Question

Calculate the rms speed of an oxygen gas molecule, o2, at 31.0 ∘c . express your answer numerically in meters per second.

Answer 1

Answer: The rms speed of an oxygen gas molecule is 486.77 m/s.

Explanation: The expression for the rms speed of the molecule is given by:

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

where,

R = Gas constant = 8.314 J/K mol  = $8.314\text{ kg }m^2s^{-2}K^{-1}mol^{-1}$

T = Temperature of the gas = 31°C = (273 + 31)K = 304K

M = Molar mass of the gas = $32g/mol=0.032kg/mol$    (Gas is Oxygen)

Putting values in above equation, we get:

$v_{rms}=\sqrt{\frac{3\times 8.314\text{ kg }m^2s^{-2}K^{-1}mol^{-1}\times 304K}{0.032kg/mol}$

$v_{rms}=486.77m/s$