Question

It is a hot, sunny day. The outdoor temperature is 32°c (90°f), and the water temperature is 21°c (70°f). A scuba diver submerses himself while holding a balloon filled with 1.5 l of air. If the balloon shrinks to a volume of 0.5 l at a depth of 20 m, what is the pressure at this depth? (assume that the outdoor pressure is 1 atm.)

Answer 1

initially when balloon is in atmosphere then we have

$PV = nRT$

initial temperature is T = 32 degree C

Initial pressure P = 1 atm

Initial volume V = 1.5 L

now we have number of moles as

$n = \frac{1*1.5}{R*(32 + 273)}

similarly when it is submerged into the water

Temperature will be T = 21 degree C

Volume will be 0.5 L

now for same number of moles

[tex]PV = nRT$

$P*0.5 = \frac{1*1.5}{R * 305}* R* (21+273)$

$P = 2.89 atm$

so pressure at this depth will be 2.89 atm