At a depth of 20.0 meters, a scuba diver exhales a bubble of air with a volume of 0.5 mL. The volume of the bubble when it reaches the surface of the water at 1.0 atm of pressure is 1.5 mL. Assume the temperature of the air in the bubble stays the same.
What was the pressure on the diver when he exhaled the bubble of air?
A. 0.3 atm
B. 1.0 atm
C. 2.0 atm
D. 3.0 atm
To find the pressure on the diver when he exhaled the bubble of air, we can use Boyle’s Law, which states that ( P_1V_1 = P_2V_2 ), where ( P ) is pressure and ( V ) is volume.
Let’s assign the values:
– ( V_1 = 0.5 , text{mL} ) (volume at depth)
– ( P_1 = ? ) (pressure at depth)
– ( V_2 = 1.5 , text{mL} ) (volume at surface)
– ( P_2 = 1.0 , text{atm} ) (pressure at surface)
Rearranging Boyle’s Law gives us:
[
P_1 = frac{P_2V_2}{V_1}
]
Substituting the known values:
[
P_1 = frac{(1.0 , text{atm})(1.5 , text{mL})}{0.5 , text{mL}} = frac{1.5 , text{atm mL}}{0.5 , text{mL}} = 3.0 , text{atm}
]
Thus, the pressure on the diver when he exhaled the bubble of air was D. 3.0 atm