Question

A diving bell is a container open at the bottom. As the bell descends, the water level inside changes so the the pressure inside equals the pressure outside. Initially, the volume of air is 8.58 m3 at 1.020 atm and 20 degrees celsius. What is the volume at 1.212 atm and 20 degrees Celsius?

Answer 1

We assume it as ideal gas

PV = nRT

now from initial conditions we find out moles of air trapped inside,which will essentially remain constant

P = pressure

V = volume

n = no. of molesR = gas constant(8.314 Pa.m3/mol.K)

T = temperature in kelvin

Initially,

P = 1.020 atm = 1.020 x 101325 Pa = 103351.5 Pa

V = 8.58 m³

n = ?

T = 20+273 = 293 K

n = PV/RT = 364.021 moles

volume at 1.212 atm and 20 ^oC

1.212 atm = 101325 x 1.212 = 122805.9 Pa

T = 20+273 = 293 K

V = (nRT)/P = (364.021 mol*8.314 Pa.m³/mol.K *293 K)/122805.9 Pa = 7.22 m³

Ans = 7.22 m³