A bubble of gas released at the bottom of a lake increases to four times its original volume when it reaches the surface. Assuming that atmospheric pressure is equivalent to the pressure exerted by a column of water 10 m high, what is the depth of the lake?

Answer

To solve the problem of determining the depth of the lake based on the behavior of a gas bubble, we can use the principles of gas laws and hydrostatic pressure. Here’s a step-by-step solution:

### Step 1: Understand the Problem
A bubble of gas is released at the bottom of a lake and expands to four times its original volume when it reaches the surface. We need to find the depth of the lake, given that atmospheric pressure is equivalent to the pressure exerted by a 10 m high column of water.

### Step 2: Apply Boyle's Law
Since the temperature is constant, we can apply Boyle's Law, which states that \( P_1 V_1 = P_2 V_2 \).

- Let \( V_1 = V \) (initial volume at the bottom)
- Let \( V_2 = 4V \) (final volume at the surface)

Let \( P_1 \) be the pressure at the bottom of the lake and \( P_2 \) be the atmospheric pressure at the surface.

### Step 3: Express the Pressures
At the surface, the pressure \( P_2 \) is equal to atmospheric pressure, which we can take as 1 atm. The pressure at the bottom of the lake \( P_1 \) can be expressed as:

\[ P_1 = P_2 + \rho g h \]

Where:
- \( \rho \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \)),
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)),
- \( h \) is the depth of the lake.

### Step 4: Set Up the Equation
Using Boyle's Law, we can set up the equation:

\[ P_1 V = P_2 (4V) \]

Dividing both sides by \( V \) (assuming \( V \neq 0 \)) gives:

\[ P_1 = 4P_2 \]

### Step 5: Substitute the Pressures
Now we can substitute \( P_2 \):

\[ P_1 = 4 \times 1 \, \text{atm} = 4 \, \text{atm} \]

### Step 6: Relate Pressures to Depth
Now, substituting \( P_1 \) into the pressure equation:

\[ 4 \, \text{atm} = 1 \, \text{atm} + \rho g h \]

Rearranging gives:

\[ \rho g h = 4 \, \text{atm} - 1 \, \text{atm} = 3 \, \text{atm} \]

### Step 7: Convert Atmospheric Pressure to Pascals
1 atm is equivalent to \( 101325 \, \text{Pa} \), so:

\[ 3 \, \text{atm} = 3 \times 101325 \, \text{Pa} = 303975 \, \text{Pa} \]

### Step 8: Solve for Depth \( h \)
Now substituting the values of \( \rho \) and \( g \):

\[ 1000 \, \text{kg/m}^3 \times 10 \, \text{m/s}^2 \times h = 303975 \, \text{Pa} \]

This simplifies to:

\[ 10000 \, \text{kg/(m}^2\text{s}^2) \times h = 303975 \, \text{Pa} \]

Thus:

\[ h = \frac{303975}{10000} = 30.3975 \, \text{m} \]

### Final Answer
The depth of the lake is approximately **30.4 meters**.

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