Question

Two gases $X$ (Mol. Wt. $M_x$) and $Y$ (Mol. Wt. $M_Y; M_Y > M_x$) 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. $E_{x}=E_{Y}=\frac{3}{2}RT$
• D. $E_{x}=E_{Y}=\frac{3}{2}k_{B}T$

Answer 1

Answer: (D)

$E_{x}=E_{Y}=\frac{3}{2}k_{B}T$

Answer 2

Given, molecular weight of $X=M_{X}$
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,
$C=\sqrt{\frac{3 R T}{M}} \text { or } C \propto \sqrt{\frac{1}{M}}$
$\therefore \frac{C_{X}}{C_{Y}}=\frac{M_{Y}}{M_{X}}$
Since $M_{Y} > M_{X}$
$\therefore C_{X} > C_{Y}$
KE molecules $=\frac{3}{2} n R T$
$\Rightarrow E_{X} \neq E_{Y} \neq \frac{3}{2} R T$
Further, $E_{X}=E_{Y}=\frac{3}{2} k_{B} T$