Physics · Kinetic Theory of Gases

JEE Advanced 2023 — Paper 2 — Question 17

An ideal gas is in thermodynamic equilibrium. The number of degrees of freedom of a molecule of the gas is $n$ . The internal energy of one mole of the gas is $U_{n}$ and the speed of sound in the gas is $v_{n}$ . At a fixed temperature and pressure, which of the following is the correct option ?

Option A:
$\mathrm{v}_{3}<\mathrm{v}_{6}$ and $\mathrm{U}_{3}>\mathrm{U}_{6}$
Option B:
$v_{5}>v_{3}$ and $U_{3}>U_{5}$
Option C:
$\mathrm{v}_{5}>\mathrm{v}_{7}$ and $\mathrm{U}_{5}<\mathrm{U}_{7}$
Correct
Option D:
$\mathrm{v}_{6}<\mathrm{v}_{7}$ and $\mathrm{U}_{6}<\mathrm{U}_{7}$

Answer: C

Step-by-step solution

U_{n}=\frac{n}{2} R T …(i) \end{gathered}$$ $\mathrm{U}_{\mathrm{n}} \propto \mathrm{n}, $ where n is degree of freedom As $\mathrm{n} \rightarrow$ increases, hence; $\mathrm{U}_{\mathrm{n}} \rightarrow$ increases $$\begin{gathered} V_{n}=\sqrt{\frac{\left(1+\frac{2}{n}\right) R T}{M}} …(ii) \end{gathered}$$ where $n$ is degree of freedom As $\mathrm{n} \rightarrow$ increases, hence; $\mathrm{V}_{\mathrm{n}} \rightarrow$ decreases

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Exam: JEE Advanced 2023
Paper: Paper 2
Subject: Physics
Chapter: Kinetic Theory of Gases
Topic: Equipartition Law of Energy and Degrees of Freedom