Animated Solution for Physics - Thermodynamics: The temperature of $3.00\text{ mol}$ of an ideal diatomic gas is increased by $40.0^{\circ }\text{C}$ without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of workdone by the gas is $x/10$, then the value of $x$ (round off to the nearest integer) is ……… . (Given, $R=8.31{\text{ J mol}}^{-1}{\text{ K}}^{-1}$)

Thermodynamics is often perceived as a dense jungle of formulas, constants, and specific numerical values. However, some of the most elegant problems in physics are designed to test your conceptual clarity rather than your ability to crunch numbers. This problem is a perfect example of a "distractor trap"—where you are given specific values like $3.00\text{ mol}$ and $40.0^{\circ }\text{C}$, but the underlying physics renders them completely irrelevant.

Let's embark on a journey to understand why this happens, exploring the concepts of degrees of freedom, internal energy, and work done in an isobaric process.

Degrees of Freedom

The problem introduces an ideal diatomic gas. The behavior of any gas at a microscopic level is governed by its degrees of freedom ($f$), which represent the number of independent ways a molecule can possess energy.

For a diatomic molecule (imagine a dumbbell shape like $O_2$ or $N_2$), it can move in three independent directions in space, giving it 3 translational degrees of freedom. Additionally, it can rotate around two independent axes perpendicular to the bond connecting the atoms, contributing 2 rotational degrees of freedom.

The problem explicitly states: "The molecules in the gas rotate but do not oscillate." This is a crucial constraint. It tells us that the vibrational modes are "frozen" (which is typical at room temperatures). Therefore, the total degrees of freedom is simply the sum of translational and rotational modes:

Internal Energy and Work

Now, we need to evaluate two macroscopic thermodynamic quantities: the change in internal energy ($\Delta U$) and the work done ($W$).

1. Change in Internal Energy ($\Delta U$) According to the equipartition theorem, the internal energy of an ideal gas depends solely on its temperature and degrees of freedom. The change in internal energy is given by:

Where $C_V$ is the molar heat capacity at constant volume. For an ideal gas, $C_V=\frac{f}{2}R$. Substituting our value of $f=5$, we get $C_V=\frac{5}{2}R$. Thus, the expression for the change in internal energy becomes:

2. Work Done ($W$) The problem states that the temperature is increased "without changing the pressure of the gas." This defines an isobaric process. The work done by a gas expanding at a constant pressure $p$ is:

While we don't know the change in volume ($\Delta V$), we can use the ideal gas law, $pV=nRT$. Since pressure $p$, number of moles $n$, and the gas constant $R$ are all constants in this scenario, a change in volume is directly proportional to a change in temperature:

Therefore, the work done can be elegantly rewritten as:

The Trap Revealed

We are asked to find the ratio of the change in internal energy to the work done. Let's set up the fraction using our derived expressions:

Look closely at this expression. The number of moles ($n$), the universal gas constant ($R$), and the change in temperature ($\Delta T$) appear in both the numerator and the denominator. They cancel out perfectly!

This is the beautiful trap. The $3.00\text{ mol}$ and $40.0^{\circ }\text{C}$ were completely unnecessary. The ratio depends only on the atomicity of the gas (its degrees of freedom).

To find the value of $x$, we equate our result to the given format $\frac{x}{10}$:

Multiplying the numerator and denominator of our fraction by 5 yields:

Comparing the numerators, we arrive at our final answer:

This problem serves as a powerful reminder: always set up your equations algebraically before plugging in numerical values. You might just find that the universe (or the examiner) has provided a beautiful shortcut.