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Derive Meyer’s relation.

 Proof of Cp- Cv = R (Based on first law of thermodynamics)
Let us consider 1 mole of an ideal gas in equilibrium at a pressure P, volume V and absolute temperature T. Let Cv and Cp 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 CvdT. 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= CvdT and dW = 0, so that
                            CvdT = 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 CpdT. 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 = Cp dT and dW = P dV. Hence from the first law of thermodynamics, we obtain
                                      Cp 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
                                    (Cp – Cv) 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
                                        (Cp – Cv) dT = R dT
Or                                    Cp – Cv = R



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