This example problem demonstrates how to calculate the root mean square velocity of particles in an ideal gas.

### Root Mean Square Velocity Problem

What is the average velocity or root mean square velocity of a molecule in a sample of oxygen at 0 °C?**Solution**

Gases consist of atoms or molecules that move at different speeds in random directions. The root mean square velocity (RMS velocity) is a way to find a single velocity value for the particles.

The average velocity of gas particles is found using the root mean square velocity formula

μ_{rms} = (3RT/M)^{½}

where

μ_{rms} = root mean square velocity in m/sec

R = ideal gas constant = 8.3145 (kg·m^{2}/sec^{2})/K·mol

T = absolute temperature in Kelvin

M = mass of a mole of the gas in **kilograms**.

Really, the RMS calculation gives you **root mean square speed**, not velocity. This is because velocity is a vector quantity, which has magnitude and direction. The RMS calculation only gives the magnitude or speed.

The temperature must be converted to Kelvin and the molar mass must be found in kg to complete this problem.**Step 1** Find the absolute temperature using the Celsius to Kelvin conversion formula:

T = °C + 273

T = 0 + 273

T = 273 K**Step 2** Find molar mass in kg:

From the periodic table, molar mass of oxygen = 16 g/mol.

Oxygen gas (O_{2}) is comprised of two oxygen atoms bonded together. Therefore:

molar mass of O_{2} = 2 x 16

molar mass of O_{2} = 32 g/mol

Convert this to kg/mol:

molar mass of O_{2} = 32 g/mol x 1 kg/1000 g

molar mass of O_{2} = 3.2 x 10^{-2} kg/mol**Step 3** - Find μ_{rms}

μ_{rms} = (3RT/M)^{½}

μ_{rms} = [3(8.3145 (kg·m^{2}/sec^{2})/K·mol)(273 K)/3.2 x 10^{-2} kg/mol]^{½}

μ_{rms} = (2.128 x 10^{5} m^{2}/sec^{2})^{½}

μ_{rms} = 461 m/sec**Answer:**

The average velocity or root mean square velocity of a molecule in a sample of oxygen at 0 °C is 461 m/sec.