Adiabatic process is isentropic, but I can't get it why is it so. If we go by statistical method

$$S = K\ln W,$$

where $W$ is the thermodynamic probability, and if we look into adiabatic expansion the volume increases, we can say that the thermodynamic probability increases, and hence the entropy of the system should increase.

Moreover, if we look into isothermal expansion of an ideal gas in vacuum, there would be no work done, and so no heat would enter or move out of the system. But since there is a change in volume, we can write

$$∆S = nR\ln\frac{V_2}{V_1}.$$


You left out the adjective “reversible.” The entropy change for an adiabatic reversible process is zero. In an adiabatic reversible process, the temperature also changes.

  • $\begingroup$ I get it that with the decrease in temperature due to expansion we cannot achieve the higher energy states the entropy increase due to increase in volume would be cancelled by the decrease in entropy due to decrease in temperature but temperature would also change in irreversible adiabatic expansion. $\endgroup$ – Nikhil Pundir Dec 9 '20 at 14:48
  • $\begingroup$ No it wouldn’t. For an ideal gas, viscous heat generation in the irreversible expansion would cancel out the temperature decrease. $\endgroup$ – Chet Miller Dec 9 '20 at 14:57
  • $\begingroup$ Reversible is important as you've mentioned. $\endgroup$ – Kashmiri Dec 9 '20 at 16:59

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