The rms speed of a molecule of a diatomic gas | vedclass

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(C) The $rms$ 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 abs

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$2v$

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(C) The $rms$ 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.Initially,for the diatomic molecule: $v = \sqrt{\frac{3RT}{M}}$.When the temperature is doubled,the new temperature $T' = 2T$.When the diatomic molecule dissociates into two atoms,the molar mass of the atom becomes $M' = \frac{M}{2}$.The new $rms$ speed $v'$ is given by: $v' = \sqrt{\frac{3RT'}{M'}} = \sqrt{\frac{3R(2T)}{M/2}} = \sqrt{\frac{12RT}{M}} = 2 \sqrt{\frac{3RT}{M}}$.Substituting the initial speed $v$: $v' = 2v$.

The $rms$ speed of a molecule of a diatomic gas is $v$. When the temperature is doubled,the molecule dissociates into two atoms. The new $rms$ speed of the atom is:

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