The volume of air in a diving bell changes as it descends into deep water. If the air in a diving bell occupies 9.50 L at 1.02 atm and 20.1°C, what will be the volume of the air at 1.55 atm and 15.0°C?

V₂ = 6.14 L

Explanation:

Initial volume = 9.50 L

Initial temperature = 20.1°C (20.1 + 273 = 293.1 K)

Initial pressure = 1.02 atm

Final temperature = 15°C (15+273 = 288 K)

Final volume = ?

Final pressure = 1.55 atm

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂ / T₁ P₂

V₂ = 1.02 atm. 9.50 L. 288 K / 293.1 K. 1.55 atm

V₂ = 2790.72 L / 454.305

V₂ = 6.14 L