Question

At 400 $K$, the root mean square ($rms$) speed of a gas $X$ (molar mass $=40$) is equal to the most probable speed of a gas $Y$ at 60 $K$. The molar mass of the gas $Y$ in $g$ $mo{l}^{-1}$ is:

• A. 2
• B. 4
• C. 40
• D. 20

Answer 1

Answer: (B)

4

As we know,
$u_{rms}=\sqrt{3RT/M}$ and
$u_{mp}=\sqrt{2RT/M}$
$U_{rms}$ at 400 $K$ = $u_{MP}$ at 60 $K$
$\sqrt{\frac{3R{T}_1}{M_1}}=\sqrt{\frac{2R{T}_2}{M_2}}$
$\frac{3T_1}{M_1}=\frac{2T_2}{M_2}$
$\frac{3\times 400}{40}=\frac{2\times 60}{M_2}$
$M_2=4gmo{l}^{-1}$