Q 11-11-154JEE MainJEE Main 2019 (11 Jan, Shift 1)Easy
A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is $TV^x = $ constant, then $x$ is:
Answer: (B) $\dfrac25$
For an adiabatic process $TV^{\gamma - 1} = $ constant. A rigid diatomic gas has 5 degrees of freedom, so $\gamma = \dfrac75$ and
$$x = \gamma - 1 = \frac25$$
Solution by Sreeraj P, M.Sc Physics