At 256 K,the rms speed of SO_2 gas molecules | vedclass

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(B) The relationship between $rms$ speed $(v_{rms})$ and most probable velocity $(v_{mp})$ is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$ a

Vedclass · Accepted Answer

$2.58 	imes 10^2$

Vedclass · Answer

(B) The relationship between $rms$ speed $(v_{rms})$ and most probable velocity $(v_{mp})$ is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$ and $v_{mp} = \sqrt{\frac{2RT}{M}}$.Dividing the two expressions,we get: $\frac{v_{mp}}{v_{rms}} = \sqrt{\frac{2}{3}}$.Given $v_{rms} = 3.16 	imes 10^2 \ ms^{-1}$,we calculate $v_{mp} = v_{rms} 	imes \sqrt{\frac{2}{3}}$.$v_{mp} = 3.16 	imes 10^2 	imes \sqrt{0.6667} \approx 3.16 	imes 10^2 	imes 0.8165$.$v_{mp} \approx 2.58 	imes 10^2 \ ms^{-1}$.

At $256 \ K$,the $rms$ speed of $SO_2$ gas molecules is $3.16 \times 10^2 \ ms^{-1}$. What is the most probable velocity (in $ms^{-1}$) of the same gas at the same temperature?

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