Question

a bag is blown up to a volume of 3 liters. a student jumps into lake with the bag and sinks 15 meters. Assuming air temp is the same as water temp, what is the volume of this bag at this depth

Answer 1

The number of mols of gas remains constant, then: n1=n2

Also the temperature in both circumstances. T1=T2

$n_{1}\cdot R\cdot T_{1} =n_{2}\cdot R\cdot T_{2}$

$P\cdot V =n\cdot R\cdot T$

Then: $P_{1}\cdot V_{1} = P_{2}\cdot V_{2}$

$V_{2} = \frac{P_{1}\cdot V_{1}}{P_{2}}$

V1= 3l

P1= atmospheric pressure = 1 atm = 10,33 m H2O

P2= pressure at 15 m underwater= atmospheric pressure + water depth = (10, 33 + 15) m H2O = 25,33 m H2O

$V_{2} = \frac{10,33\cdot 3}{25,33} = 1,22 l$