Question
Prove the Mayer's relation $C_p-C_v=\frac{R}{J}$

Answer

The first law of thermodynamics states that a system's heat absorption equals its internal energy increase and work done.(i.e) $dQ = dU + dW.$
Relation between $C_p$ and $C_v:$ Consider one mole of an ideal gas. Let the gas be heated at a constant value so that its temperature increases by $dT$
if $Q_1 =$ heat supplied to $1$ mole of gas at constant volume, then $Q_1 = C_vdT ...(i)$
Now, let the gas be heated at constant pressure until its temperature rises by $dT.$
If $Q_2 =$ heat supplied to $1$ mole of gas at constant pressure, then $Q_2 = C_pdT ...(ii)$
When gas is heated at constant volume, it will not perform external work. According to the first law of thermodynamics, the heat supplied will just increase the internal energy of the gas.
Therefore, equation $(i)$ becomes, $dU = C_vdT ...(iii)$
When heat is given under continuous pressure, it increases internal energy and allows the gas to perform work $(dW).$ As the volume of the gas increases $(dV),$ the work done by the gas $(dW)$ is equal to the $\text{PDV}...(iv).$ According to the basic law of thermodynamics.
$Q_2=Q_1+d W$
$Q_2=d U+d W$
$C_p d T=C_v d T+p d V$
$\left(C_p-C_v\right) d T=P d V$
or $C_p-C_v=\frac{p d V}{d T} ...(vi)$
if $C_p$ and $C_V$ are measured in heat units and $R$ in units of work, then
$C_p-C_v=\frac{R}{J}$

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