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(B) Let $m$ be the total mass of the gas. The kinetic energy of the gas due to the motion of the vessel is $K = \frac{1}{2}mv^2$.When the vessel is su

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$\frac{Mv^2(\gamma - 1)}{2R}$

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(B) Let $m$ be the total mass of the gas. The kinetic energy of the gas due to the motion of the vessel is $K = \frac{1}{2}mv^2$.When the vessel is suddenly stopped,this kinetic energy is converted into the internal energy of the gas,causing a rise in temperature $\Delta T$.Using the relation for internal energy change,$\Delta U = \mu C_V \Delta T$,where $\mu = \frac{m}{M}$ is the number of moles.Equating the kinetic energy to the change in internal energy: $\frac{1}{2}mv^2 = \frac{m}{M} C_V \Delta T$.Since $C_V = \frac{R}{\gamma - 1}$,we substitute this into the equation:$\frac{1}{2}mv^2 = \frac{m}{M} \left( \frac{R}{\gamma - 1} \right) \Delta T$.Canceling $m$ from both sides and solving for $\Delta T$:$\Delta T = \frac{Mv^2(\gamma - 1)}{2R}$.

$A$ certain amount of an ideal gas is contained in a closed vessel. The vessel is moving with a constant velocity $v$. The molecular mass of the gas is $M$. What is the rise in temperature of the gas when the vessel is suddenly stopped? (Given $\gamma = C_P/C_V$)

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