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(C) The rms velocity of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}.$ Since $v_{rms} \propto \sqrt{T},$ if the velocity is doubled,the t

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(C) The rms velocity of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}.$ Since $v_{rms} \propto \sqrt{T},$ if the velocity is doubled,the temperature must increase by a factor of $4.$ Initial temperature $T_1 = 27 + 273 = 300\,K.$ Final temperature $T_2 = 4 	imes 300 = 1200\,K.$ Nitrogen $(N_2)$ is a diatomic gas,so the degrees of freedom $f = 5.$ The number of moles $n = \frac{15}{28}.$ The heat transferred at constant volume is $\Delta Q = n C_v \Delta T = n \left( \frac{f}{2}R ight) (T_2 - T_1).$ Substituting the values: $\Delta Q = \left( \frac{15}{28} ight) 	imes \left( \frac{5}{2} 	imes 8.3 ight) 	imes (1200 - 300).$ $\Delta Q = \left( \frac{15}{28} ight) 	imes 20.75 	imes 900 \approx 10000\,J = 10\,kJ.$

$A$ $15\,g$ mass of nitrogen gas is enclosed in a vessel at a temperature $27\,^{\circ}C.$ The amount of heat transferred to the gas,so that the rms velocity of the molecules is doubled,is about ...... $kJ$ [Take $R = 8.3\,J/K\,mole$ ]

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