Relation between Cpand 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 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

Cp - Cv = R…... (5)

This equation is known as Meyer’s relation.

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