The specific heats, CP and Cv of gas of diatomic molecules, A, is given (in units of J mol-1K-1 by 29 and 22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30 and 21. If they are treated as ideal gases, then

For any ideal gas γ=CpCv=1+2fFor the first case   A 2922=1+2f   Solving for f we get f= 44/7=   6  So it has 3 translational , 2 rotational and 1 vibrational  degrees of freedomFor the second case  B  solving for f we get f= 14/3 = 5so it has 3 translational  and 2 rotational degrees of freedom