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 Certain amounts of an ideal gas are contained in a closed vessel. The vessel is moving with a constant velocity v. The molecular mass of gas is M. The rise in temperature of the gas when the vessel is suddenly stopped is -

Option: 1

\frac{M_{r}^{2}}{2 R(r+1)}


Option: 2

\frac{M v^{2}}{R}


Option: 3

\frac{M v^{2}}{R}


Option: 4

None of these


Answers (1)

best_answer

If m is the total mass of the gas then its kinetic energy =\frac{1}{2}Mv^2 

When the vessel is suddenly stopped the total kinetic energy will increase the temperature of the gas. 

\frac{1}{2}Mv^2=\mu C_v \Delta T=\frac{m}{\mu }C_v\Delta T

[C_v=\frac{R}{\gamma -1}]

\begin{gathered} \Rightarrow \frac{m}{M} \frac{R}{\gamma-1} \Delta T=\frac{1}{2} m v^{2} \\ \Rightarrow \Delta T=\frac{M v^{2}(\gamma-1)}{2 R} \end{gathered}

Posted by

Rishabh

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