Kinetic Theory of Gases JEE 2023: Why Option C Is Correct

Q: How do you calculate gamma from degrees of freedom?

For an ideal gas with n active degrees of freedom, the molar heat capacities are Cᵥ = nR/2 and Cₚ = (n + 2)R/2. Therefore, gamma = Cₚ/Cᵥ = 1 + 2/n, which decreases as n increases.

Q: Does equal temperature and pressure mean equal sound speed in gases?

No: ideal-gas sound speed is v = √(gamma RT/M), so gamma and molar mass also matter. Equal temperature and pressure do not guarantee equal molar mass. Comparing sound speeds using gamma alone requires common temperature and molar mass.

Q: Why is option D wrong in the JEE 2023 kinetic theory question?

Option D correctly states U₆ v₇. More stored internal energy does not imply faster sound.

What is the correct answer to the Kinetic Theory of Gases JEE 2023 question?

Option C is correct in the kinetic theory of gases JEE 2023 question: at fixed temperature and molar mass, more degrees of freedom mean higher molar internal energy but lower sound speed. This is a single-correct MCQ from JEE Advanced 2023, Paper 2, Physics.

Restated: consider an ideal gas in thermal equilibrium. Its molecular degrees of freedom, energy of one mole and sound speed are denoted by:

n, U_{n}, v_{n}

At the same temperature and pressure, select the valid pair:

A) & v_{3} < v_{6} and U_{3} > U_{6} B) & v_{5} > v_{3} and U_{3} > U_{5} C) & v_{5} > v_{7} and U_{5} < U_{7} D) & v_{6} < v_{7} and U_{6} < U_{7}

The question bank rates this medium, with a 90-second target. Neither label is an official exam standard. The supplied official solution also assumes common molar mass, which the restated question does not explicitly specify.

How do degrees of freedom determine the internal-energy ordering?

At a fixed positive absolute temperature, molar internal energy increases with the number of active degrees of freedom. Equipartition assigns the following energy to each active quadratic degree of freedom per molecule: $\frac{k_{B} T}{2}$

One mole contains Avogadro’s number of molecules. Multiplying the energy per molecule by that number gives:

U_{n} = N_{A} \frac{n k_{B} T}{2} = \frac{n R T}{2}, N_{A} k_{B} = R

For the four cases in the question:

U_{3} = \frac{3 R T}{2}, U_{5} = \frac{5 R T}{2}, U_{6} = 3 R T, U_{7} = \frac{7 R T}{2}

The temperature and gas constant are common, so:

U_{3} < U_{5} < U_{6} < U_{7}

More energy-storage modes allow the gas to contain more internal energy at the same temperature. Higher total internal energy does not imply higher temperature when the degree count changes.

How do you derive gamma and compare the sound speeds?

Sound speed decreases as the degree count increases when temperature and molar mass are common. The supplied official solution uses both conditions, although the restated question explicitly specifies only common temperature and pressure.

Start with the molar heat capacities:

C_{V} = \frac{n R}{2}, C_{P} = C_{V} + R = \frac{(n + 2) R}{2}

Their ratio is:

γ = \frac{C_{P}}{C_{V}} = \frac{n + 2}{n} = 1 + \frac{2}{n}

For an ideal gas, sound speed is:

v_{n} = \sqrt{\frac{γ R T}{M}}

The denominator contains the molar mass. The degree-count symbol denotes molecular degrees of freedom, not the amount of gas:

M = molar mass, n = degrees of freedom per molecule

Substitution gives:

v_{n} = \sqrt{(1 + \frac{2}{n}) \frac{R T}{M}}

The four heat-capacity ratios are:

γ_{3} = \frac{5}{3}, γ_{5} = \frac{7}{5}, γ_{6} = \frac{4}{3}, γ_{7} = \frac{9}{7}

The reciprocal term decreases as the degree count rises: $\frac{2}{n}$

Since the gas constant, temperature and molar mass are common, the sound-speed ordering is:

v_{3} > v_{5} > v_{6} > v_{7}

Equal temperature and pressure alone do not guarantee equal molar mass. Do not apply this ordering indiscriminately to different gases. Pressure does not appear explicitly once the ideal-gas sound-speed formula is written in terms of temperature and molar mass.

Why does only option C satisfy both inequalities?

Only C has both its sound-speed comparison and its internal-energy comparison correct. An option containing two assertions is valid only when both assertions hold; one correct inequality cannot rescue the other.

Using the two orderings:

The correct answer is:

C: v_{5} > v_{7} and U_{5} < U_{7}

At fixed temperature and molar mass, increasing the degree count raises molar internal energy. It lowers gamma and sound speed.

Why is “more stored energy means faster sound” wrong in option D?

The invalid step is assuming that sound speed is proportional to total molar internal energy. Option D can arise from this mistaken argument: seven degrees of freedom store more energy than six, so sound should also travel faster.

That argument produces:

U_{6} < U_{7} ⟹ v_{6} < v_{7} (invalid implication)

Extra degrees of freedom change the heat capacities and their ratio. They do not raise the temperature in this fixed-temperature comparison.

The correct sound-speed test is:

γ_{6} = \frac{4}{3} > \frac{9}{7} = γ_{7} ⟹ v_{6} > v_{7} at common T, M

Repair the method by making two independent comparisons:

U_{n} = \frac{n R T}{2}, v_{n}^{2} = \frac{γ R T}{M}

Order internal energy using the first expression and sound speed using the second. Only then inspect the options.

How do you solve three related kinetic theory questions?

Use the heat-capacity ratio to find the degree count, then compare internal energy and sound speed independently. These are original related practice questions, not verified past-paper questions.

1. An ideal gas has the following heat-capacity ratio. Find its degrees of freedom and the internal energy of one mole at the given temperature.

γ = \frac{7}{5}, temperature = T

Rearrange the heat-capacity relation, then use equipartition:

n = \frac{2}{γ - 1} = \frac{2}{7 / 5 - 1} = 5, U = \frac{5 R T}{2}

2. Compare one mole each of the three-degree and five-degree cases at common temperature and molar mass. Find the energy and sound-speed ratios.

Divide the corresponding expressions:

\frac{U_{5}}{U_{3}} = \frac{5}{3} \approx 1.667

\frac{v_{5}}{v_{3}} = \sqrt{\frac{7 / 5}{5 / 3}} = \frac{\sqrt{21}}{5} \approx 0.917

3. Two ideal gases are at the same temperature, with these properties. Find their sound-speed ratio.

n_{A} = 5, M_{A} = 2 M, n_{B} = 7, M_{B} = M

Keep both molar masses in the calculation:

\frac{v_{A}}{v_{B}} = \sqrt{\frac{γ_{A} M_{B}}{γ_{B} M_{A}}} = \sqrt{\frac{7 / 5}{2 \times 9 / 7}} = \sqrt{\frac{49}{90}} \approx 0.738 < 1

Gas A has fewer degrees of freedom but lower sound speed because its molar mass is larger. This shows why the original ordering requires common molar mass.

Before comparing sound speeds, write down which quantities are fixed. Cancel the molar masses only if they are equal.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Keep going with How to Study Determinants JEE: Operations and Practice.

Frequently asked questions

What is the correct answer to the kinetic theory of gases JEE 2023 question?

Option C is correct: v₅ > v₇ and U₅ < U₇, assuming common temperature and molar mass. Molar internal energy increases with degrees of freedom, while gamma and sound speed decrease under these conditions.

How do you calculate gamma from degrees of freedom?