Question

Derive the Mayer’s relation.

Answer 1

(i) Consider one mole of an ideal gas that is enclosed in a cylinder by a light, frictionless airtight piston.

(ii) Let P, V, and T be the pressure, volume, and temperature respectively of the gas.

(iii) If the gas is heated so that its temperature rises by dT, but the volume remains constant, then the amount of heat supplied to the gas (dQ₁) is used to increase the internal energy of the gas (dE).

Since the volume of the gas is constant, no work is done in moving the piston. 

∴ dQ₁ = dE = C_V dT ..............(1)

where C_V is the molar specific heat of the gas at constant volume. 

(iv) On the other hand, if the gas is heated to the same temperature, at constant pressure, the volume of the gas increases by an amount say dV. The amount of heat supplied to the gas is used to increase the internal energy of the gas as well as to move the piston backward to allow expansion of gas. The work is done to move the piston dW = PdV.

∴ dQ₂ = dE + dW = C_p dT ..............(2) 

Where C_P is the molar specific heat of the gas at constant pressure.

(v) From equations (1) and (2),

∴ C_p dT = C_V dT + dW

∴ (C_p - C_v)dT = PdV ..............(3)

(vi) For one mole of gas,

PV = RT

∴ P dV = R dT, since pressure is constant.

Substituting equation (3), we get

(C_p - C_v) dT = R dT

∴ C_p - C_v = R

This is known as Mayer’s relation between C_P and C_V. 

(vii) Also, C_P = M₀S_P and C_V = M₀S_V, where M₀ is the molar mass of the gas and S_P and S_V are respective principal specific heats. Thus, M₀S_P - M₀S_V = R/J

Where J is the mechanical equivalent of heat.

S_P - S_v = \(\frac R{M_0J}\)

Answer 2

Consider one mole of an ideal gas. Let dQ be the amount of heat is given to the system to raise the temperature by dT, and change in internal energy be dV. Then, from first law of Thermodynamics,
dQ = dU + PdV …(1)
If heat is supplied to one mole at constant volume, i.e.,
V = constant,
then, dV = 0
From equation (2),
dQ = dU …(3)
Molar specific heat (C_V ) at constant volume,

From equation (4),
dQ = dU = Cv dT …(5)
If one mole of gas is supplied heat at constant pressure, i.e., from equation (3) of molar specific heat Cp at constant pressure,

The above expression is called Mayer’s relation where R = 8.31 J.mol^-1 K^-1. This relation is for one mole of gas.