Question

A cylinder with a moveable piston holds 3.60 mol of argon at a constant temperature of 295 K. As the gas is compressed isothermally, its pressure increases from 101 kPa to 151 kPa. (a) Find the final volume of the gas.

Answer 1

Universal gas constant = R = 8.314 J/(mol.K)

Number of moles of argon gas = n = 3.6 mol

Initial temperature of the argon gas = T₁ = 295 K

Initial pressure of the argon gas = P₁ = 101 kPa = 101 x 10³ Pa

Initial volume of the argon gas = V₁

By Ideal Gas Law,

P₁V₁ = nRT₁

(101x10³)V₁ = (3.6)(8.314)(295)

V₁ = 8.742 x 10^-2 m³

Final pressure of the argon gas = P₂ = 151 kPa = 151 x 10³ Pa

Final volume of the argon gas = V₂

The process is isothermal therefore the temperature remains constant.

P₁V₁ = P₂V₂

(101x10³)(8.742x10^-2) = (151x10³)V₂

V₂ = 5.85 x 10^-2 m³

a) Final volume of the argon gas = 5.85 x 10^-2 m³