Two gases $X$ (molecular weight $M_{X}$ ) and $Y$ (molecular weight $M_{Y}; M_{Y} > M_{X}$ ) are at the same temperature $T$ in two different containers. Their root mean square velocities are $C_{X}$ and $C_{Y}$ respectively. If the average kinetic energies per molecule of two gases $X$ and $Y$ are $E_{X}$ and $E_{Y}$ respectively then which of the following relation(s) is(are) true?

E_{X} > E_{Y}

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C_{X} > C_{Y}

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E_{X} = E_{Y} = (3 / 2) R T

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

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Solution

The correct options are
C $C_{X} > C_{Y}$
D $E_{X} = E_{Y} = (3 / 2) k_{B} T$
Given, molecular weight of $X = M_{X}$

Molecular weight of $Y = M_{Y}$ and $M_{Y} > M_{X}$

Root mean square velocities $= C_{X}$ and $C_{Y}$ respectively.

Average KE/molecule $= E_{X}$ and $E_{Y}$ respectively.

We know that,

Root mean square velocity,

$C = \sqrt{\frac{3 R T}{M}}$ or $C \propto \sqrt{\frac{1}{M}}$

$∴ \frac{C_{X}}{C_{Y}} = \frac{M_{Y}}{M_{X}}$

Since $M_{Y} > M_{X}$

$∴ C_{X} > C_{Y}$

KE/molecule $= \frac{3}{2} n R T$

$\Rightarrow E_{X} \neq E_{Y} \neq \frac{3}{2} R T$

Further, $E_{X} = E_{Y} = \frac{3}{2} k_{B} T$

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