If the rms speed of the molecules of a gas at | vedclass

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(C) The root mean square (rms) speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.This implies that $v_{rms} \propto \sqr

Vedclass · Accepted Answer

$55$

Vedclass · Answer

(C) The root mean square (rms) speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.This implies that $v_{rms} \propto \sqrt{T}$, where $T$ is the absolute temperature in Kelvin.Given $T_1 = 77^{\circ} C = 77 + 273 = 350 \,K$ and $v_1 = 50 \,ms^{-1}$.Given $T_2 = 150.5^{\circ} C = 150.5 + 273 = 423.5 \,K$.Using the ratio $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$, we get:$v_2 = v_1 	imes \sqrt{\frac{T_2}{T_1}} = 50 	imes \sqrt{\frac{423.5}{350}}$.$v_2 = 50 	imes \sqrt{1.21} = 50 	imes 1.1 = 55 \,ms^{-1}$.Thus, the correct option is $C$.

If the rms speed of the molecules of a gas at a temperature of $77^{\circ} C$ is $50 \,ms^{-1}$, then the rms speed of the same gas molecules at a temperature of $150.5^{\circ} C$ is (in $\,ms^{-1}$)

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