Question

A sample of 0.440 mol of Ar gas has a volume of 6.55L and a pressure of 9.00 x 10^4 pa. What is the average velocity of the argon atoms? What is the temperature of the Ar?

Answer 1

Solution:

Volume "V" of gas = 6.55 L = 6.55 x 10^-3 m³

Pressure "P" = 9 x10⁴ Pa

Number of moles "n" of argon = 0.44 mol

Molar mass 'M"of argon = 39.948 g/mol = 39.948 x 10^-3 kg/mol

From kinetic theory of gases Pressur, volume and root mean square velocity "vrms"are related through the relation:

PV = (1/3)nMv_rms² ; Where M is the molar mass and n is the number of moles of a gas.

9 x10⁴ Pa x 6.55 x 10^-3 m³ = (1/3) x 0.44 mol x 39.948 x 10^-3 kgmol^-1 x v_rms²

v_rms = 317.2 m/s

We know that

v_rms = (3RT/M)^1/2

317.2 m/s = [3 x 8.314 JK-1mol-1 x T/(39.948 x 10^-3 kgmol^-1)]^1/2

317.2 m/s = [ 624.36167 x T]^1/2

T = 161.15 K

Hence, average velocity can be calculated from the following expression

v_avg = (8RT/M)^1/2

v_avg = [8 x 8.314 JK-1mol-1 x 161.15 K /(39.948 x 10^-3 kgmol^-1 x )]^1/2

v_avg = 292.242 m/s