Interpret the processes from the usual $P$–$V$ diagram

The two shown thermodynamic paths are understood as:

• Process A: steeper than an adiabat / lies between isochoric and adiabatic behavior in such a way that its molar heat capacity is between $C_V$ and $C_P$.
• Process B: another path whose slope places its heat capacity between that of process A and the isobaric value.

To decide the ordering, use the general expression for molar heat capacity along any quasi-static process.

General formula for molar heat capacity along a process

For 1 mole of an ideal gas,

Hence the heat capacity for the process is

Also, using $PV=RT$,

Depending on the slope of the path on the $P$–$V$ diagram, the value of $\dfrac{dV}{dT}$ changes, and therefore the heat capacity changes.

Reference values

• At constant volume: $$dV=0 \implies C=C_V.$$
• At constant pressure: $$dP=0 \implies P\,dV=R\,dT \implies C=C_V+R=C_P.$$

Therefore, for any intermediate process between isochoric and isobaric behavior,

Compare process A and process B

From the figure, process B is closer to the isobaric case than process A, so its heat capacity is larger. Thus,

Check options

• Option A: $C_P > C_B > C_A > C_V$ ✅
• Option B: $C_P > C_V > C_A = C_B$ ❌
• Option C: $C_A=0$ and $C_B=\infty$ ❌
• Option D: $C_A=\infty,\ C_B=0$ ❌

Final answer

The correct statement is

i.e. Option A.