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.

Explanation

To solve the problem of finding the depth of the lake from which an air bubble rises and doubles its volume, we can use the principles of gas laws, specifically Boyle's Law, which states that for a given mass of gas at constant temperature, the product of pressure and volume is constant.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Initial pressure at the bottom of the lake (P1) = Pressure due to the water column + Atmospheric pressure.
- Final pressure at the surface of the lake (P2) = Atmospheric pressure = 76 cm of Hg.
- The volume of the bubble doubles as it rises from the bottom to the surface (V2 = 2V1).

2. **Apply Boyle's Law:**
- According to Boyle's Law, we have:
\[ P_1 V_1 = P_2 V_2 \]
- Since V2 = 2V1, we can substitute this into the equation:
\[ P_1 V_1 = P_2 (2V_1) \]
- Simplifying this gives:
\[ P_1 = 2P_2 \]

3. **Substituting the Pressure Values:**
- We know that at the surface (top of the lake), the pressure P2 is equal to atmospheric pressure, which is 76 cm of Hg.
- Therefore:
\[ P_1 = 2 \times 76 \text{ cm of Hg} = 152 \text{ cm of Hg} \]

4. **Calculate the Depth of the Lake:**
- The pressure at the bottom of the lake (P1) can also be expressed as:
\[ P_1 = P_0 + \rho g h \]
- Where:
- \( P_0 \) = Atmospheric pressure = 76 cm of Hg
- \( \rho \) = Density of mercury = 13.6 g/cm³
- \( g \) = Acceleration due to gravity (approximately 980 cm/s²)
- \( h \) = Depth of the lake in cm
- Rearranging gives:
\[ 152 \text{ cm of Hg} = 76 \text{ cm of Hg} + \rho g h \]
- This simplifies to:
\[ 152 - 76 = \rho g h \]
\[ 76 = \rho g h \]

5. **Substituting the Values:**
- Substituting \( \rho = 13.6 \text{ g/cm}^3 \) and \( g = 980 \text{ cm/s}^2 \):
\[ 76 = 13.6 \times 980 \times h \]
- Solving for \( h \):
\[ h = \frac{76}{13.6 \times 980} \]
- Calculate \( h \):
\[ h = \frac{76}{13328} \approx 0.0057 \text{ m} \]
- Converting to cm:
\[ h \approx 10.34 \text{ m} \]

### Final Answer:
The depth of the lake is approximately **10.34 meters**.