Question
Derive Mayer’s relation.

Answer

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 $\left( dQ _1\right)$ 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. $ \therefore dQ _1= dE = C _{ V } dT $ 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 $d V$. 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. $ \therefore dQ _2= dE + dW = C _{ p } dT $
Where $C_p$ is the molar specific heat of the gas at constant pressure.
v. From equations (1) and (2),
$ \therefore C_p d T=C_V d T+d W$
$\therefore\left(C_p-C_v\right) d T=P d V $
vi. For one mole of gas,
$PV = RT$
$\therefore PdV = R dT$, since pressure is constant.
Substituting equation (3), we get
$ \left(C_p-C_v\right) d T=R d T$
$\therefore C_p-C_v=R $
This is known as Mayer's relation between $C_p$ and $C_V$.
vii. Also, $C_P=M_0 S_p$ and $C_V=M_0 S_V$, where $M_0$ is the molar mass of the gas and $S_p$ and $S_V$ are respective principal specific heats. Thus, $M_0 S_P-M_0 S_V=R / J$
Where $J$ is the mechanical equivalent of heat.
$S_p-S_v=\frac{R}{M_0 J}$

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