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(B) The kinetic energy of the gas is converted into internal energy of the gas when the vessel is suddenly brought to rest.Let $m$ be the mass of the

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$\frac{(\gamma - 1)}{2R} M v^2 	ext{ K}$

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(B) The kinetic energy of the gas is converted into internal energy of the gas when the vessel is suddenly brought to rest.Let $m$ be the mass of the gas. The kinetic energy is $K.E. = \frac{1}{2} m v^2$.The change in internal energy is given by $\Delta U = n C_v \Delta T$,where $n = \frac{m}{M}$ is the number of moles and $C_v = \frac{R}{\gamma - 1}$.Equating the kinetic energy to the change in internal energy:$\frac{1}{2} m v^2 = \frac{m}{M} \left( \frac{R}{\gamma - 1} \right) \Delta T$Canceling $m$ from both sides:$\frac{1}{2} v^2 = \frac{R}{M(\gamma - 1)} \Delta T$Solving for $\Delta T$:$\Delta T = \frac{(\gamma - 1) M v^2}{2R}$

$A$ thermally insulated vessel contains an ideal gas of molecular mass $M$ and ratio of specific heats $\gamma$. It is moving with speed $v$ and is suddenly brought to rest. Assuming no heat is lost to the surroundings,its temperature increases by

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