Let (C_{rms})_{H_2} be the r.m.s. speed of H_ | vedclass

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(B) The formula for r.m.s. speed is $(C_{rms}) = \sqrt{\frac{3RT}{M}}$ and for most probable speed is $(C_{mp}) = \sqrt{\frac{2RT}{M}}$.Given $(C_{rms

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$112.5$

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(B) The formula for r.m.s. speed is $(C_{rms}) = \sqrt{\frac{3RT}{M}}$ and for most probable speed is $(C_{mp}) = \sqrt{\frac{2RT}{M}}$.Given $(C_{rms})_{H_2} = \sqrt{\frac{3 	imes R 	imes 150}{2}}$.According to the problem, $(C_{mp})_{He} = \frac{1}{2} (C_{rms})_{H_2}$.Substituting the values: $\sqrt{\frac{2RT}{4}} = \frac{1}{2} \sqrt{\frac{3 	imes R 	imes 150}{2}}$.Squaring both sides: $\frac{2RT}{4} = \frac{1}{4} 	imes \frac{3 	imes R 	imes 150}{2}$.$\frac{RT}{2} = \frac{450R}{8}$.$T = \frac{450}{4} = 112.5 \ K$.

Let $(C_{rms})_{H_2}$ be the r.m.s. speed of $H_2$ at $150 \ K$. At what temperature will the most probable speed of helium $[(C_{mp})_{He}]$ be half of $(C_{rms})_{H_2}$ (in $K$)?

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