Q140 / 150 · MHT CET 2024 (11 May Shift 1) · Physics
An insulated container contains a diatomic gas of molar mass ' $m$ '. The container is moving with velocity ' V ', if it is stopped suddenly, the change in temperature is ( $\mathrm{R}=$ gas constant)
- A$\frac{m V^2}{3 R}$
- Correct answerB$\frac{\mathrm{mV}^2}{5 \mathrm{R}}$
- C$\frac{\mathrm{mV}}{7 \mathrm{R}}$
- D$\frac{5 m V}{3 R}$
Solution
Kinetic energy of gas $=n\left(\frac{1}{2} m V^2\right)$
The change in internal energy when the box is stopped suddenly is,
$\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{~T}$
For diatomic gas, $\mathrm{C}_{\mathrm{v}}=\frac{5}{2} \mathrm{R}$ As the energy is conserved, $\Delta \mathrm{U}=\mathrm{K} . \mathrm{E} .$
So, from above equations, $\begin{array}{ll} & n \frac{5}{2} R \Delta T=n\left(\frac{1}{2} \mathrm{mV}^2\right) \\ \therefore \quad & \Delta \mathrm{T}=\frac{\mathrm{mV}^2}{5 \mathrm{R}} \end{array}$
For diatomic gas, $\mathrm{C}_{\mathrm{v}}=\frac{5}{2} \mathrm{R}$ As the energy is conserved, $\Delta \mathrm{U}=\mathrm{K} . \mathrm{E} .$
So, from above equations, $\begin{array}{ll} & n \frac{5}{2} R \Delta T=n\left(\frac{1}{2} \mathrm{mV}^2\right) \\ \therefore \quad & \Delta \mathrm{T}=\frac{\mathrm{mV}^2}{5 \mathrm{R}} \end{array}$
Hence, the correct answer is option: 'B'.