A polyatomic gas follows a law $T^2 V^\alpha=$ constant. Find $\alpha$ for which the heat exchange of gas in the process becomes zero.

• A

  $\alpha=\frac{3}{2}$
• Correct answer
  B

  $\alpha=\frac{2}{3}$
• C

  $\alpha=\frac{4}{3}$
• D

  $\alpha=\frac{3}{4}$

Solution

The process in which there is no heat exchange of gas, it is called adiabatic process. In this process, $T V^{\gamma-1}=k$ [constant $]$
Comparing Eqs. (i) and (ii), we get $$ \alpha=2(\gamma-1) $$ Using $\gamma=\frac{4}{3}$, for polyatomic gas $$ \alpha=2\left(\frac{4}{3}-1\right)=\frac{2}{3} $$

Hence, the correct answer is option: 'B'.