Consider two ideal diatomic gases A and B at some temperature T. Molecules of the gas A are rigid an

Q: Why does a vibrational mode add 2 degrees of freedom instead of 1?

A vibrational mode involves two types of energy: the kinetic energy of the oscillating atoms and the potential energy of the interatomic "spring" force. Each contributes one degree of freedom, making it 2 in total.

Q: Why didn't we use the masses $m$ and $m/4$ in our calculation?

The question asks for the ratio of molar specific heats ($C_V$). Molar specific heat is defined per mole of the gas and depends solely on the degrees of freedom ($f$) and the universal gas constant ($R$). The mass of individual molecules is irrelevant here.

Q: What are the 5 degrees of freedom for a rigid diatomic gas?

A rigid diatomic molecule can move in 3 independent directions (3 translational degrees of freedom) and can rotate about 2 independent axes perpendicular to the bond connecting the atoms (2 rotational degrees of freedom).

Q: Does temperature affect the degrees of freedom?

Yes! At low temperatures, diatomic gases behave rigidly ($f=5$). At higher temperatures, the vibrational modes become active, increasing the degrees of freedom to 7. The question explicitly stated Gas B has an active vibrational mode.

Q: What if the question asked for the ratio of specific heat capacities per unit mass?

If it asked for specific heat capacity per unit mass (often denoted by lowercase $c_v$), you would divide the molar specific heat by the molar mass ($M$). In that case, the given masses $m$ and $m/4$ would be crucial for the calculation!

Question

Consider two ideal diatomic gases $A$ and $B$ at some temperature $T$. Molecules of the gas $A$ are rigid and have a mass $m$. Molecules of the gas $B$ have an additional vibrational mode and have a mass $m/4$. The ratio of the specific heats ($C_{V}^{A}$ and $C_{V}^{B}$) of gas $A$ and $B$ respectively is

Options:

(A) 5 : 9

(B) 7 : 9

(C) 3 : 5

(D) 5 : 7

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at two different diatomic gases, Gas $A$ and Gas $B$, enclosed in a container at a temperature $T$.

Gas $A$ is described as **rigid**. You can visualize its molecules like tiny dumbbells made of two atoms connected by an unbreakable, unbendable rod.

Gas $B$, on the other hand, is a bit more energetic. Its molecules have an active **vibrational mode**. Instead of a rigid rod, imagine the two atoms are connected by a spring, constantly oscillating back and forth.

The question also throws in a detail about their masses: Gas $A$ has mass $m$, and Gas $B$ has mass $m/4$. Our goal is to find the ratio of their molar specific heats at constant volume, $\frac{C_V^A}{C_V^B}$.

### The Master Equation

To solve this, we need to bridge the gap between the microscopic behavior of the molecules and their macroscopic specific heat. The key lies in the **Law of Equipartition of Energy**.

This law tells us that the molar specific heat at constant volume, $C_V$, is directly proportional to the degrees of freedom, $f$, of the gas molecules. The master equation is:
$$C_V = \frac{f}{2}R$$
where $R$ is the universal gas constant.

This means our entire problem boils down to simply finding the degrees of freedom for Gas $A$ and Gas $B$.

### Calculating Degrees of Freedom

Let's start with Gas $A$. Since it is a rigid diatomic molecule, it can move freely in 3-dimensional space, giving it **3 translational degrees of freedom**. It can also tumble and rotate around two independent axes perpendicular to its bond, giving it **2 rotational degrees of freedom**. 
$$f_A = 3 + 2 = 5$$

Substituting this into our master equation, we get:
$$C_V^A = \frac{5}{2}R$$

Now, let's look at Gas $B$. It has the same 5 degrees of freedom as Gas $A$, but it also has an active vibrational mode. Here is a crucial concept: **one vibrational mode always adds 2 degrees of freedom**. Why? Because vibration involves both the kinetic energy of the moving atoms and the potential energy stored in the "spring" between them.
$$f_B = 5 + 2 = 7$$

So, the specific heat for Gas $B$ is:
$$C_V^B = \frac{7}{2}R$$

### Final Calculation and The Distractor

We are now ready for the final calculation. We just need to take the ratio of $C_V^A$ to $C_V^B$:
$$\frac{C_V^A}{C_V^B} = \frac{\frac{5}{2}R}{\frac{7}{2}R}$$

The $\frac{R}{2}$ terms cancel out beautifully, leaving us with our final answer:
$$\frac{C_V^A}{C_V^B} = \frac{5}{7}$$

But wait, what about the masses $m$ and $m/4$? **This is where mistakes happen...** The masses were a cleverly disguised distractor! Molar specific heat ($C_V$) is defined per mole of the gas, and it depends *only* on the degrees of freedom. The actual mass of the individual molecules does not change the amount of energy required to raise the temperature of one mole of the gas. Always trust your core concepts and don't let extra information intimidate you!

Select Answer:

Visualized Solution

$A$ and $B$ Gases

Molar Specific Heat $C_{V}$

Degrees of Freedom for Gas $A$

Degrees of Freedom for Gas $B$

Ratio of Specific Heats

The Mass Distractor

The Sigma Insight: Kinetic Theory of Gases

Analyzing the Setup

The Master Equation

Calculating Degrees of Freedom

Final Calculation and The Distractor

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