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(C) Since the vessel is thermally insulated,no heat is exchanged with the surroundings.The kinetic energy of the gas is converted into internal energy

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Since the vessel is thermally insulated,no heat is exchanged with the surroundings.The kinetic energy of the gas is converted into internal energy of the gas when it is suddenly brought to rest.Loss in kinetic energy = Gain in internal energy$\frac{1}{2} m v^2 = n C_V \Delta T$Here,$n = \frac{m}{M}$ is the number of moles and $C_V = \frac{R}{\gamma - 1}$ is the molar specific heat at constant volume.Substituting these values:$\frac{1}{2} m v^2 = \left( \frac{m}{M} \right) \left( \frac{R}{\gamma - 1} \right) \Delta T$Canceling $m$ from both sides:$\frac{1}{2} v^2 = \frac{R \Delta T}{M(\gamma - 1)}$Solving for $\Delta T$:$\Delta T = \frac{M v^2 (\gamma - 1)}{2R} \ K$

$A$ 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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