For one mole of a monoatomic gas (molar mass | vedclass

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(C) The average kinetic energy $(E)$ for one mole of an ideal gas is given by $E = \frac{3}{2}RT$.We know that $U_{rms} = \sqrt{\frac{3RT}{M}}$.Squari

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$U_{rms} = \sqrt{\frac{2E}{M}}$

Vedclass · Answer

(C) The average kinetic energy $(E)$ for one mole of an ideal gas is given by $E = \frac{3}{2}RT$.We know that $U_{rms} = \sqrt{\frac{3RT}{M}}$.Squaring both sides,we get $U_{rms}^2 = \frac{3RT}{M}$,which implies $RT = \frac{U_{rms}^2 	imes M}{3}$.Substituting the value of $RT$ into the kinetic energy equation: $E = \frac{3}{2} 	imes \left( \frac{U_{rms}^2 	imes M}{3} ight) = \frac{1}{2} M U_{rms}^2$.Rearranging for $U_{rms}$: $U_{rms}^2 = \frac{2E}{M}$,so $U_{rms} = \sqrt{\frac{2E}{M}}$.

For one mole of a monoatomic gas (molar mass $= M$),the root mean square velocity is $U_{rms}$. What is the relationship between the average kinetic energy $(E)$ and $U_{rms}$?

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