I am not able to completely understand the derivation of the thermodynamic relation $C_p - C_v = R$ [closed]

Question

$$\Delta Q = \Delta U + P\Delta V = \Delta U + \Delta W$$

At constant volume,

$$\Delta V = 0$$ $$\Delta W = P \Delta V = 0$$ $$\Delta Q = \Delta U$$ $$\Delta Q = n C_V \Delta T = \Delta U$$

At constant pressure,

$$P\Delta V = nR\Delta T$$ $$\Delta W=P \Delta V = n R \Delta T$$ $$\Delta Q = \Delta U+nR \Delta T$$ $$\Delta Q=n C_P \Delta T = \Delta U + nR\Delta T$$

Now for n=1 mole,

$\Delta U = C_V \Delta T$
&
$C_P \Delta T = \Delta U + R\Delta T$

$\implies$

$\frac{\Delta U}{\Delta T} = C_V$ $\to$ (1)
&
$C_P - \frac{\Delta U}{\Delta T} = R$ $\to$ (2)

from (1) & (2),

$C_P - C_V = R$

This is the derivation usually found everywhere.

Here, I am unable to understand how $ \Delta U/ \Delta T = C_v$ in (1) replaces/substitutes $ \Delta U/ \Delta T$ in (2).

Answer 1

$ \Delta Q_{@constantP}$ - $ \Delta Q_{@constantV}$ = $ \Delta U_{@constantP}$ - $ \Delta U_{@constantV}$ + $P \Delta V$

Dividing both sides by $(n \Delta T)$

[Same change in temperature ($ \Delta T$) is brought in both the processes.]

$C_P - C_V$ = $ \Delta U/(n \Delta T)_{@ constant P}$ - $ \Delta U/(n \Delta T)_{@ constant V}$ + $P \Delta V/(n \Delta T)$

Now, the key point is that, for ideal gases, internal energy (U) depends on temperature only, so the $\Delta U$ in any process, whether at constant pressure or constant volume, will be the same if the change in temperature ($ \Delta T$) attained in both the processes is the same.
Hence, in the above result, the two terms for the change in internal energy w.r.t. the change in temperature ($ \Delta U/n \Delta T$) will cancel each other, and you get,

$C_P - C_V = P \Delta V/n \Delta T$ = R

[ $P \Delta V/n \Delta T = R$ ]

Answer 2

$dQ = dU + dW = dU + PdV$,
therefore, at constant Volume, Work done by the gas is zero, so $dQ = dU$
&
From Law Of Equipartition Of Energy,
for 'n' moles of an ideal gas $dU = \frac{f}{2}nRdT$

$\implies$
$dU = dQ = \frac{f}{2}nRdT$ (at constant volume)
$nC_VdT = \frac{f}{2}nRdT$
hence,
$C_V = \frac{f}{2}R$;

also,
$nCpdT = \frac{f}{2}nRdT + PdV$ (at constant Pressure)

[$PdV +VdP = nRdT$; (Differentiating $PV = nRT$)]

$nCpdT = \frac{f}{2}nRdT + nRdT$
$Cp = \frac{f}{2}R + R$
hence,
$Cp = Cv + R$