Two gases X (molecular weight M_{X}) and Y (m | vedclass

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(B, D) The root mean square velocity is given by $C = \sqrt{\frac{3RT}{M}}$. Since $C \propto \frac{1}{\sqrt{M}}$ and $M_{Y} > M_{X}$, it follows that

Vedclass · Accepted Answer

$E_{X} = E_{Y} = (3 / 2) k_{B} T$

Vedclass · Answer

(B, D) The root mean square velocity is given by $C = \sqrt{\frac{3RT}{M}}$. Since $C \propto \frac{1}{\sqrt{M}}$ and $M_{Y} > M_{X}$, it follows that $C_{X} > C_{Y}$.The average kinetic energy per molecule of an ideal gas is given by $E = \frac{3}{2} k_{B} T$, where $k_{B}$ is the Boltzmann constant.Since both gases are at the same temperature $T$, their average kinetic energies per molecule are equal, i.e.,$E_{X} = E_{Y} = \frac{3}{2} k_{B} T$.Thus, both $C_{X} > C_{Y}$ and $E_{X} = E_{Y} = \frac{3}{2} k_{B} T$ are correct relations.

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