Question

The molecules of a given mass of a gas have root mean square speeds of $100m{s}^{-1}$ at $27^{\circ }C$ and 1 atmospheric pressure. The root mean square speeds of the molecules of the gas at $127^{\circ }C$ and 2 atmospheric pressure is

• A. $\frac{200}{\sqrt{3}}$
• B. $\frac{100}{\sqrt{3}}$
• C. $\frac{400}{3}$
• D. $\frac{200}{3}$

Answer 1

Answer: (A)

Here, $v_{rms1}=100m{s}^{-1},T_1=27{}^{\circ }C$
$=(27+273)K=300K$
$P_1=1atm{v}_{rms2}=$?,
$T_2=127^{\circ }C=(127+273)K=400K$
$P_2=2atm$
From $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2};\frac{V_1}{V_2}=\frac{P_2}{P_1}\cdot \frac{T_1}{T_2}$
$=2\times \frac{300}{400}=\frac{3}{2}$
Again $P_1=\frac{1}{3}\frac{M}{V_1}{v}_{rm{s}_1}^2$
and $P_2=\frac{1}{3}\frac{M}{V_2}{v}_{rm{s}_2}^2$
$\therefore \frac{v_{rm{s}_2}^2}{v_{rm{s}_1}^2}\cdot \frac{V_1}{V_2}=\frac{P_2}{P_1}$
$v_{rm{s}_2}^2=v_{rm{s}_1}^2\times \frac{P_2}{P_1}\times \frac{V_2}{V_1}$
$=(100{)}^2\times 2\times \frac{2}{3}$
$v_{rm{s}_2}=\frac{200}{\sqrt{3}}m{s}^{-1}$