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A vessel contains 14 g of nitrogen gas at a temperature of $27^{0} C$ . The amount of heat to be transferred to the gas to double the r.m.s. speed of its molecules will be: (Take $R = 8.32 J {mol}^{- 1} K^{- 1}$ )

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2229 J

5616 J

9360 J

13,104 J

answer is C.

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Detailed Solution

$v_{rms} \propto \sqrt{T} v_{rms} \propto \sqrt{300 K}, v_{{rms}_{f}} = 2 v_{{rms}_{i}} T_{f} = 1200 K, T_{i} = 300 K, n = \frac{14}{28} = \frac{1}{2} Q = {nC}_{v} ∆ T = \frac{1}{2} \times \frac{5 R}{2} \times 900 Q = 9360 J$

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