The rms speed of an oxygen molecule in a vess | vedclass

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Question

(A) The root mean square (rms) speed of a gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.The average speed of a gas molecule is given by $v

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(A) The root mean square (rms) speed of a gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.The average speed of a gas molecule is given by $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.According to the problem,$v_{rms} = \left(1+\frac{5}{x}ight)^{\frac{1}{2}} v_{avg}$.Substituting the formulas: $\sqrt{\frac{3RT}{M}} = \left(1+\frac{5}{x}ight)^{\frac{1}{2}} \sqrt{\frac{8RT}{\pi M}}$.Squaring both sides: $\frac{3RT}{M} = \left(1+\frac{5}{x}ight) \frac{8RT}{\pi M}$.Canceling $\frac{RT}{M}$ from both sides: $3 = \left(1+\frac{5}{x}ight) \frac{8}{\pi}$.Given $\pi = \frac{22}{7}$,we have $3 = \left(1+\frac{5}{x}ight) \frac{8}{22/7} = \left(1+\frac{5}{x}ight) \frac{8 	imes 7}{22} = \left(1+\frac{5}{x}ight) \frac{56}{22} = \left(1+\frac{5}{x}ight) \frac{28}{11}$.$\Rightarrow 1+\frac{5}{x} = 3 	imes \frac{11}{28} = \frac{33}{28}$.$\Rightarrow \frac{5}{x} = \frac{33}{28} - 1 = \frac{5}{28}$.$\Rightarrow x = 28$.

The rms speed of an oxygen molecule in a vessel at a particular temperature is $\left(1+\frac{5}{x}\right)^{\frac{1}{2}} v$,where $v$ is the average speed of the molecule. The value of $x$ will be: (Take $\pi=\frac{22}{7}$)

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