$dQ = dU + dW$;$dQ = dU + dW = dU + PdV$,
At
therefore, at constant Volume, Work done by the gas is zero, so
$dQ = dU$;
$dQ = f/2 nRdT;$&
$dQ/dT = f/2 nR $;From Law Of Equipartition Of Energy,
$nCvdT = f/2 nRdT $;for 'n' moles of an ideal gas
$Cv = f/2R$;$dU = \frac{f}{2}nRdT$
$\implies$
$nCpdT = f/2 nRdT + PdV$$dU = dQ = \frac{f}{2}nRdT$ (at constant volume)
$nC_VdT = \frac{f}{2}nRdT$
hence,
$C_V = \frac{f}{2}R$;
also,
$nCpdT = \frac{f}{2}nRdT + PdV$ (Atat constant Pressure)
[$PdV +VdP = nRdT$; (Differentiating $PV = nRT$)
]
$nCpdT = \frac{f}{2}nRdT + nRdT$
$nCpdT = f/2nRdT + nRdT$
$Cp = \frac{f}{2}R + R$
$Cp = f/2 R + R$
hence,
$Cp = Cv + R$