A) \[{{E}_{X}}>{{E}_{Y}}\]
B) \[{{C}_{X}}>{{C}_{Y}}\]
C) \[{{E}_{X}}={{E}_{Y}}=(3/2)RT\]
D) \[{{E}_{X}}={{E}_{Y}}=(3/2){{k}_{B}}T\]
Correct Answer: B
Solution :
(b, d) Given, molecular weight of \[\text{X}\,\text{=}\,{{\text{M}}_{x}}\] Mol. wt. 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, \[C=\sqrt{\frac{3RT}{M}}\]or \[C\propto \sqrt{\frac{1}{M}}\] \[\therefore \] \[\frac{{{C}_{X}}}{{{C}_{Y}}}=\frac{{{M}_{Y}}}{{{M}_{X}}}\] Since \[{{M}_{Y}}>{{M}_{X}}\] \[\therefore \] \[{{C}_{X}}>{{C}_{Y}}\] KE/ molecules \[=\frac{3}{2}nRT\] \[\Rightarrow \] \[{{E}_{x}}\ne {{E}_{Y}}\ne \frac{3}{2}RT\] Further, \[{{E}_{X}}={{E}_{Y}}=\frac{3}{2}{{k}_{B}}T\]You need to login to perform this action.
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