Question

A sample of 2 moles of an ideal diatomic gas experiences a temperature increase of 60 K at constant volume. (a) Find the increase in internal energy if only translational and rotational motions are possible. 8.73 103 J (b) Find the increase in internal energy if translational, rotational, and vibrational motions are possible. 0.01222 (c) What fraction of the energy calculated in (b) is translational kinetic energy?

Answer 1

Answer:

A. To find increase in internal energy

We use

CV= f/2* R

=20.77J/molK

So change in internal energy is

nCvdt

So substituting

= 2 x 20.77x 60K

= 2493J

B.

We know that molar specific heat increase by R/2

So CV= (f/2+1)R

Which is 29.1J/milk

So

Increase in internal energy will be

nCvdt

= 2* 29.1*60

= 3490.2J

C. the translational energy is 3/2R

And the total molar specific heat capacity is 7/2R

So the fraction of energy in b due to K.E will be

3/7 * 3490.2

=1495.8J