Q56 / 200 · WBJEE 2014 · Chemistry

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
    $E_{X}>E_{Y}$
  • Correct answer
    B
    $C_{X}>C_{Y}$
  • C
    $E_{X}=E_{Y}=(3 / 2) R T$
  • Correct answer
    D
    $E_{X}=E_{Y}=(3 / 2) k_{B} T$

Solution

Given, molecular weight of $X=M_{X}$ Mol. wt. of $Y=M_{Y}$ and $M_{Y}>M_{x}$ Root mean square velocitios $=C_{X}$ and $C_{Y}$ respectively Average $\mathrm{KE}$/molecule $=E_{X}$ and $E_{Y}$ respectively We know that. Root mean square velocity. $$ C=\sqrt{\frac{3 R T}{M}} \text { or } C \propto \sqrt{\frac{1}{M}} $$ $\therefore \quad \frac{C_{X}}{C_{Y}}=\frac{M_{Y}}{M_{X}}$ since $\quad M_{Y}>M_{X}$ $\therefore$ $$ C_{X}>C_{Y} $$ KE/molecules $=\frac{3}{2} n R T$ $\Rightarrow \quad E_{X} \not \subset E_{Y} \not = \frac{3}{2} R T$ Further. $\quad E_{X}=E_{Y}=\frac{3}{2} \boldsymbol{k}_{\boldsymbol{B}} \boldsymbol{T}$

Hence, the correct answer is options: 'B', 'D'.

Relevant Questions