Q 11-11-135JEE MainJEE Main 2020 (5 Sep, Shift 2)Easy
In an adiabatic process, the density of a diatomic gas becomes $32$ times its initial value. The final pressure of the gas is found to be $n$ times the initial pressure. The value of $n$ is:
Answer: (C) $128$
For an adiabatic process $PV^\gamma$ = constant, and $V \propto 1/\rho$, so $P \propto \rho^\gamma$.
For a diatomic gas $\gamma = \tfrac75$:
$$n = 32^{7/5} = (2^5)^{7/5} = 2^7 = 128$$
Solution by Sreeraj P, M.Sc Physics