State Boyle's Law: P 1 V 1 = P 2 V 2 .
Substitute given values: P 1 = 3.0 atm, V 1 = 5.0 mL, P 2 = 1.0 atm.
Solve for V 2 : V 2 = P 2 P 1 V 1 = 1.0 atm ( 3.0 atm ) ( 5.0 mL ) = 15.0 mL .
The final volume of the air bubble is 15.0 mL .
Explanation
Problem Analysis and Given Data We are given an air bubble at the bottom of a lake with an initial volume V 1 = 5.0 mL and pressure P 1 = 3.0 atm. As the bubble rises to the surface, the pressure changes to P 2 = 1.0 atm. We need to find the final volume V 2 of the air bubble using Boyle's Law, which states that P 1 V 1 = P 2 V 2 .
Stating Boyle's Law and Given Values Boyle's Law is expressed as: P 1 V 1 = P 2 V 2 We are given: P 1 = 3.0 atm V 1 = 5.0 mL P 2 = 1.0 atm We need to find V 2 .
Calculating the Final Volume To find V 2 , we rearrange Boyle's Law: V 2 = P 2 P 1 V 1 Now, we substitute the given values: V 2 = 1.0 atm ( 3.0 atm ) ( 5.0 mL ) V 2 = 1.0 atm 15.0 atm ⋅ mL V 2 = 15.0 mL
Final Answer and Conclusion The final volume of the air bubble when it reaches the surface is 15.0 mL. Since the initial volume was 5.0 mL, the volume has increased.
Examples
Boyle's Law is useful in understanding how scuba diving equipment works. As a diver descends, the pressure increases, and the air in their tanks becomes compressed. Understanding the relationship between pressure and volume helps divers manage their air supply and avoid potential hazards such as lung overexpansion upon ascent. Similarly, in weather forecasting, Boyle's Law helps predict changes in atmospheric volume based on pressure variations, aiding in more accurate weather predictions. This principle is also applied in various engineering applications, such as designing pneumatic systems where controlling air volume and pressure is crucial for efficient operation.
As the air bubble rises from a pressure of 3.0 atm to 1.0 atm, its volume increases from 5.0 mL to 15.0 mL according to Boyle's Law. The decrease in pressure allows the bubble to expand significantly. Therefore, the final volume of the bubble is 15.0 mL.
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