$$0.056$$ kg of Nitrogen is enclosed in a vessel at a temperature of $$127°$$C. The amount of heat required to double the speed of its molecules is ______ kcal.
(Take $$R = 2$$ cal mole$$^{-1}$$ K$$^{-1}$$)

We have 0.056 kg = 56 g of Nitrogen ($$N_2$$). The molar mass of $$N_2$$ is 28 g/mol.

Calculate the number of moles:: $$n = \frac{56}{28} = 2 \text{ moles}$$

We need to double the speed of the molecules. Since the rms speed is given by $$v_{rms} = \sqrt{\frac{3RT}{M}}$$, speed is proportional to $$\sqrt{T}$$. To double the speed:: $$\frac{v_2}{v_1} = 2 = \sqrt{\frac{T_2}{T_1}}$$

$$\frac{T_2}{T_1} = 4$$

$$T_2 = 4 \times 400 = 1600 \text{ K}$$

The change in temperature is:: $$\Delta T = T_2 - T_1 = 1600 - 400 = 1200 \text{ K}$$

$$N_2$$ is a diatomic gas, so $$C_v = \frac{5}{2}R$$. The heat required at constant volume is:: $$Q = n C_v \Delta T = n \times \frac{5}{2}R \times \Delta T$$

$$Q = 2 \times \frac{5}{2} \times 2 \times 1200$$

$$Q = 2 \times 5 \times 1200 = 12000 \text{ cal} = 12 \text{ kcal}$$

The answer is 12 kcal.