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

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