Root Square Mean Velocity Example Problem

Gases are made up of individual atoms or molecules freely moving in random directions with a wide variety of speeds. Kinetic molecular theory tries to explain the properties of gases by investigating the behavior of individual atoms or molecules making up the gas. This example problem shows how to find the average or root mean square velocity (rms) of particles in a gas sample for a given temperature.

Root Mean Square Problem

What is the root mean square velocity of the molecules in a sample of oxygen gas at 0 °C and 100 °C?

Solution:

Root mean square velocity is the average velocity of the molecules that make up a gas. This value can be found using the formula:

v_rms = [3RT/M]^1/2

where
v_rms = average velocity or root mean square velocity
R = ideal gas constant
T = absolute temperature
M = molar mass

The first step is to convert the temperatures to absolute temperatures. In other words, convert to the Kelvin temperature scale:

K = 273 + °C
T₁ = 273 + 0 °C = 273 K
T₂ = 273 + 100 °C = 373 K

The second step is to find the molecular mass of the gas molecules.

Use the gas constant 8.3145 J/mol·K to get the units we need. Remember 1 J = 1 kg·m²/s². Substitute these units into the gas constant:

R = 8.3145 kg·m²/s²/K·mol

Oxygen gas is made up of two oxygen atoms bonded together. The molecular mass of a single oxygen atom is 16 g/mol. The molecular mass of O₂ is 32 g/mol.

The units on R use kg, so the molar mass must also use kg.

32 g/mol x 1 kg/1000 g = 0.032 kg/mol

Use these values to find the v_rms.

0 °C:
v_rms = [3RT/M]^1/2
v_rms = [3(8.3145 kg·m²/s²/K·mol)(273 K)/(0.032 kg/mol)]^1/2
v_rms = [212799 m²/s²]^1/2
v_rms = 461.3 m/s

100 °C
v_rms = [3RT/M]^1/2
v_rms = [3(8.3145 kg·m²/s²/K·mol)(373 K)/(0.032 kg/mol)]^1/2
v_rms = [290748 m²/s²]^1/2
v_rms = 539.2 m/s

Answer:

The average or root mean square velocity of the oxygen gas molecules at 0 °C is 461.3 m/s and 539.2 m/s at 100 °C.