Question

The molecules of a given mass of a gas have root mean square speeds of 100 ms^-1 at 27°C and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at 127°C and 2.0 atmospheric pressure?

Answer 1

\(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\)

\(\frac{V_1}{V_2}=\frac{P_2T_1}{P_1T_2}\)

\(=\frac{2\times300}{400}\)

\(=\frac{3}{2}\)

\(P_1=\frac{1}{3}\frac{M}{V_1}c_1^{-2};\) 

\(P_2=\frac{1}{3}\frac{M}{V^2}c_2^{-2}\)

∴ \(c_2^2=c_1^2\times\frac{V_2}{V_1}\times\frac{P_2}{P_1}\)

\(=(100)^2\times\frac{2}{3}\times2\)

\(c_2=\frac{200}{\sqrt3}ms^{-1}\)