A box of negligible mass containing 2 moles of an ideal gas of molar mass M and adiabatic exponent $γ$ , moves with constant speed V on a smooth horizontal surface. If the box suddenly stops, then change in temperature of gas will be

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$\frac{(γ - 1) {MV}^{2}}{2 R}$

$\frac{(γ - 1) {MV}^{2}}{4 R}$

$\frac{{γMV}^{2}}{2 R}$

$\frac{{MV}^{2}}{2 (γ - 1) R}$

answer is D.

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Detailed Solution

$\begin{array}{l} ΔU = \frac{1}{2} {mV}^{2} \\ {nc}_{v} ΔT = \frac{1}{2} {mv}^{2} \\ \frac{m}{M} \frac{R}{γ - 1} ΔT = \frac{1}{2} {mv}^{2} \\ ΔT = \frac{(γ - 1) {MV}^{2}}{2 R} \end{array}$

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