Animated Solution for Physics - Thermodynamics: The specific heats, $C_p$ and $C_V$ of a gas of diatomic molecules, $A$ are given (in units of ${\text{J mol}}^{-1}{\text{K}}^{-1}$) 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

Imagine you have two mysterious containers, each holding a different diatomic gas. You can't see the molecules, but you are given their specific heat capacities at constant pressure ($C_p$) and constant volume ($C_V$). How can you tell if the molecules inside are vibrating like tiny springs or spinning like rigid dumbbells? The secret lies in the Law of Equipartition of Energy and the concept of degrees of freedom.

To peek into the microscopic world, we use a powerful thermodynamic tool: the ratio of specific heats, denoted by $\gamma$ (gamma).

This ratio is intimately connected to the degrees of freedom ($f$) of the gas molecules through the elegant relation:

A standard rigid diatomic molecule (like a dumbbell) can move in 3 dimensions (translational) and rotate around 2 axes (rotational), giving it $f=3+2=5$ degrees of freedom. However, if the bond between the atoms is flexible, it can vibrate, adding 2 more degrees of freedom (one for kinetic energy, one for potential energy), making $f=7$. Let's use this to interrogate our gases!

For Gas A, we are given $C_p=29$ and $C_V=22$. Let's plug these into our master equation:

Subtracting 1 from both sides:

Solving for $f_A$, we get:

Since $f_A$ is significantly greater than 5, it tells us a fascinating physical reality: the molecules in Gas A are not perfectly rigid. The extra degrees of freedom indicate that vibrational modes are active.

Now, let's turn our attention to Gas B, where $C_p=30$ and $C_V=21$. Substituting these values:

Simplifying the fraction $\frac{30}{21}$ to $\frac{10}{7}$:

Solving for $f_B$, we find:

This value is very close to 5. In the realm of kinetic theory, a degree of freedom near 5 confirms that the molecule behaves as a rigid rotor. It translates and rotates, but it does not vibrate.

By simply analyzing the macroscopic specific heats, we've deduced the microscopic behavior of the gases. Gas A has active vibrational modes, while Gas B is rigid and lacks them. This perfectly aligns with option (a). Thermodynamics is truly a window into the invisible!