Q 11-12-055JEE MainJEE Main 2024 (9 Apr, Shift 2)Easy
The temperature of a gas is $-78\,^\circ\text{C}$ and the average translational kinetic energy of its molecules is $K$. The temperature at which the average translational kinetic energy of the molecules of the same gas becomes $2K$ is:
Answer: (B) $117\,^\circ\text{C}$
Average translational kinetic energy $\frac32k_BT$ is proportional to the absolute temperature.
$T_1 = -78 + 273 = 195\ \text{K}$, so $T_2 = 2\times195 = 390\ \text{K} = 117\,^\circ\text{C}$.
Solution by Sreeraj P, M.Sc Physics