Root Square Mean Velocity Example Problem

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:vrms [3RT/M]1/2wherevrms average velocity or root mean square velocityR ideal gas constantT absolute temperatureM molar massThe first step is to convert the temperatures to absolute temperatures. In other words, convert to the Kelvin temperature scale:K 273  °CT1 273 0  °C 273 KT2 273 100  °C 373 KThe 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 ·molOxygen 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 al so use kg.32 g/mol x 1 kg/1000 g 0.032 kg/molUse these values to find the vrms. 0  °C:vrms [3RT/M]1/2vrms [3(8.3145 kg ·m2/s2/K ·mol)(273 K)/(0.032 kg/mol)]1/2vrms [212799 m2/s2]1/2vrms 461.3 m/s100  °Cvrms [3RT/M]1/2vrms [3(8.3145 kg ·m2/s2/K ·mol)(373 K)/(0.032 kg/mol)]1/2vrms [290748 m2/s2]1/2vrms 539.2 m/sAnswer: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.