An insulated box containing a diatomic gas of molar mass M is moving with a velocity v. The box is suddenly stopped. The resulting change in temperature is (R is the gas constant)

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$\frac{M v^{2}}{2 R}$

$\frac{M v^{2}}{3 R}$

$\frac{2 M v^{2}}{5 R}$

$\frac{M v^{2}}{5 R}$

answer is C.

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Detailed Solution

Let n be the number of moles of the gas in the box. The kinetic energy of the gas = n $(\frac{1}{2} M v^{2})$ . When the box is suddenly stopped, this energy is used up in changing the internal energy, as a result of which the temperature of the gas rises. The change in internal energy is given by

$∆ U = n C_{v} ∆ T$

For a diatomic gas $C_{v} = \frac{5}{2} R .$

Hence $n \times \frac{5 R}{2} \times ∆ T = n \times \frac{1}{2} M v^{2}$