Derivation of Mayer's Formula | Specific Heats Cp and Cv

Mayer's Formula is defined as the amount of heat required to raise the temperature of a 1 mole of a gas through 1°C when its volume is kept constant.

The specific heat at constant volume Cv

It is defined as the amount of heat required to raise the temperature of a 1 mole of a gas through 1°C when its volume is kept constant. It is denoted by (Cv) and given by

`C_V=\left(\frac{\triangle Q}{\triangle T}\right)_V`

The specific heat at constant pressure Cp

It is defined as the amount of heat required to raise the temperature of 1 mole of the gas through 1°C when its pressure is kept constant. It is denoted by (Cp) and given by

`C_P=\left(\frac{\triangle Q}{\triangle T}\right)_P`

Derivation of Mayer's Formula

Consider two isothermal AB and CD drawn for 1 mole of an ideal gas at close temperature T and (T+ΔT). Let the initial state of the gas be represented by point M on lower isothermal at T. Let it now be heated at constant volume until its temperature rises to (T+ΔT) and its new pressure corresponding to point L on the upper isothermal CD.

Mayer's formula

As the volume of the gas is kept constant no external work has been done by the gas and hence the change in internal energy in the process M→L is given by first law of thermodynamics

`U_L-U_M=\triangle Q-\triangle W`

`=C_V\triangle T` .... (ΔW = 0 )......(1)

Again starting from the same initial condition of the gas represented by M, let it now be heated at constant pressure until its temperature rises to (T+ΔT) and its new volume (V+ΔV) corresponds to point N on upper isothermal CD. Hence the change in internal energy in the process M→N is

`U_N-U_M=\triangle Q-\triangle W`

`=C_V\triangle T-P\triangle V` .....(2)

Because in this case ΔQ = Cp.ΔT and work performed by the gas ΔW = PΔV

Since the gas is supposed to be a perfect one, the perfect equation (PV=RT) holds good. In the initial position at M, the pressure, volume and temperature of the gas were respectively P, V and T while at N volume becomes (V+ΔV) and temperature (T+ΔT). We have

PV = RT.........(3)

P (V+ΔV) = R (T+ΔT)..........(4)

Subtracting Eq. (3) from (4), we have

P. ∆V = R. ∆T

Putting this value of P. ΔV in Eq. (2) we get

`U_N-U_M=\triangle T-R\triangle T` .......(5)

As the internal energy of a perfect gas depends only upon its temperature, the same change in internal energy will take place in both the processes M⟶L and M⟶N because temperature changes from T to (T+ΔT) in both the processes. Thus

`U_L-U_M=U_N-U_M`

`C_V\triangle T=C_P\triangle T-R.\triangle T` [using eq. 1) and (5)]

Dividing throughout by ΔT, we have

`C_V=C_P-R`

`C_P-C_V=R`

This formula is known as Mayer's Formula. All the three quantities (Cp), (Cv) and R in this equation should be expressed in the same units either in joule/mole°C or in cal/mole°C.

Read also

  1. Ensembles in Statistical Mechanics
  2. What is Entropy?
  3. Introduction to Statistical Mechanics

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