Answer :

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, Cp 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, UAB = UACB … (ii), where UAB = change in internal energy for path AB, UACB = 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, WAB > WACB … (iii), where WAB = work done for process following path AB, WACB = work done for process following path ACB.


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


UAB + WAB > UACB + WACB … (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, QAB > QACB … (v),


We know, QAB = nCdT, QAC = nCpdT, QCB = nCvdT,


where QAB = heat supplied in process AB


QAC = heat supplied in process AC


QCB = heat supplied in process CB


n = number of moles


C = molar heat capacity


Cp = molar heat capacity at constant pressure


Cv = molar heat capacity at constant volume


dT = change in temperature.


Also, QACB = QAC + QCB … (vi), where


QACB = heat supplied in process ACB


QAC = heat supplied in process AC


QCB = heat supplied in process CB


Hence, from (v), we get,


nCdT > nCpdT + nCvdT


Dividing by ndT on both sides :


C > Cp + Cv => C > Cv (option c)

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