MHT CET · Physics · Kinetic Theory of Gases
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 \(\frac{m V^2}{3 R}\)
- B \(\frac{\mathrm{mV}^2}{5 \mathrm{R}}\)
- C \(\frac{\mathrm{mV}}{7 \mathrm{R}}\)
- D \(\frac{5 m V}{3 R}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{mV}^2}{5 \mathrm{R}}\)
Step-by-step Solution
Detailed explanation
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,
\(\begin{array}{ll}
& n \frac{5}{2} R \Delta T=n\left(\frac{1}{2} \mathrm{mV}^2\right) \\
\therefore \quad & \Delta \mathrm{T}=\frac{\mathrm{mV}^2}{5 \mathrm{R}}
\end{array}\)
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,
\(\begin{array}{ll}
& n \frac{5}{2} R \Delta T=n\left(\frac{1}{2} \mathrm{mV}^2\right) \\
\therefore \quad & \Delta \mathrm{T}=\frac{\mathrm{mV}^2}{5 \mathrm{R}}
\end{array}\)
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