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(C) The $r.m.s.$ velocity of gas molecules is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.This shows that $v_{rms} \propto \sqrt{T}$,where

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$\frac{400}{\sqrt{3}}$

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(C) The $r.m.s.$ velocity of gas molecules is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.This shows that $v_{rms} \propto \sqrt{T}$,where $T$ is the absolute temperature in Kelvin.Note that the $r.m.s.$ velocity is independent of the pressure of the gas.Given:Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \, K$.Initial $r.m.s.$ velocity $v_1 = 200 \, m/s$.Final temperature $T_2 = 127^{\circ}C = 127 + 273 = 400 \, K$.Using the proportionality $v_2 / v_1 = \sqrt{T_2 / T_1}$:$v_2 = v_1 	imes \sqrt{\frac{T_2}{T_1}}$$v_2 = 200 	imes \sqrt{\frac{400}{300}}$$v_2 = 200 	imes \sqrt{\frac{4}{3}}$$v_2 = 200 	imes \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}} \, m/s$.

The molecules of a given mass of a gas have a $r.m.s.$ velocity of $200 \, m/s$ at $27^{\circ}C$ and $1.0 \times 10^5 \, N/m^2$ pressure. When the temperature is $127^{\circ}C$ and pressure is $0.5 \times 10^5 \, N/m^2$,the $r.m.s.$ velocity in $m/s$ will be

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