Question

a scuba diver exhales 2.05 l of air while swimming at a depth of 98.70 m where the sum of atmospheric pressure and water pressure is 10.54 atm. by the time this exhaled air rises to the surface, where the pressure is 1.00 atm, what is its volume?

Answer 1

The final volume of the air assuming constant temperature is 21.607 Liters using Boyle's gas law.

What is the Boyle's gas law equation? 

The empirical relation states that under constant temperature, the pressure (p) of a given amount of gas changes inversely with its volume (v); that is, pv = k.

Given,

Initial volume = 2.05 Liters.

Height =  98.70 meters

Final pressure = 1.00 atm.

Initial pressure =10.54atm.

PV = k

P₁V₁=P₂V₂

Substituting the given parameters into the formula,

10.54 × 2.05 = 1 × V₂

V₂ = 21.607

Final volume = 21.607 Liters.

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