Ask a Teacher

Let us consider 1 mole of an ideal gas in equilibrium at a pressure P, volume V and absolute temperature T. Let C_v and C_p be the molar specific heats of the gas at constant volume and at constant pressure respectively.

Let the gas be heated at constant volume so that its temperature is raised by an infinitesimal amount dT. The heat supplied will be C_vdT. As the volume remains constant, the external work done is zero. From the differential form of the first law of thermodynamics, we have

                                 dQ = dU + dW

Where dQ is the heat supplied, dU the increase in internal energy of the gas and dW the work done (all expressed in same units).

But, here dQ= C_vdT and dW = 0, so that

                            C_vdT = dU

Let the same gas be now heated at constant pressure P. until the temperature is raised by the same amount dT. The heat supplied will be C_pdT. Now the gas would expand and do external work against the pressure P. If dV be the change in volume of the gas, the external work would be PdV. Thus, for this process, dQ = C_p dT and dW = P dV. Hence from the first law of thermodynamics, we obtain

                                      C_p dT = dU + dV

The temperature –change is the same in both processes. Since internal energy U of an ideal gas depends only on the temperature, the change in internal energy dU is same in both processes. Then, eliminating dU from equation (i) and (ii), we get

                                    (C_p – C_v) dT = P dV

Now, the equation of state for 1 mole of an ideal gas is

                                    PV = RT

Where R is the universal gas constant. Differentiating it, keeping P constant, we get

                                 P dV = R dT

Substituting this value of P dV in equation (iii), we get

                                        (C_p – C_v) dT = R dT

Or                                    C_p –C_v = R

This is called 'Mayer's Relation.