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 answer
    A
    $\frac{7}{5}, \frac{5}{3}, \frac{9}{7}$
  • 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'.

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