Chemistry - Ideal Gas Question with Solution | TestHub

For option (A) From Fig (i), if is clear that $\left({\mathrm{u}}_{\mathrm{mps}}\right)\mathrm{Y}>\left({\mathrm{u}}_{\mathrm{mps}}\right)\mathrm{X}$

or, $\sqrt{\frac{2RT}{M_Y}}>\sqrt{\frac{2RT}{M_X}}$

since both X and Y are at same temp.

$$\therefore M_Y<M_X$$

since X is ${\mathrm{CH}}_4$ whose molar Mass is 16 .

$\therefore \mathrm{Y}$ can't be ${\mathrm{CO}}_2$ because. ${\mathrm{M}}_{{\mathrm{CO}}_2}=44$

∴ (A) is wrong.

for option (B), see Fig (ii)

(ii)

in lower speed region shaded area in figure, fraction of molecules of $\mathrm{X}>$ fraction of molecules of Y . As shaded area is greater for X as compared to Y . In higher speed region, shaded area in fig. fraction of molecules of $\mathrm{Y}>$ fraction of molecules of X . Therefore (B) is wrong.

for option 'C'  

under identical conditions of P & T.

rate of effusion ∝ $\frac{1}{\sqrt{M}}$

∴ M_Y < M_X (from (i))

∴ rate of effusion of Y > X.

∴ (C) is correct.

for option 'D'

Molar KE = $\frac{3}{2}RT$

∴ Both X and Y are at same tempt. = 200 K

∴ Molar (KE)_X = Molar(KE)_Y .

∴ (D) is correct