Physics · Thermodynamics

JEE Main 2025 — 2 April, Morning Shift — Question 66

$\gamma_{A}$ is the specific heat ratio of monoatomic gas $A$ having 3 translational degrees of freedom. $\gamma_{B}$ is the specific heat ratio of polyatomic gas $B$ having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If $\frac{\gamma_{A}}{\gamma_{B}}=\left(1+\frac{1}{n}\right)$ , then the value of $n$ is $\qquad$ .

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

$\gamma=1+\frac{2}{f}$

$f_{A}=3, \Rightarrow \gamma_{A}=\frac{5}{3}$

$f_{B}=3+3+2=8, \quad \gamma_{B}=1+\frac{2}{8}=\frac{5}{4}$

$\frac{\gamma_{A}}{\gamma_{B}}=\frac{4}{3}$

$\therefore n=3$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Morning Shift
Subject: Physics
Chapter: Thermodynamics
Topic: Different Thermodynamic Processes

In an adiabatic process, which of the following statements is true?

Answer: 3

Step-by-step solution

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