Why change in entropy in reversible adiabatic process is zero?

Question

In the adiabatic expansion step of carnot cycle, adiabatic change is carried out in an insulated system. In this process volume gets increased but entropy remains constant.

Now if we look at the concept (as far as I know) of entropy, it determines the number of arrangement in which gas particles can remain within the container/box. Since the volume increased, shouldn't this number of arrangement increase and a change in entropy take place? I know there is no exchange of heat but as far as I know entropy is concerned with the number of arrangement. So why entropy of an isentropic process remains constant?

Answer 1

The temperature also decreases due to the system doing work on the surrounding. So you have more volume but less energy throughout that volume. Entropy is not only concerned with spatial configurations.

Answer 2

In classical statistical mechanics the entropy is related to the volume of phase space, that is the space of the positions and momentum. The volume of this space is related to the number of microstates, and it's proportional to the volume of the container, but also to the volume in momentum space. In this case the space volume increases, but the temperature (that is a measure of the mean kinetic energy) decreases, affecting the momentum space.

Answer 3

For an ideal gas $PV = n RT$, the expression of entrop can be calculate:

\begin{align} dS =& \frac{dU}{T} + \frac{PdV}{T};\\ =& n C_v \frac{dT}{T} + nR\frac{dV}{V};\\ \end{align}

Carry out the integal from state 1 to state 2: \begin{align} S_2 - S_1 =& n C_v \left(\ln T_2 - \ln T_1\right) + nR \left(\ln V_2 - \ln V_1\right);\\ =& nC_v \ln \frac{T_2}{T_1} + nR\ln\frac{V_2}{V_1}. \tag{1} \end{align}

In an adiabatic expanion, $V_2 > V_1$, the second term in Eq.(1) is positive; but the temperature becomes lower $T_2 < T_1$, the first term is negative. It is a good practice to show explicitly that these tow terms cancel each other in an adiabatic expansion process.