Question

An air bubble having volume 1 cm³ at depth 40 m inside water comes to the surface. What will be the volume of the bubble at the surface?

• A. 5 cm³
• B. 2 cm³
• C. 4 cm³
• D. 3 cm³

Answer 1

The Correct Option is A

To solve this problem, we need to apply the principles of buoyancy and the ideal gas law. The problem states that an air bubble with a volume of 1 cm³ is at a depth of 40 meters in water. We need to find its volume when it reaches the surface.

Step 1: Understand the Concept

The volume of the bubble changes due to the change in pressure as it rises to the surface. At depth, the bubble is subjected to the water column pressure plus atmospheric pressure. As it rises, the pressure decreases, and according to Boyle's Law (assuming constant temperature), the volume will increase.

Step 2: Apply Boyle's Law

Boyle's Law states that for a given amount of gas at constant temperature, the product of pressure and volume is constant. This is expressed as:

\(P_1 V_1 = P_2 V_2\)

• \(P_1\): Pressure at depth (sum of atmospheric + water pressure)
• \(V_1\): Initial volume of the bubble (1 cm³)
• \(P_2\): Pressure at the surface (atmospheric pressure)
• \(V_2\): Volume at the surface

Step 3: Calculate the Pressures

Given:

• \(\rho = \text{density of water} = 1000 \, \text{kg/m}^3\)
• h = 40 \, \text{m}
• Atmospheric pressure, P_1 = P_{\text{atm}} + \rho g h

 

\(P_1 = 1.01 \times 10^5 + 1000 \times 9.8 \times 40\)

\(P_1 = 1.01 \times 10^5 + 392000\)

\(P_1 = 493000 \, \text{Pa}\)

Pressure at the surface:

\(P_2 = P_{\text{atm}} = 1.01 \times 10^5 \, \text{Pa}\)

Step 4: Solve for \(V_2\)

Using Boyle's Law:

\(P_1 V_1 = P_2 V_2\)

\(V_2 = \frac{493000}{101000}\)

\(V_2 \approx 4.88 \, \text{cm}^3\)

Since we need the nearest whole number, \(V_2 = 5 \, \text{cm}^3\).

Conclusion:

The volume of the air bubble when it reaches the surface will be approximately 5 cm³. This corresponds to the correct option: 5 cm³.