Consider n mole of an ideal gas is enclosed in a cylinder fitted with movable and function less piston.

Let, dθ be the amount of heat supplied to the gas at constant volume. So, that its temperature rises by dT. Then,

dθ = nCvdT……eqn(1)

Again, from first law of thermodynamics, we have,

dθ = du + d w

Since,

dw = Pdv

∴ dQ = dv + Pdv

At constant volume, dv = 0

So, dQ = du……..eqn(2)

From eqn (1) and (2)

Du = nCvdT………eqn(3)

Let, the gas is heated at constant pressure so that its temperature rises by same amount dT.

Then,

dQ’ = nCpdT……..(4)

From first law of thermodynamics, we have

dQ’ = nCpdT

Or, nCpdT = du + pdv……(5)

From (3) and (5) we have,

nCpdt = nCvdT + pdv……(6)

Again from n mole of ideal gas we have

Pv = nRT

Differentiating both sides with respect to T at constant pressure,

PdvdT=nRdTdTPdvdT=nRdTdT

Or, Pdv = nRdT

From equation (6) and (7) we have

nCpdT = nCvdT =nRdT

Or, (Cp –Cv) ndT =R ndT

Or, [Cp – Cv = R]

This is the relation between the two specific heats of gas.