Question 573: Thermodynamics & KTG

Question

$\gamma_{\mathrm{A}}$ 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 ________ .

Accepted Answer

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

Thermodynamics & KTG

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