An air bubble of `20 cm^3` volume is at the bottom of a lake `40 m` deep where the temperature is `4^@ C`. The bubble rises to the surface which is at a temperature of `20 ^@ C`. Take the temperature to be the same as that of the surrounding water and find its volume just before it reaches the surface.

Explanation

To solve the problem of the air bubble rising from the bottom of a lake to the surface, we will use the ideal gas law and the principles of thermodynamics. Here's a step-by-step solution:

### Step 1: Understand the Given Information
- Initial volume of the bubble, \( V_1 = 20 \, \text{cm}^3 \)
- Depth of the lake, \( h = 40 \, \text{m} \)
- Temperature at the bottom, \( T_1 = 4^\circ C = 273 + 4 = 277 \, \text{K} \)
- Temperature at the surface, \( T_2 = 20^\circ C = 273 + 20 = 293 \, \text{K} \)

### Step 2: Calculate the Pressure at the Bottom of the Lake
The pressure at the bottom of the lake can be calculated using the formula:
\[ P_1 = P_0 + \rho g h \]
Where:
- \( P_0 \) is the atmospheric pressure (approximately \( 1.01 \times 10^5 \, \text{Pa} \))
- \( \rho \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \))
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))

Calculating \( P_1 \):
\[ P_1 = 1.01 \times 10^5 + (1000 \times 9.81 \times 40) \]
\[ P_1 = 1.01 \times 10^5 + 392400 \]
\[ P_1 = 492400 \, \text{Pa} \]

### Step 3: Calculate the Pressure at the Surface
At the surface of the lake, the pressure is simply the atmospheric pressure:
\[ P_2 = P_0 = 1.01 \times 10^5 \, \text{Pa} \]

### Step 4: Use the Ideal Gas Law to Find the Volume at the Surface
Using the relationship from the ideal gas law, we can express the volumes and pressures at the two states:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Rearranging to find \( V_2 \):
\[ V_2 = V_1 \cdot \frac{P_1}{P_2} \cdot \frac{T_2}{T_1} \]

### Step 5: Substitute the Values
Substituting the known values:
\[ V_2 = 20 \, \text{cm}^3 \cdot \frac{492400}{1.01 \times 10^5} \cdot \frac{293}{277} \]

Calculating \( \frac{P_1}{P_2} \):
\[ \frac{492400}{1.01 \times 10^5} \approx 4.87 \]

Calculating \( \frac{T_2}{T_1} \):
\[ \frac{293}{277} \approx 1.06 \]

Now substituting these ratios back into the volume equation:
\[ V_2 = 20 \cdot 4.87 \cdot 1.06 \]
\[ V_2 \approx 20 \cdot 5.16 \approx 103.2 \, \text{cm}^3 \]

### Step 6: Final Answer
The volume of the bubble just before it reaches the surface is approximately:
\[ \boxed{105 \, \text{cm}^3} \]