Question

Derive Mayer's relation Cp - Cy=R

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Answer 1

Answer:

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

Answer 2

Answer:

Relation between Cp and Cv (Meyer’s relation)

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 heat dQ is 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 to P + dP and T + dT respectively. This heat energy dQ is used to increase the internal energy dU of 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 of the enclosed gas is equal to the atmospheric pressure P. The temperature of the gas decreases due to the expansion of the gas.

Meyer's Relation

Now a quantity of heat dQ’ is supplied to the gas till its temperature becomes T + dT. This heat energy is not only used to increase the internal energy dU of the gas but also to do external work dW in 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

…... (5)