Question

Derive the relation between cp and cv for an ideal gas

Answer 1

Relationship between CP and CV for an Ideal Gas

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 internal energy, that is, qV = n CV∆T= ∆U

We know that for one mole (n=1) of 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

Or CP = CV + R

Or CP – CV= R

Answer 2

Explanation:

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