91Ó°ÊÓ

The figure shown point i is the initial state of ideal gas at temperatureT which undergoes a reversible process.

Entropy change is a phenomenon that quantifies how disorder or randomness has changed in a thermodynamic system. We can compare the entropy changes of the gas at different points from specific heat and temperature from the point i using the relation between change in entropy, specific heat, and temperature at the given point.

Formulae:

The entropy change of a gas for isobaric process, $∆S_{gas}=n{C}_{P}In\frac{T_f}{T_i}$ …(¾±)

The entropy change of a gas for isochoric process, $∆S_{gas}=n{C}_{V}In\frac{T_f}{T_i}$ …(¾±¾±)

In $P-V$ plot, there is initial point i with temperature T. There are two isothermal states with temperatures $T+∆T\mathrm{and}T-∆T$

There are four processes in which two of them are at a higher temperature and two of them are at a lower temperature. The points a and c are at constant volume.

So, the volume at final and initial position can be given as follows: $V_f=V_i$.

The process in which heat is absorbed leads to an increase in the temperature and entropy of the gas. So, the change of entropy of the gas is positive. $∆S>0$.

The process that releases energy in the form of heat leads to decrease in entropy. i.e., $∆S<0$

The molar specific heat at constant pressure is greater than constant volume, i.e. $C_P>C_V$.

The points and are at higher temperatures.

So the change of entropy is larger for the isobaric process considering equations (i) and (ii). Thus, the relation of entropy change in both processes is given as: $∆S_P>S_V$.

So, entropy change is greater at point b and d than at point a and c.

Since b is at a higher temperature than that of d and a is at a higher temperature than that of c. So, the relation of entropy changes can be given as:

$$\begin{gathered} ∆S_b>∆S_d \\ ∆S_a>∆S_c \end{gathered}$$

Therefore, the ranking of the entropy changes of the gas is $b>a>c>d$.