Kinetic Theory

JEE Main level, numerical answer, medium

70 of 168

Question 70 of 168

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

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Answer: 3

γAγB=fA+2fA×fBfB+2=3+23×(6+2)(6+2)+2=53×810=4030∵4030=1+1n⇒4030−1=1n\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}
⇒n=3\Rightarrow \mathrm{n}=3

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