Q 11-12-138JEE MainJEE Main 2019 (10 Apr, Shift 2)Medium
One mole of an ideal gas passes through a process where pressure and volume obey the relation $P = P_0\left[1 - \dfrac12\left(\dfrac{V_0}{V}\right)^2\right]$. Here $P_0$ and $V_0$ are constants. Calculate the change in the temperature of the gas if its volume changes from $V_0$ to $2V_0$.
Answer: (B) $\dfrac54\dfrac{P_0V_0}{R}$
For one mole $T = \dfrac{PV}{R} = \dfrac{P_0}{R}\left(V - \dfrac{V_0^2}{2V}\right)$.
At $V_0$: $T_1 = \dfrac{P_0}{R}\cdot\dfrac{V_0}{2}$. At $2V_0$: $T_2 = \dfrac{P_0}{R}\left(2V_0 - \dfrac{V_0}{4}\right) = \dfrac{7P_0V_0}{4R}$.
$$\Delta T = \frac{7}{4}\frac{P_0V_0}{R} - \frac12\frac{P_0V_0}{R} = \frac54\frac{P_0V_0}{R}$$
Solution by Sreeraj P, M.Sc Physics