The rms velocity of oxygen molecules at 27^{c | vedclass

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(B) The formula for rms velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.For oxygen $(O_2)$,$T_1 = 27 + 273 = 300 \ K$ and $M_1 = 32 \ g/mol$. Given $v_{r

Vedclass · Accepted Answer

$1600$

Vedclass · Answer

(B) The formula for rms velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.For oxygen $(O_2)$,$T_1 = 27 + 273 = 300 \ K$ and $M_1 = 32 \ g/mol$. Given $v_{rms,1} = 800 \ m/s$.For methane $(CH_4)$,$T_2 = 600 \ K$ and $M_2 = 16 \ g/mol$. Let $v_{rms,2} = x$.Taking the ratio: $\frac{v_{rms,1}}{v_{rms,2}} = \sqrt{\frac{T_1}{M_1} 	imes \frac{M_2}{T_2}}$.Substituting the values: $\frac{800}{x} = \sqrt{\frac{300}{32} 	imes \frac{16}{600}} = \sqrt{\frac{1}{2} 	imes \frac{1}{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Thus,$x = 800 	imes 2 = 1600 \ m/s$.Therefore,the correct option is $B$.

The rms velocity of oxygen molecules at $27^{\circ} C$ is around $800 \ m/s$. The rms velocity of methane molecules at $600 \ K$ temperature is around ...... (in $m/s$)

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