Question

Derive relationship between Cp and Cv for 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

$\huge{Hey mate..!!}$

What is the realation between CP and CV for an ideal gas?

Answer




We know that,

H = U + PV

by differentiating above equation with respect to temperature at constant pressure we get,

dH/dT = dU/dT + P dV/dT ————(1)

for one mole of gas,

PV = RT

differentiating this , temperature at constant pressure,we get

PdV/dT = R

dH/dT =Cp , dU/dT = Cv and PdV/dT=R hence equation (1) becomes.

Cp = Cv + R

Cp-Cv =R

this relation is known as Meyer’s relation

$<b><marquee>$Hope its help u...