MHT CET · Physics · Kinetic Theory of Gases
An insulated container contains a monoatomic 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 \(\frac{\mathrm{mV}^2}{5 \mathrm{R}}\)
- B \(\frac{\mathrm{mV}^2}{3 \mathrm{R}}\)
- C \(\frac{\mathrm{mV}^2}{7 \mathrm{R}}\)
- D \(\frac{\mathrm{mV}^2}{9 \mathrm{R}}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{mV}^2}{3 \mathrm{R}}\)
Step-by-step Solution
Detailed explanation
If the container stops suddenly loss in kinetic energy of gas \(=\frac{1}{2}(\mathrm{mn}) \mathrm{V}^2\)
For monoatomic gas, the change in internal energy of the gas is given by,
\(\Delta \mathrm{U}=\frac{3}{2} \mathrm{nR} \Delta \mathrm{~T}\)
The change in kinetic energy is converted into internal energy,
\(\begin{aligned}
& \therefore \quad \frac{3}{2} n R \Delta T=\frac{1}{2} \mathrm{mnV}^2 \\
& \therefore \quad \Delta \mathrm{~T}=\frac{\mathrm{mV}^2}{3 \mathrm{R}}
\end{aligned}\)
For monoatomic gas, the change in internal energy of the gas is given by,
\(\Delta \mathrm{U}=\frac{3}{2} \mathrm{nR} \Delta \mathrm{~T}\)
The change in kinetic energy is converted into internal energy,
\(\begin{aligned}
& \therefore \quad \frac{3}{2} n R \Delta T=\frac{1}{2} \mathrm{mnV}^2 \\
& \therefore \quad \Delta \mathrm{~T}=\frac{\mathrm{mV}^2}{3 \mathrm{R}}
\end{aligned}\)
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