An air bubble of volume `1.0 cm^(3)` rises from the bottom of a lake 40 m deep at a temperature of `12^(@) C`. To what volume does it grow when it reaches the surface, which is at a temperature of `35^(@) C`. ? Given `1 atm = 1.01 xx 10^(5) Pa`.

Options

A. `10.6xx10^(-6)m^(3)`
B. `5.3xx10^(-6)m^(3)`
C. `2.8xx10^(-6)m^(3)`
D. `15.6xx10^(-6)m^(3)`

Explanation

To solve the problem of the air bubble rising from the bottom of a lake, we can use the ideal gas law, which states that for an ideal gas, the relationship between pressure (P), volume (V), and temperature (T) is given by:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial volume of the bubble, \( V_1 = 1.0 \, \text{cm}^3 = 1.0 \times 10^{-6} \, \text{m}^3 \)
- Depth of the lake, \( h = 40 \, \text{m} \)
- Temperature at the bottom, \( T_1 = 12^\circ C = 12 + 273 = 285 \, \text{K} \)
- Temperature at the surface, \( T_2 = 35^\circ C = 35 + 273 = 308 \, \text{K} \)
- Atmospheric pressure, \( P_{\text{atm}} = 1.01 \times 10^5 \, \text{Pa} \)
- Density of water, \( \rho = 10^3 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Pressure at the Bottom of the Lake:**
The pressure at the bottom of the lake can be calculated using the formula:
\[ P_1 = P_{\text{atm}} + \rho g h \]
Substituting the values:
\[ P_1 = 1.01 \times 10^5 + (10^3)(10)(40) = 1.01 \times 10^5 + 4 \times 10^5 = 5.01 \times 10^5 \, \text{Pa} \]

3. **Identify the Pressure at the Surface:**
At the surface, the pressure is simply the atmospheric pressure:
\[ P_2 = P_{\text{atm}} = 1.01 \times 10^5 \, \text{Pa} \]

4. **Use the Ideal Gas Law to Find the Final Volume \( V_2 \):**
Rearranging the ideal gas law gives:
\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]
Substituting the known values:
\[ V_2 = \frac{(5.01 \times 10^5)(1.0 \times 10^{-6})(308)}{(1.01 \times 10^5)(285)} \]

5. **Calculate \( V_2 \):**
Performing the calculations step-by-step:
- Calculate the numerator:
\[ 5.01 \times 10^5 \times 1.0 \times 10^{-6} \times 308 = 154.708 \, \text{Pa m}^3 \]
- Calculate the denominator:
\[ 1.01 \times 10^5 \times 285 = 28885.5 \, \text{Pa} \]
- Now, divide the numerator by the denominator:
\[ V_2 = \frac{154.708}{28885.5} \approx 5.35 \times 10^{-6} \, \text{m}^3 \]

6. **Convert \( V_2 \) to cm³:**
Since \( 1 \, \text{m}^3 = 10^6 \, \text{cm}^3 \):
\[ V_2 \approx 5.35 \, \text{cm}^3 \]

### Final Answer:
The volume of the air bubble when it reaches the surface is approximately **5.35 cm³**.