Answer to Question #142747 in Chemistry for Sebastian Luis

Question

Consider two 1-L samples of gas: one is H2 and the other is
O2. Both are at 1 atm and 25°C. How do the samples compare in
terms of (a) mass, (b) density, (c) RMS, and (d) time
for a given fraction of molecules to effuse?

Answer 1

According to the ideal gas law $pV = nRT$ , at equal temperature, pressure and volume, the gases contain the same number of the moles $n$. Therefore, their masses compare as their molar masses $M$ : 32 g/mol for O₂ and 2.02 g/mol for H₂. Thus, the oxygen gas has $\approx16$ times higher mass than the hydrogen gas.

From this, the density, defined as $d = \frac{m}{V}$ , is different, and that of the oxygen gas is $\approx16$ times higher than that of the hydrogen.

From the kinetic molecular theory, the root-mean-square (RMS) velocity depends on the molar mass of the gas:

$v_{rms} = \sqrt{\frac{3RT}{M}}$ .

From this equation, the RMS velocity of oxygen is lower than that of hydrogen, as its molecules are heavier (have larger weight). Approximately, the RMS velocity of the oxygen gas molecules is 4 times lower than that of the hydrogen gas molecules.

Finally, according to the Graham's law, the rate of effusion is inversely proportional to the square root of the molar mass of the gas molecule (logical, in view of the previous expression from the kinetic molecular theory):

$\frac{r_1}{r_2} = \sqrt\frac{M_2}{M_1}$ .

Thus, the oxygen gas molecules effusion time will be $\approx4$ times longer than that of the hydrogen gas molecules under the same conditions.