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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

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

Vedclass · Answer

(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 of the gas is $K.E. = \frac{1}{2} m v^2$.The change in internal energy of an ideal gas is given by $\Delta U = n C_V \Delta T$.Here,$n = \frac{m}{M}$ and $C_V = \frac{R}{\gamma - 1}$.Equating the kinetic energy to the change in internal energy: $\frac{1}{2} m v^2 = \frac{m}{M} \cdot \frac{R}{\gamma - 1} \cdot \Delta T$.Solving for $\Delta T$: $\Delta T = \frac{(\gamma - 1) M v^2}{2R}$.Thus,the temperature increases by $\frac{(\gamma - 1) M v^2}{2R} 	ext{ K}$.

$A$ 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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