Entropy for irreversible adiabatic process

Question

For an adiabatic process we know $dQ=0$. So the entropy change should also be zero. Please explain how can we calculate the entropy change.

Answer 1

For an adiabatic reversible process the entropy change is zero. For an irreversible adiabatic process, it is not. If you want to calculate the entropy change for an adiabatic irreversible process, you need to devise a reversible process between the same initial and final states as for the irreversible process, and then calculate the entropy change for that. dq will not be equal to zero for the reversible path.

Answer 2

Definition
Thermodynamic entropy of a system (see this thread for other kinds of entropy) is defined in terms of heat transfer in a reversible process as: $$ dS = \frac{\delta Q_{rev}}{T} $$ (note that the temperature is the same for the system and the environment, since we are in equilibrium, but in the actual definition by Clausius, $T$ is the temperature of the environment - $T_{ext}$.)

Entropy as a state function
Once defined in this way, the entropy can be shown to be a function of state, which means that for a system in thermodynamic equilibrium the entropy is unambiguously defined by the independent state variables. E.g., for an ideal gas with $f$ degrees of freedom the entropy is given by Sackur-Tetrode equation, so that $$ S = S_0 + k_B N\log V + \frac{f}{2}k_B N \log T $$ From this we see that the surfaces of constant entropy are determined by equation $$ VT^{\frac{f}{2}}=\text{const}\Leftrightarrow TV^{\frac{2}{f}}=\text{const}, $$ which is actually the equation of an adiabatic curve.

This means that an adiabatic expansion/compression should be accompanied by appropriate temperature (i.e., the internal energy) change for the entropy to remain constant. This energy is provided via work done by the external forces during the expansion: $$ dU=\frac{f}{2}Nk_BdT=PdV. $$ If the expansion/compression is quasistatic, than at every point the gas is in equilibrium and the ideal gas equation $PV=Nk_BT$ applies. We can then write $$ dU=\frac{f}{2}Nk_BdT=PdV=\frac{Nk_BT}{V}dV\Rightarrow \frac{f}{2}\frac{dT}{T}=\frac{dV}{V}\Rightarrow \frac{f}{2}\log T = \log V + \text{const}, $$ i.e., we reproduce the equation for the adiabatic curve without resorting to the molecular theory, from purely thermodynamic arguments.

However, our key assumption in this derivation was that the process is reversible and the ideal gas equation applies. This may not be the case - e.g., the piston in a cylinder may move too quickly for some molecules to strike it, so that it does less work than the equilibrium calculation predicts. As an extreme case we could imagine volume increasing instantaneously from $V_0$ to $V_1$, so that no molecules have time to collide with the container walls during the expansion, and no work is done. Then the energy of the gas does not change, but the volume increases. This means that, when the gas expands to fill all the new volume, it will have the same temperature but different volume, i.e. $$ S_1 - S_0=k_B N \log\frac{V_1}{V_0}, $$ i.e., the entropy has increased.

Clausius inequality
How does this relate to the original definition of entropy? It is answered by Clausius inequality: $$ dS\geq \frac{\delta Q}{T_{ext}} $$ That is, once we determined the surfaces of equal entropy using equilibrium processes, any spontaneous non-equilibrium change will mean movement not along these surfaces, but from the surfaces of lower entropy to those of higher entropy. In case of an adiabatic process this simply means movement between different adiabatic curves: $$ dS \geq 0. $$