Question

The average kinetic energy of the molecules in a gas sample depends only on the temperature, t. but given the same kinetic energies, a lighter molecule will move faster than a heavier molecule.what is the rms speed of o2 molecules at 423 k?

Answer 1

The root mean square (rms) speed of a gas is given by
$v_{rms} = \sqrt{ \frac{3RT}{M_m} }$
where R is the gas constant, T the absolute temperature of the gas and $M_m$ is the molar mass of the gas.

For molecular oxygen $O_2$, the molar mass is $M_m = 0.032 kg/mol$, so since the temperature of the gas in the problem is T=423 K, the rms speed of its molecules is
$v_{rms}= \sqrt{ \frac{3 (8.31 J mol^{-1} K^{-1})(423 K)}{0.032 kg/mol} }= 574.1 m/s$