If V_{rms} for O_2 molecules is V at temperat | vedclass

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(B) The root mean square velocity is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M_w}}$,where $R$ is the universal gas constant,$T$ is the temper

Vedclass · Accepted Answer

$2V$

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(B) The root mean square velocity is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M_w}}$,where $R$ is the universal gas constant,$T$ is the temperature,and $M_w$ is the molar mass.From this,we can see that $V_{rms} \propto \sqrt{\frac{T}{M_w}}$.For $O_2$ molecules,the molar mass is $M_w = 32 \ g/mol$. Given $V_{rms} = V$ at temperature $T$.For oxygen atoms $(O)$,the molar mass is $M_w' = 16 \ g/mol$. The new temperature is $T' = 2T$.Now,calculating the new $V_{rms}'$:$V_{rms}' = V 	imes \sqrt{\frac{T'}{T} 	imes \frac{M_w}{M_w'}}$$V_{rms}' = V 	imes \sqrt{\frac{2T}{T} 	imes \frac{32}{16}}$$V_{rms}' = V 	imes \sqrt{2 	imes 2} = V 	imes \sqrt{4} = 2V$.

If $V_{rms}$ for $O_2$ molecules is $V$ at temperature $T \ K$. Then $V_{rms}$ for oxygen atoms at temperature $2T$ will be

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