Question

Can anyone help me with this? I'm so stuck!
A popular size of scuba-diving tank holds the amount of compressed air that would occupy 71.2 ft3 at a normal surface pressure of 1 atm. The air in the tank is at a pressure of about 2250 lb/in.2, so the tank itself can have a volume much less than 71.2 ft3. How large does the tank need to be to hold 71.2 ft3? (Hint: Use Boyle’s Law: PV 5 k. Remember that 1 atm 5 14.7 lb/in.2.)

Answer 1

71.2 ft3 = 2.016 m^3
1 atm = 101325 Pa
2250 lb/in^2 = 158.2 kg/cm^2 = 15503600 Pa

V' = P V/P'

V' = (101325)(2.016)/15503600

V' = 0.0131757 m^3

Answer 2

You lost me there lol, could you explain it more?

Answer 3

First convert 71.2 cu. ft into cu. in. or vice versa.
Then P1V1=P2V2

Answer 4

I kinda understand that part. But when it asks how large does the tank need to be, does it mean volume?