Question

The molecules of a given mass of gas have root mean square speeds of 100 m s–1 at 27 °C and 1 atmospheric pressure. The root mean square speeds of the molecules of the gas at 127 °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)

$\frac{200}{\sqrt{3}}$

Answer 2

Here

$V_{rms1} = 100ms^{- 1},T_{1} = 27{^\circ}C = (27 + 273)L = 300K$$P_{2} = 2atm$

From$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}};\frac{V_{1}}{V_{2}} = \frac{P_{2}}{P_{1}}\frac{T_{1}}{T_{2}} = 2 \times \frac{300}{400}\frac{3}{2}$

Again $P_{1} = \frac{1}{3}\frac{M}{V_{1}}{v^{2}}_{rms1}$and $P_{2} = \frac{1}{3}\frac{M}{V}{v^{2}}_{rms_{2}}$

$\therefore\frac{{V^{2}}_{rms_{2}}}{{V^{2}}_{rms_{1}}} \times \frac{V_{1}}{V_{2}} = \frac{P_{2}}{P_{1}}$

${V^{2}}_{rms_{2}} = {v^{2}}_{rms_{1}} \times \frac{P_{2}}{P_{1}} \times \frac{V_{2}}{V_{1}} = (100)^{2} \times 2 \times \frac{2}{3}$

$V_{rms_{2}} = \frac{200}{\sqrt{3}}ms^{- 1}$