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WB JEE Medical WB JEE Medical Solved Paper-2014

  • question_answer
    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?

    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, \[{{E}_{X}}={{E}_{Y}}=\frac{3}{2}{{k}_{B}}T\]


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