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:

For an ideal gas the molar heat capacities at constant pressure and volume are related by the very first identity we always quote

$$C_p-C_v=R,$$

where $$R$$ is the universal gas constant. In addition, the total molar internal energy is written in the form

$$U=\dfrac{f}{2}RT,$$

with $$f$$ denoting the total number of quadratic degrees of freedom that are actually “active” at the temperature of the experiment. Because

$$C_v=\left(\dfrac{\partial U}{\partial T}\right)_V=\dfrac{f}{2}R,$$

we obtain immediately the useful pair of working formulae

$$C_v=\dfrac{f}{2}R\qquad\text{and}\qquad C_p=C_v+R=\left(\dfrac{f}{2}+1\right)R.$$

Finally we define the ratio of heat capacities

$$\gamma=\dfrac{C_p}{C_v}=1+\dfrac{R}{C_v}=\dfrac{f+2}{f}.$$

Now we recall the textbook values for a diatomic molecule:

• 3 translational degrees of freedom.
• 2 rotational degrees of freedom (because the molecule is linear).
• Each normal vibrational mode contributes two quadratic terms (one kinetic, one potential) and therefore counts as two in the energy‐counting scheme.

Hence

• Without any vibrational excitation (a “rigid” diatomic) we have $$f=3+2=5.$$ This gives $$C_v=\dfrac{5}{2}R\approx20.8\ \text{J mol}^{-1}\text{K}^{-1},\qquad C_p=\dfrac{7}{2}R\approx29.1\ \text{J mol}^{-1}\text{K}^{-1},\qquad\gamma=\dfrac{7}{5}=1.40.$$
• If the single vibrational mode is fully excited, two more terms are added, so $$f=5+2=7.$$ Then $$C_v=\dfrac{7}{2}R\approx29.1\ \text{J mol}^{-1}\text{K}^{-1},\qquad C_p=\dfrac{9}{2}R\approx37.4\ \text{J mol}^{-1}\text{K}^{-1},\qquad\gamma=\dfrac{9}{7}=1.29.$$

With the theory organised, let us examine the two experimental gases.

Gas A. The given numbers are

$$C_p(A)=29\ \text{J mol}^{-1}\text{K}^{-1},\qquad C_v(A)=22\ \text{J mol}^{-1}\text{K}^{-1}.$$

We use the difference to obtain the effective $$R$$ that belongs to the same set of rounded data:

$$R_A=C_p-C_v=29-22=7\ \text{J mol}^{-1}\text{K}^{-1}.$$

With this value the effective number of degrees of freedom is

$$f_A=\dfrac{2C_v}{R_A}=\dfrac{2\times22}{7}=6.29\ (\text{approximately}).$$

This figure is distinctly larger than 5 and lies quite close to the ideal value 7 that one gets when a single vibrational mode is active. We confirm the same tendency by the ratio

$$\gamma_A=\dfrac{C_p}{C_v}=\dfrac{29}{22}=1.318,$$

which is very near the theoretical 1.29 for a diatomic possessing one vibrational mode, and markedly below 1.40 for a rigid diatomic. Hence Gas A does possess its vibrational mode.

Gas B. The quoted values are

$$C_p(B)=30\ \text{J mol}^{-1}\text{K}^{-1},\qquad C_v(B)=21\ \text{J mol}^{-1}\text{K}^{-1}.$$

First compute the rounded difference:

$$R_B=C_p-C_v=30-21=9\ \text{J mol}^{-1}\text{K}^{-1}.$$

Now derive the corresponding $$f$$:

$$f_B=\dfrac{2C_v}{R_B}=\dfrac{2\times21}{9}=4.67\ (\text{approximately}).$$

This result is very close to 5 and far from 7, so the only degrees of freedom effectively operating are the three translational and the two rotational ones; the vibrational mode is evidently still “frozen out.” The heat-capacity ratio reinforces the same conclusion:

$$\gamma_B=\dfrac{C_p}{C_v}=\dfrac{30}{21}=1.429,$$

essentially the theoretical 1.40 expected for a rigid diatomic with no vibrational contribution.

Putting the two discussions together we find that Gas A has its single vibrational mode active, whereas Gas B behaves as a rigid diatomic molecule with no vibrational contribution.

Hence, the correct answer is Option B.