(N/A) For $1$ mole of an ideal gas,the first law of thermodynamics is $\Delta Q = \Delta U + \Delta W$.
At constant volume,$\Delta W = 0$,so $\Delta Q = \Delta U$. Since $\Delta U = C_V \Delta T$,we have $C_V = \frac{\Delta U}{\Delta T}$.
For an ideal gas,the internal energy $U$ depends only on temperature,so $\Delta U = C_V \Delta T$ holds for any process.
At constant pressure,$\Delta Q = \Delta U + P \Delta V$. Dividing by $\Delta T$,we get $\frac{\Delta Q}{\Delta T} = \frac{\Delta U}{\Delta T} + P \frac{\Delta V}{\Delta T}$.
This gives $C_P = C_V + P \frac{\Delta V}{\Delta T}$.
From the ideal gas equation for $1$ mole,$PV = RT$. Differentiating at constant pressure,$P \Delta V = R \Delta T$,which implies $P \frac{\Delta V}{\Delta T} = R$.
Substituting this into the expression for $C_P$,we get $C_P = C_V + R$.
Therefore,the relation is $C_P - C_V = R$.