Question

The molecules of a given mass of a gas have root mean square speeds of $100\, m\, s ^{-1}$ 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, $v_{r m s 1}=100 \,m s ^{-1}, T_{1}=27{ }^{\circ} C$
$=(27+273) K =300 K$
$P_{1}=1 \,atm \,v_{r m s 2}=$?,
$T_{2}=127^{\circ} C =(127+273) K =400 \,K$
$P_{2}=2 \,atm$
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_{r m s_{1}}^{2} $
and $P_{2}=\frac{1}{3} \frac{M}{V_{2}} v_{r m s_{2}}^{2}$
$\therefore \frac{v_{r m s_{2}}^{2}}{v_{r m s_{1}}^{2}} \cdot \frac{V_{1}}{V_{2}}=\frac{P_{2}}{P_{1}}$
$v_{r m s_{2}}^{2}=v_{r m s_{1}}^{2} \times \frac{P_{2}}{P_{1}} \times \frac{V_{2}}{V_{1}}$
$=(100)^{2} \times 2 \times \frac{2}{3}$
$v_{r m s_{2}}=\frac{200}{\sqrt{3}} m s ^{-1}$