Question

Derivation for Cp-Cv=R​

Answer 1

From the equation q = n C ∆T, we can say:

At constant pressure P, we have

qP = n CP∆T

This value is equal to the change in enthalpy, that is,

qP = n CP∆T = ∆H

Similarly, at constant volume V, we have

qV = n CV∆T

This value is equal to the change in internal energy, that is,

qV = n CV∆T = ∆U

We know that for one mole (n=1) of an ideal gas,

∆H = ∆U + ∆(pV ) = ∆U + ∆(RT) = ∆U + R ∆T

Therefore, ∆H = ∆U + R ∆T

Substituting the values of ∆H and ∆U from above in the former equation,

CP∆T = CV∆T + R ∆T

CP = CV + R

CP – CV = R

Answer 2

Answer:

From first law of thermodynamics :

Δ q = Δ u + p Δ v

For constant volume i.e. Δ v = 0

Δ q = Δ u

Now divide by Δ T both side

Δ q / Δ T = Δ u / Δ T    For constant volume :

$\displaystyle{\text{C}_v}$ = Δ u / Δ T .... ( i )

Again :

For constant pressure :

Δ q = Δ u + p Δ v

Divide by Δ T both side :

Δ q /  Δ T = Δ u /  Δ T + p Δ v /  Δ T

From ( i ) we have $\displaystyle{\text{C}_v}$ = Δ u / Δ T

$\displaystyle{\text{C}_p}$ = $\displaystyle{\text{C}_v}$ + p Δ v /  Δ T

From ideal equal we have :

p Δ v = n R Δ T  

For n = 1

R = p Δ v /  Δ T

$\displaystyle{\text{C}_p}$ = $\displaystyle{\text{C}_v}$ + R

$\displaystyle{\text{C}_p}$ - $\displaystyle{\text{C}_v}$ = R

Hence , proved.