The temperature at which oxygen molecules hav | vedclass

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(D) The root mean square speed $(V_{rms})$ is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M}}$.Given that $V_{rms(O_2)} = V_{rms(He)}$,we equate

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$2400$

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(D) The root mean square speed $(V_{rms})$ is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M}}$.Given that $V_{rms(O_2)} = V_{rms(He)}$,we equate the expressions:$\sqrt{\frac{3RT_{O_2}}{M_{O_2}}} = \sqrt{\frac{3RT_{He}}{M_{He}}}$Squaring both sides and simplifying,we get $\frac{T_{O_2}}{M_{O_2}} = \frac{T_{He}}{M_{He}}$.Here,$M_{O_2} = 32 \ g/mol$,$M_{He} = 4 \ g/mol$,and $T_{He} = 300 \ K$.Substituting the values: $\frac{T_{O_2}}{32} = \frac{300}{4}$.$T_{O_2} = \frac{300 	imes 32}{4} = 300 	imes 8 = 2400 \ K$.

The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at $300 \ K$ is (Atomic masses: $He = 4 \ u$,$O = 16 \ u$) (in $K$)

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