As an air bubble rises from the bottom of a lake to the surface, its volume is doubled. Find the depth of the lake. Take atmospheric pressure = 76 cm of Hg.

Answer

To solve the problem of finding the depth of the lake based on the behavior of an air bubble rising from the bottom to the surface, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature. Here’s a step-by-step solution:

### Step 1: Understand Boyle's Law
According to Boyle's Law, \( P_1 V_1 = P_2 V_2 \), where:
- \( P_1 \) = initial pressure at the bottom of the lake
- \( V_1 \) = initial volume of the bubble
- \( P_2 \) = pressure at the surface of the lake
- \( V_2 \) = final volume of the bubble

### Step 2: Set Up the Problem
Given that the volume of the bubble doubles as it rises, we can express this as:
\[ V_2 = 2 V_1 \]

### Step 3: Apply Boyle's Law
Substituting into Boyle's Law:
\[ P_1 V_1 = P_2 (2 V_1) \]
We can simplify this to:
\[ P_1 = 2 P_2 \]

### Step 4: Define Pressures
At the surface of the lake, the pressure \( P_2 \) is equal to atmospheric pressure, which is given as:
\[ P_2 = P_0 = 76 \text{ cm of Hg} \]

The pressure at the depth \( h \) of the lake is given by:
\[ P_1 = P_0 + \rho g h \]
Where:
- \( \rho \) = density of water (approximately \( 13.6 \text{ g/cm}^3 \) or \( 13600 \text{ kg/m}^3 \))
- \( g \) = acceleration due to gravity (approximately \( 9.81 \text{ m/s}^2 \))
- \( h \) = depth of the lake in meters

### Step 5: Substitute and Solve for Depth
From the relationship \( P_1 = 2 P_2 \):
\[ P_0 + \rho g h = 2 P_0 \]
Substituting \( P_0 \):
\[ 76 \text{ cm of Hg} + \rho g h = 2 \times 76 \text{ cm of Hg} \]
This simplifies to:
\[ \rho g h = 76 \text{ cm of Hg} \]

### Step 6: Convert Pressure to SI Units
To convert \( 76 \text{ cm of Hg} \) to Pascals:
\[ 76 \text{ cm of Hg} = 76 \times 13.6 \text{ g/cm}^3 \times 9.81 \text{ m/s}^2 \]
Calculating this gives:
\[ 76 \text{ cm of Hg} = 76 \times 13600 \text{ kg/m}^3 \times 9.81 \text{ m/s}^2 \]
\[ = 10133.6 \text{ Pa} \]

### Step 7: Calculate Depth
Now, we can find \( h \):
\[ h = \frac{76 \text{ cm of Hg}}{\rho g} \]
Substituting the values:
\[ h = \frac{10133.6 \text{ Pa}}{13600 \text{ kg/m}^3 \times 9.81 \text{ m/s}^2} \]
Calculating this gives:
\[ h \approx 10.04 \text{ m} \]

### Final Answer
The depth of the lake is approximately **10.04 meters**.

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