Question

27. A gas performs Q work when it expand at constant pressure. During this process heat absorbed by the gas is 4Q. The average number of degrees of freedom for the gas is

Answer 1

In a isobaric process,

U= 4Q - Q=n Cv dT = 3Q

Work = n R dT

So, U/Work = Cv/R

3=Cv/R

R = Cp−Cv3= CvCp−CvCp

Cv = 4/3 = γγ = 1+2f

where f is degree of freedom,

4/3 = 1+2f

f = 6

Answer 2

Answer:

f =6

this is the right answer

Explanation:

u hope it helps