amount of work done.

In an irreversible expansion an ideal gas always does less work than in a reversible expansion. So if your expansion is irreversible the gas will cool less than it would in a reversible expansion.

So although an irreversible expansion would not be decribed by your equation, it does not contradict the conclusion that you cannot reach absolute zero.

Response to comment:

Ah, OK, I misread your question slightly. The reason the equation you cite holds for all processes is that every quantity in it is a state variable, so overall the equation is an equation of state. That means it doesn't matter how you got from $(V_1, T_1)$ to $(V_2, T_2)$, reversible or irreversible, the entropy change will be the same.


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