A vessel contains 14 g of nitrogen gas at a temperature of $$27^\circ C$$. 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$$^{-1}$$)

We have 14 g of nitrogen gas ($$N_2$$, molecular mass = 28 g/mol), so the number of moles is $$n = \frac{14}{28} = 0.5$$ mol. The initial temperature is $$T_1 = 27^\circ C = 300$$ K.

The r.m.s. speed of gas molecules is given by $$v_{rms} = \sqrt{\frac{3RT}{M}}$$. Since $$v_{rms} \propto \sqrt{T}$$, to double the r.m.s. speed, we need the temperature to become 4 times the original. So the final temperature is $$T_2 = 4 \times 300 = 1200$$ K.

The change in temperature is $$\Delta T = 1200 - 300 = 900$$ K.

Nitrogen is a diatomic gas, so the molar heat capacity at constant volume is $$C_v = \frac{5}{2}R$$. The heat transferred (at constant volume, since the vessel is rigid) is:

$$Q = n C_v \Delta T = 0.5 \times \frac{5}{2} \times 8.32 \times 900$$

$$Q = 0.5 \times 2.5 \times 8.32 \times 900 = 1.25 \times 8.32 \times 900 = 10.4 \times 900 = 9360 \text{ J}$$

Hence, the correct answer is Option C.