3. A submarine containing 15,000L of air at an air pressure of 1.2atm breaks apart deep within the
ocean, where pressures reach 250atm. What volume of air would escape out of the submarine at
that depth?
a. Identify the law to be used:
b. Solve the problem:
ocean, where pressures reach 250atm. What volume of air would escape out of the submarine at
that depth?
a. Identify the law to be used:
b. Solve the problem:
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Boyle's Law
P1V1 = P2V2
where in, P1 = first pressure
V1 = first volume
P2 = second pressure
V2 = second volume
Given: P1 = 1.2atm
V1 = 15000L
P2 = 250atm
Unknown: V2 = ?
Formula: V2 = P1V1 / P2
Note: To get the formula, apply division and cancellation of like terms using the original formula of Boyles Law.
P1V1 = P2V2 ← divide both sides by P2
P1V1 / P2 = P2V2 / P2 ← cancel like terms
P1V1 / P2 = V2 ← final formula for V2
Note: Easier method is by Transposition
P1V1 = P2V2 ← transpose P2 to the other
side
P1V1 / P2 = V2 ← final formula for V2
Solution: V2 = P1V1 / P2
= (1.2atm)(15000L) / 250atm
= 18000atm•L / 250atm
= 72L
Note: "atm" is cancelled out, so the unit for volume will remain which is "L"
Reminder: Cancellation of units is applicable when the other same unit is on the denominator.
Final Answer: V2 = 72L