$$\gamma_A$$ 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 ______.

For any ideal gas, the molar specific heats at constant volume and at constant pressure are given by the equipartition theorem.

If a molecule has $$f$$ degrees of freedom, then
$$C_V = \frac{f}{2}\,R$$
$$C_P = C_V + R = \frac{f+2}{2}\,R$$
Therefore, the specific-heat ratio is
$$\gamma = \frac{C_P}{C_V} = \frac{f+2}{f}$$ $$-(1)$$

Case A: Mono-atomic gas A

Such a molecule can translate along the three axes only, so
$$f_A = 3$$

Using $$(1)$$,
$$\gamma_A = \frac{3+2}{3} = \frac{5}{3}$$ $$-(2)$$

Case B: Poly-atomic gas B

Given data for one molecule of B:
• Translational degrees = 3
• Rotational degrees = 3
• Vibrational modes = 1

A single vibrational mode contributes two degrees of freedom (one kinetic + one potential). Hence
$$f_{\text{vib}} = 2$$

Total degrees of freedom for B:
$$f_B = 3 + 3 + 2 = 8$$

Using $$(1)$$ with $$f_B = 8$$,
$$\gamma_B = \frac{8+2}{8} = \frac{10}{8} = \frac{5}{4}$$ $$-(3)$$

The question states that
$$\frac{\gamma_A}{\gamma_B} = 1 + \frac{1}{n}$$ $$-(4)$$

Substitute $$(2)$$ and $$(3)$$ into $$(4)$$:
$$\frac{\,\frac{5}{3}\,}{\,\frac{5}{4}\,} = 1 + \frac{1}{n}$$
$$\frac{5}{3} \times \frac{4}{5} = 1 + \frac{1}{n}$$
$$\frac{4}{3} = 1 + \frac{1}{n}$$

Isolate $$\frac{1}{n}$$:
$$\frac{1}{n} = \frac{4}{3} - 1 = \frac{1}{3}$$
$$n = 3$$

Hence, the required value of $$n$$ is $$3$$.