JEE Mains · Physics · STD 11 - Kinetic theory of gases
- JEE Mains 2025
- Easy
- Numerical
\(\gamma_{\mathrm{A}}\) 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 \(\frac{\gamma_{\mathrm{A}}}{\gamma_{\mathrm{B}}}=\left(1+\frac{1}{\mathrm{n}}\right)\), then the value of \(n\) is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) 3
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \frac{\gamma_A}{\gamma_B}=\frac{\mathrm{f}_{\mathrm{A}}+2}{\mathrm{f}_{\mathrm{A}}} \times \frac{\mathrm{f}_{\mathrm{B}}}{\mathrm{f}_{\mathrm{B}}+2} \\ & =\frac{3+2}{3} \times \frac{(6+2)}{(6+2)+2} \\ & =\frac{5}{3} \times \frac{8}{10}=\frac{40}{30} \\ & \because \frac{40}{30}=1+\frac{1}{\mathrm{n}} \\ & \Rightarrow \frac{40}{30}-1=\frac{1}{\mathrm{n}}\end{aligned}\)
\(\Rightarrow \mathrm{n}=3\)
\(\Rightarrow \mathrm{n}=3\)
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