MHT CET · Physics · Thermodynamics
An insulated container contains a monoatomic gas of molar mass ' \(\mathrm{m}\) '. The container is moving with velocity ' \(\mathrm{V}\) '. If it is stopped suddenly, the change in temperature of a gas is \([R\) is gas constant]
- A \(\frac{\mathrm{MV}^2}{\mathrm{R}}\)
- B \(\frac{\mathrm{MV}^2}{2 \mathrm{R}}\)
- C \(\frac{\mathrm{MV}^2}{3 \mathrm{R}}\)
- D \(\frac{3 \mathrm{MV}^2}{2 \mathrm{R}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{MV}^2}{3 \mathrm{R}}\)
Step-by-step Solution
Detailed explanation
Kinetic energy toss of the gas is
\(\Delta \mathrm{E}=\frac{1}{2} \mathrm{MV}^2, \mathrm{n}\)...(i)
(where \(\mathrm{n}\) is the no. of moles of the gas) Heat gained by the gás due to temperature change \(\Delta \mathrm{T}\) is
\(\Delta Q=n C V \Delta T\)
But, \(\mathrm{C}_{\mathrm{V}}=\frac{3}{2} \mathrm{R}\) …(gas is monoatomic)
\(\therefore \quad \Delta Q=\frac{3}{2} R \cdot n \Delta T\)...(ii)
Equating (i) and (ii),
\(\begin{aligned}
& \Delta \mathrm{E}=\Delta \mathrm{Q} \\
& \frac{1}{2} \mathrm{MV}^2 \cdot \mathrm{n}=\frac{3}{2} \mathrm{R} \cdot \mathrm{n} \Delta \mathrm{T} \\
& \Delta \mathrm{T}=\frac{\mathrm{MV}^2}{3 \mathrm{R}}
\end{aligned}\)
\(\Delta \mathrm{E}=\frac{1}{2} \mathrm{MV}^2, \mathrm{n}\)...(i)
(where \(\mathrm{n}\) is the no. of moles of the gas) Heat gained by the gás due to temperature change \(\Delta \mathrm{T}\) is
\(\Delta Q=n C V \Delta T\)
But, \(\mathrm{C}_{\mathrm{V}}=\frac{3}{2} \mathrm{R}\) …(gas is monoatomic)
\(\therefore \quad \Delta Q=\frac{3}{2} R \cdot n \Delta T\)...(ii)
Equating (i) and (ii),
\(\begin{aligned}
& \Delta \mathrm{E}=\Delta \mathrm{Q} \\
& \frac{1}{2} \mathrm{MV}^2 \cdot \mathrm{n}=\frac{3}{2} \mathrm{R} \cdot \mathrm{n} \Delta \mathrm{T} \\
& \Delta \mathrm{T}=\frac{\mathrm{MV}^2}{3 \mathrm{R}}
\end{aligned}\)
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