Certain amount of an ideal gas is contained in a closed vessel. The vessel is moving with a constant velcity $v$ . The molecular mass of gas is M. The rise in temperature of the gas when the vessel is suddenly stopped is $γ = C_{P} / C_{V}$

M v^{2} (γ - 1) / 2 R (γ + 1)

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M v^{2} (γ - 1) / 2 R

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M v^{2} / 2 R (γ + 1)

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M v^{2} / 2 R (γ - 1)

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Solution

The correct option is B $M v^{2} (γ - 1) / 2 R$
If $m$ is the mass of the gas, then its kinetic energy
$= \frac{1}{2} m v^{2}$

When the vessel is suddenly stopped, total kinetic energy will increase the temperature of the gas (because process will be adiabatic), i.e.,
$= \frac{1}{2} m v^{2} = n C_{v} Δ T$
$= \frac{m}{M} C_{v} Δ T$
$⟹ \frac{m}{M} \frac{R}{(γ - 1)} Δ T = \frac{1}{2} m v^{2}$
$(A s C_{v} = \frac{R}{(γ - 1)})$

$⟹ Δ T = \frac{M v^{2} (γ - 1)}{2 R}$

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Certain amount of an ideal gas is contained in a closed vessel. The vessel is moving with a constant velcity v. The molecular mass of gas is M. The rise in temperature of the gas when the vessel is suddenly stopped is γ=CP/CV

Certain amount of an ideal gas is contained in a closed vessel. The vessel is moving with a constant velcity $v$ . The molecular mass of gas is M. The rise in temperature of the gas when the vessel is suddenly stopped is $γ = C_{P} / C_{V}$