Let us consider the figure given below :

We consider the process AB to be composed of two processes, AC(at constant pressure) and CB(at constant volume), such that AB = AC + CB. C is the molar heat capacity of AB, C_p is the molar heat capacity of AC(constant pressure) and Cv is the molar heat capacity of CB(constant volume).

From the first law of thermodynamics, we know that Q = U + W … (i),

Where Q = heat supplied, U = change in internal energy, W = work done on the system.

Since the change in internal energy is independent of the path, it will have the same value for paths AB and ACB.

Hence, U_AB = U_ACB … (ii), where U_AB = change in internal energy for path AB, U_ACB = change in internal energy for path ACB.

Now, the work done by a process is given by the area under the PV diagram. We can clearly see that the area under process AB is greater than the sum of areas under processes AC and CB.

Hence, W_AB > W_ACB … (iii), where W_AB = work done for process following path AB, W_ACB = work done for process following path ACB.

Adding (ii) and (iii), we get :

U_AB + W_AB > U_ACB + W_ACB … (iv)

But, from the first law of thermodynamics, Q = U + W, where Q = heat supplied, U = change in internal energy, W = work done on the system.

Hence, we get, Q_AB > Q_ACB … (v),

We know, Q_AB = nCdT, Q_AC = nC_pdT, Q_CB = nC_vdT,

where Q_AB = heat supplied in process AB

Q_AC = heat supplied in process AC

Q_CB = heat supplied in process CB

n = number of moles

C = molar heat capacity

C_p = molar heat capacity at constant pressure

C_v = molar heat capacity at constant volume

dT = change in temperature.

Also, Q_ACB = Q_AC + Q_CB … (vi), where

Q_ACB = heat supplied in process ACB

Q_AC = heat supplied in process AC

Q_CB = heat supplied in process CB

Hence, from (v), we get,

nCdT > nC_pdT + nC_vdT

Dividing by ndT on both sides :

C > C_p + C_v => C > C_v (option c)