Question

$\gamma_A$ 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 \underline{}.

Hint:

For problems involving specific heat ratios, break down the equation and substitute values for the translational, rotational, and vibrational degrees of freedom separately. This will help you arrive at the correct value of $n$.

Answer 1

Correct Answer: 3

\[ \frac{\gamma_A}{\gamma_B} = \frac{f_A + 2}{f_A} : \text{for monoatomic gas A} \] \[ \frac{\gamma_B}{\gamma_B} = \frac{f_B + 2}{f_B} : \text{for polyatomic gas B} \] For monoatomic gas A: \[ f_A = 3 \quad \text{(translational degrees of freedom)} \] For polyatomic gas B: \[ f_B = 3 + 3 + 1 = 7 \quad \text{(translational, rotational, and vibrational modes)} \] Substituting these values into the formula: \[ \frac{\gamma_A}{\gamma_B} = \frac{3+2}{3} : \frac{7+2}{7} = \frac{5}{3} : \frac{9}{7} \] \[ \frac{5}{3} : \frac{9}{7} = \left( 1 + \frac{1}{n} \right) \] \[ \frac{5}{3} \cdot \frac{7}{9} = 1 + \frac{1}{n} \] \[ \frac{35}{27} = 1 + \frac{1}{n} \] \[ \frac{35}{27} - 1 = \frac{1}{n} \] \[ \frac{8}{27} = \frac{1}{n} \Rightarrow n = 3 \]