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 (Fig.).

**Relation between C_{p} and C_{v} (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 (Fig.).

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 ? C _{v} ? 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.

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 *′* *=* C _{p}dT*

*C _{p}dT = C_{v}dT + dW ????(2)*

Work done, *dW* = force ? distance

= *P ? A ? dx*

*dW = P dV (since A ? dx = dV, change in volume)*

*C _{p}dT = C_{v}dT + 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),

*C _{p}dT = C_{v}dT + RdT*

*C _{p} = C_{v} + R*

* C _{p} - C_{v}
= R*

This equation is known as Meyer?s
relation

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11th 12th std standard Class Physics sciense Higher secondary school College Notes : Relation between Cp and Cv (Meyer's relation) |

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