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

Vedclass · Accepted Answer

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

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(B) The kinetic energy of the vessel is converted into the internal energy of the gas when it is suddenly brought to rest.Let $m$ be the mass of the gas. The kinetic energy is $K = \frac{1}{2} m v^2$.This energy increases the internal energy of the gas: $\Delta U = n C_v \Delta T$.Since $n = \frac{m}{M}$ and $C_v = \frac{R}{\gamma - 1}$,we have:$\frac{1}{2} m v^2 = \frac{m}{M} \left( \frac{R}{\gamma - 1} \right) \Delta T$.Canceling $m$ from both sides:$\frac{1}{2} v^2 = \frac{R}{M(\gamma - 1)} \Delta T$.Solving for $\Delta T$:$\Delta T = \frac{M(\gamma - 1) v^2}{2R}$.

$A$ 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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