Define the two molar specific heats of a gas and deduce the relation between them.

1) Molar specific heat of a gas at constant pressure:
It is defined as the amount of heat required to raise the Temperature of 1 mole of gas through 1°C at constant pressure. It is represented by Cp.
Molar specific heat of a gas at constant volume:
It is defined as the amount of heat required to raise the Temperature of 1mole of gas through 1°C at constant volume.
2) Derivation of Cp-Cv =R

Consider one mole of ideal gas contained in a cylinder provided with a friction less piston. Let “A” be the area of the piston and “p”, ”V” and “T” be Pressure, Volume & Temperature of the gas, respectively. When the gas is heated at constant volume. So that its Temperature increases through “dt”. We know that heat supplied at constant volume is utilised to increases the internal energy of the gas.
Heat supplied at constant volume
$\begin{array}{l}(\mathrm{dQ}{)}_{\mathrm{V}}=\mathrm{dU}=1\times {\mathrm{C}}_{\mathrm{V}}\times \mathrm{dt}={\mathrm{C}}_{\mathrm{V}}\mathrm{dT}\\ \therefore \mathrm{dU}={\mathrm{C}}_{\mathrm{V}}\mathrm{dT} ...\left(1\right)\end{array}$
Let the gas be heated at constant pressure for the same increase in its Temperature dT. Let this heat be (dQ)_P. It is utilised to increase in internal energy(dU) and to do external work (dW) to move the piston against constant pressure P
$\therefore (\mathrm{dQ}{)}_{\mathrm{p}}=1\times {\mathrm{C}}_{\mathrm{p}}\times \mathrm{dT}={\mathrm{C}}_{\mathrm{p}}\mathrm{dT} ...(2)$
From the first law of Thermodynamics, we can write
$\begin{array}{l}(\mathrm{dQ}{)}_{\mathrm{p}}=\mathrm{dU}+\mathrm{dW}\\ \therefore {\mathrm{C}}_{\mathrm{p}}\mathrm{dT}=\mathrm{dU}+\mathrm{dW} ...\left(3\right)\end{array}$
The work done to move the piston through a distance
$\text{ " }\mathrm{dX}\text{ " is }\mathrm{dW}=\mathrm{pAdX}=\mathrm{pdv}[\mathrm{dv}=\mathrm{Adx}]$
Substitute dW & dU in (3) equation
$\mathrm{CpdT}=\mathrm{CvdT}+\mathrm{Pdv} ...\left(4\right)$
But according to ideal gas eqn Pv=nRT, where R is the universal gas constant, Differentiating Pv=nRT keeping “P” constant, we have
$\mathrm{Pdv}=\mathrm{RdT}\dots \left(5\right)$
From (4) & (5)
$\begin{array}{l}\Rightarrow \left({\mathrm{C}}_{\mathrm{P}}-{\mathrm{C}}_{\mathrm{V}}\right)\mathrm{dT}=\mathrm{RdT}\\ \Rightarrow \mathrm{Cp}-\mathrm{Cv}=\mathrm{R}\end{array}$