The ratio between RMS velocities of H_2 at 50 | vedclass

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(C) Root mean square speed $(v_{rms}) = \sqrt{\frac{3RT}{M}}$Given:$M_1 = 	ext{molar mass of } H_2 = 2 \ g/mol$$M_2 = 	ext{molar mass of } O_2 = 32

Vedclass · Accepted Answer

$1 : 1$

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(C) Root mean square speed $(v_{rms}) = \sqrt{\frac{3RT}{M}}$Given:$M_1 = 	ext{molar mass of } H_2 = 2 \ g/mol$$M_2 = 	ext{molar mass of } O_2 = 32 \ g/mol$$T_1 = 50 \ K, T_2 = 800 \ K$Taking the ratio of $v_{rms,1}$ and $v_{rms,2}$:$\frac{v_{rms,1}}{v_{rms,2}} = \sqrt{\frac{3RT_1}{M_1}} 	imes \sqrt{\frac{M_2}{3RT_2}} = \sqrt{\frac{M_2 	imes T_1}{M_1 	imes T_2}}$$\frac{v_{rms,1}}{v_{rms,2}} = \sqrt{\frac{32}{2} 	imes \frac{50}{800}} = \sqrt{16 	imes \frac{1}{16}} = \sqrt{1} = 1$Thus,the ratio is $1 : 1$.Hence,option $(C)$ is correct.

The ratio between $RMS$ velocities of $H_2$ at $50 \ K$ and $O_2$ at $800 \ K$ is

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