3. A balloon contains 14.0 L of air at a pressure of 760 torr. What will the volume of the air be when the balloon is taken to a depth of 10 ft in a swimming pool where the pressure is 981 torr? The temperature of the air does not change.
Given:
P1 = 760 torr
V1 = 14.0 L
T1 = T2 = constant
P2 = 981 torr
V2 = unknown
Formula: P1V1 = P2V2
Solution:
Since the temperature and amount of gas remain unchanged, Boyle's law can be used.
Boyle's gas law can be expressed as: P1V1 = P2V2
To find the volume of the air, solve the equation for V2:
V2 = P1V1 / P2
V2 = (760 torr × 14.0 L) / (981 torr) = 10.846 L = 10.8 L
V2 = 10.8 L
Answer: The volume of the air will be 10.8 liters
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