Question

The specific heats, C_p and C_v of a gas of diatomic molecules, A, are given (in units of J mol^-1 k^-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:
1. A is rigid but B has a vibrational mode.
2. A has a vibrational mode but B has none.
3. A has one vibrational mode and B has two.
4. Both A and B have a vibrational mode each.

Answer 1

Correct Answer - Option 2 : A has a vibrational mode but B has none.

Concept:

The Law of Equipartition of Energy defines the allocation of energy to each motion of the atom (translational, rotational and vibrational).

Degrees of Freedom is nothing but the number of ways in which a molecule can move.

HCl has 3 translational, 2 rotational and 1 vibrational degree of freedom = No. of degrees of freedom

In this case, the total degree of freedom is 6.

R = C_p - C_v

Where,

Cp is Specific Heat in constant pressure.

Cv is Specific heat in constant volume.

The specific heat at constant pressure (Cp) is greater than the specific heat at constant volume (Cv)

Calculation:

Given,

Specific heat, C_p of molecule A = 29

Specific heat, C_v of molecule A = 22

Specific heat, C_p of molecule B = 30

Specific heat, C_v of molecule B = 21

For A,

R = C_p - C_v = 29 – 22 =7

\({C_v} = \frac{{fR}}{2}\)

\(22 = f \times \frac{7}{2}\)

\(f = \frac{{44}}{7}\)

f = 6.3 ≈ 6

Here, 3 translational, 2 rotational and one vibrational mode exist.

For B,

\(R = {C_p} - {C_v} = 30 - 21 = 9\)

\({C_v} = \frac{{fR}}{2}\)

\(21 = f \times \frac{9}{2}\)

\(f = \frac{{42}}{9}\)

f = 4.67 ≈ 5

Here, 3 translational, 2 rotational and zero vibrational mode exist.

Thus, the correct statement regarding the molecules is A has a vibrational mode but B has none.