Boltzmann's constant is 1.38066*10^-23 J/K,

and the universal gas constant is

8.31451 J/K · mol.

If 2.9 mol of a gas is confined to a 6.6 L

vessel at a pressure of 7.1 atm, what is the average kinetic energy of a gas molecule?

Answer in units of J.

1.18266*10⁻²⁰ J

Explanation:

Applying,

E = (3/2) kT ... Equation 1

Where E = kinetic energy of the gas molecule, k = Boltzmann's constant, T = Temperature

But,

PV = nRT ... Equation 2

Where P = pressure, V = Volume, n = number of moles, R = Universal gas constant.

make T the subject of the equation

T = PV/nR ... Equation 3

Substitute equation 3 into equation 1

E = (3/2) k (PV/nR)

E = 3kPV/2nR ... Equation 4

Given: k = 1.38066*10⁻²³ J/K, V = 6.6 L = 0.0066 m³, P = 7.1 atm = (101325*7.1) N/m² = 719407.5 N/m², n = 2.9 mol, R = 8.31451 J/K. mol

Substitute into equation 4

E = (3*1.38066*10⁻²³*0.0066*719407.5) / (2*8.31451)

E = 1182.66*10⁻²³ J

E = 1.18266*10⁻²⁰ J