The rms speed of helium gas at 27,^{circ}C an | vedclass

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(B) The root mean square $(rms)$ speed of a gas molecule is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas const

Vedclass · Accepted Answer

$900$

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(B) The root mean square $(rms)$ speed of a gas molecule is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.From this formula,it is evident that $v_{rms}$ depends only on the temperature $T$ and the molar mass $M$ of the gas.It is independent of the pressure $P$ of the gas.Since the temperature remains constant at $27\,^{\circ}C$ $(300\,K)$ and the gas remains helium,the $rms$ speed will remain unchanged regardless of the change in pressure from $1\,atm$ to $2\,atm$.Therefore,the $rms$ speed is $900\,ms^{-1}$.

The $rms$ speed of helium gas at $27\,^{\circ}C$ and $1\,atm$ pressure is $900\,ms^{-1}$. Then the $rms$ speed of helium molecules at temperature $27\,^{\circ}C$ and $2\,atm$ pressure is ...... $m/s$.

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