thanks ecm,

Here's question 79:

It's based on the fact that at high tempereature you have all three groups: Translational, Rotational and vibrational energies that contribute to the internal energy; and at low temperature there is only the translational energy.

so for high temperature:

Translational: 3 Degrees of Freedom, each contributes (1/2)RT to energy

Rotational: 2 DOF(beacuse diatomic=linear shape), each contributes (1/2)RT to the energy

Vibrational

3N-5)DOF (linear shape); since N=2 we get(3*(2)-5)=1 DOF each vibrational DOF contributes (1)RT to the energy.

so at high T the internal energy is:

U=U(trans) +U(rot) +U(vib)= (3/2)RT+(2/2)RT+(1)RT=(7/2)RT

since heat capacity at constant V is Cv=dU/dT (derivative wrt to T)

so for high temps Cv1=(7/2)R.

And for low temp the internal energy is:

U=U(trans)= (3/2)RT; and since Cv=dU/dT;

Cv2=(3/2)R

so the ratio (Cv1/Cv2)=[(7/2)R/(3/2)R]=7/3 which I believe is the answer in the practice test

.

So Remember: at Low temperature only Translational energy, if you boost the temperatur a bit, you get both Translational and Rotational; then only at high temperature do you have all three( Translational, Rotational, and Vibrational).

I assume if they change the question to put it on the GRE again, they will tell you at exaclty what temperatures you start getting different energies, since it varies with the gaz.

read you soon