Certain amount of an ideal gas is contained in a closed vessel The vessel is moving with constant velocity $v$ .The rise in temperature of the gas when the vessel is suddenly stopped is (M is molecular weight of the gas and $γ = C_{p} / C_{v}$ ) :

\frac{M v^{2} (γ - 1)}{2 R}

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\frac{M v^{2} (γ + 1)}{2 R}

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\frac{M v^{2}}{2 R γ}

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\frac{M v^{2}}{2 R (γ + 1)}

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Solution

The correct option is A $\frac{M v^{2} (γ - 1)}{2 R}$
If $m$ is the total mass of the gas, then the kinetic energy is $\frac{1}{2} m v^{2}$ .
When the vessel is suddenly stopped then total kinetic energy will increase the temperature of the
gas. Hence $\frac{1}{2} m v^{2} = n C_{V} Δ T = \frac{m}{M} C_{V} Δ T$
( $C_{V} = \frac{R}{γ - 1}$ )
$⟹ Δ T = \frac{M v^{2} (γ - 1)}{2 R}$

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Certain amount of an ideal gas is contained in a closed vessel The vessel is moving with constant velocity v .The rise in temperature of the gas when the vessel is suddenly stopped is (M is molecular weight of the gas and γ=Cp/Cv) :

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Certain amount of an ideal gas is contained in a closed vessel The vessel is moving with constant velocity $v$ .The rise in temperature of the gas when the vessel is suddenly stopped is (M is molecular weight of the gas and $γ = C_{p} / C_{v}$ ) :