Q71 / 90 · JEE Main 2024 (09 Apr Shift 2) · Physics
The temperature of a gas is $-78^{\circ} \mathrm{C}$ and the average translational kinetic energy of its molecules is $\mathrm{K}$. The temperature at which the average translational kinetic energy of the molecules of the same gas becomes $2 \mathrm{~K}$ is :
- A$127^{\circ} \mathrm{C}$
- Correct answerB$117^{\circ} \mathrm{C}$
- C$-39^{\circ} \mathrm{C}$
- D$-78^{\circ} \mathrm{C}$
Solution
$\begin{aligned}& \mathrm{K} . \mathrm{E}=\frac{\mathrm{nf}_1 \mathrm{RT}}{2} \\& \mathrm{~T}_{\mathrm{i}}=-78^{\circ} \mathrm{C} \rightarrow 273+\left[-78^{\circ} \mathrm{C}\right]=195 \mathrm{~K}
\end{aligned}$
K.E $\alpha \mathrm{T}$
To double the K.E energy temp also become double
$\begin{aligned}& \mathrm{T}_{\mathrm{f}}=390 \mathrm{~K} \\
& \mathrm{~T}_{\mathrm{f}}=117^{\circ} \mathrm{C}\end{aligned}$
Hence, the correct answer is option: 'B'.