Question:
A thermally insulated vessel contains an ideal gas of molecular weight M and specific heat ratio $\gamma$. If it is moving with speed V and suddenly brought to rest, the temperature increase is:
A thermally insulated vessel contains an ideal gas of molecular weight M and specific heat ratio $\gamma$. If it is moving with speed V and suddenly brought to rest, the temperature increase is:
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Conversion of bulk kinetic energy to random molecular motion increases internal energy.
Updated On: Jun 10, 2026
- $\left[\frac{(\gamma+1)MV^2}{2(\gamma+2)R}\right] K$
- $\left[\frac{(\gamma-1)MV^2}{2\gamma R}\right] K$
- $\left[\frac{\gamma MV^2}{2R}\right] K$
- $\left[\frac{(\gamma-1)MV^2}{2R}\right] K$
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The Correct Option is D
Solution and Explanation
Step 1: Concept
Kinetic energy of the gas converts to internal energy: $\frac{1}{2}MV^2 = n C_v \Delta T$.
Step 2: Analysis
$C_v = \frac{R}{\gamma-1}$ and $n = 1$ (for molar mass $M$). $\frac{1}{2}MV^2 = \frac{R}{\gamma-1} \Delta T \implies \Delta T = \frac{(\gamma-1)MV^2}{2R}$.
Step 3: Conclusion
The increase in temperature is $\frac{(\gamma-1)MV^2}{2R}$.
Final Answer: (D)
Kinetic energy of the gas converts to internal energy: $\frac{1}{2}MV^2 = n C_v \Delta T$.
Step 2: Analysis
$C_v = \frac{R}{\gamma-1}$ and $n = 1$ (for molar mass $M$). $\frac{1}{2}MV^2 = \frac{R}{\gamma-1} \Delta T \implies \Delta T = \frac{(\gamma-1)MV^2}{2R}$.
Step 3: Conclusion
The increase in temperature is $\frac{(\gamma-1)MV^2}{2R}$.
Final Answer: (D)
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