$\gamma_{\mathrm{A}}$ 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 $\frac{\gamma_A}{\gamma_B}=\left(1+\frac{1}{n}\right)$, then the value of $n$ is ________ .

<p>To determine the value of $ n $, we start by considering the specific heat ratios for gases A and B.</p> <p>For gas A, which is monoatomic with 3 translational degrees of freedom:</p> <p><p>The degrees of freedom, $ f_A $, is 3.</p></p> <p><p>Therefore, $ \gamma_A = \frac{f_A + 2}{f_A} = \frac{3 + 2}{3} = \frac{5}{3} $.</p></p> <p>For gas B, which is polyatomic with 3 translational, 3 rotational degrees of freedom, and 1 vibrational mode:</p> <p><p>The total degrees of freedom, $ f_B $, is $ 3 + 3 + 2 \times 1 = 8 $ (since one vibrational mode contributes 2 degrees of freedom: 1 kinetic + 1 potential).</p></p> <p><p>Therefore, $ \gamma_B = \frac{f_B + 2}{f_B} = \frac{8 + 2}{8} = \frac{10}{8} = \frac{5}{4} $.</p></p> <p>Now, the relation given is:</p> <p>$ \frac{\gamma_A}{\gamma_B} = \left(1 + \frac{1}{n}\right) $</p> <p>Substitute the values of $ \gamma_A $ and $ \gamma_B $:</p> <p>$ \frac{\frac{5}{3}}{\frac{5}{4}} = \frac{5}{3} \times \frac{4}{5} = \frac{20}{15} = \frac{4}{3} $</p> <p>Equating to the given form:</p> <p>$ \frac{4}{3} = 1 + \frac{1}{n} $</p> <p>Thus, we have:</p> <p>$ \frac{4}{3} - 1 = \frac{1}{n} $</p> <p>$ \frac{1}{3} = \frac{1}{n} $</p> <p>Solving for $ n $:</p> <p>$ n = 3 $</p> <p>Hence, the value of $ n $ is 3.</p>