Q103 / 180 · NEET 2019 (Odisha) · Physics
The value of $\gamma\left(=\frac{C_p}{C_V}\right)$ for hydrogen, helium and another ideal diatomic gas $\mathrm{X}$ (whose molecules are not rigid but have an additional vibrational mode), are respectively equal to
- Correct answerA$\frac{7}{5}, \frac{5}{3}, \frac{9}{7}$
- B$\frac{5}{3}, \frac{7}{5}, \frac{9}{7}$
- C$\frac{5}{3}, \frac{7}{5} \frac{7}{5}$
- D$\frac{7}{5}, \frac{5}{3}, \frac{7}{5}$
Solution
The poisson's ratio,
$$
\gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}
$$
where, $C_p=$ molar heat capacity constant pressure and $\mathrm{C}_{\mathrm{V}}=$ molar heat capacity at constant volume Also, $C_p=C_V+R$ (from Mayer's relation) $C_V=\frac{f}{2} \mathrm{R} \quad$ (where, $\mathrm{f}=$ degree of freedom)
$$
\Rightarrow \quad \mathrm{C}_{\mathrm{p}}=\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R}
$$
So, Eq. (i) becomes,
$$
\Rightarrow \quad \gamma=1+\frac{2}{f}
$$
For hydrogen gas, which is diatomic, the degree of freedom is 5 ( 3 translational, 2 rotational).
$$
\therefore \quad \gamma=1+\frac{2}{5}=\frac{7}{5}
$$
For helium gas, which is monoatomic, the degree of freedom is 3 ( 3 translational only).
$$
\therefore \quad \gamma=1+\frac{2}{3}=\frac{5}{3}
$$
The diatomic gas $\mathrm{X}$ also have vibrational motion, so degree of freedom is 7 ( 3 translational, 2 rotational and 2 vibrational).
$$
\therefore \quad \gamma=1+\frac{2}{7}=\frac{9}{7}
$$
Hence, the correct answer is option: 'A'.