Cp(specific heat at constant pressure)Cv(specific heat at constant volume)When a gas is heated at constant volume,no external work is done and so the heat supplied is consumed only in increasing the internal energy of a gas.But if the gas is heated at constant pressure , the gas expands against the external pressure so does some external work.In this case the supplied heat is used up in increasing the internal energy of the gas and in doing some external work .Since the internal energy is depends only on temperature,for the same rise of temperature the internal energy of a mass of a gas will increase by the same amount whether the pressure or volume remains constant.But since external work is additionally done for constant pressure than at constant volume to produce the same rise in temperature of the gas.

Let us consider one mole of an ideal gas enclosed in a cylinder provided with a frictionless piston of area A. Let P, V and T be the pressure, volume and absolute temperature of gas respectively as shown in below figure.

A quantity of heatdQis supplied to the gas. To keep the volume of the gas constant, a small weight is placed over the piston. The pressure and the temperature of the gas increase toP + dP and T + dTrespectively. This heat energydQis used to increase the internal energydUof the gas. But the gas does not do any work(dW = 0).

∴dQ = dU = 1 × Cv× dT …... (1)

The additional weight is now removed from the piston. The piston now moves upwards through a distance dx, such that the pressure ofthe enclosed gas is equal to the atmospheric pressureP. The temperature of the gas decreases due to the expansion of the gas.

Now a quantity of heatdQ’is supplied to the gas till its temperature becomesT + dT. This heat energy is not only used to increase the internal energydUof the gas but also to do external workdWin moving the piston upwards.

∴ dQ’ = dU + dW

Since the expansion takes place at constant pressure,

dQ ′ = CpdT

∴ CpdT = CvdT + dW …... (2)

Work done, dW = force × distance == P × A × dx

dW = P dV (since A × dx = dV, change in volume)

∴ CpdT = CvdT + P dV …... (3)

The equation of state of an ideal gas is

PV = RT

Differentiating both the sides

PdV = RdT …... (4)

Substituting equation (4) in (3),

CpdT = CvdT + RdT

Cp= Cv+ R