A small bubble rises from the bottom of a lake, where the temperature and pressure are `8^(@)C` and `6.0 atm`, to the water's surface, where the temperature is `25^(@)C` and pressure is `1.0 atm`. Calculate the final volume of the bubble if its initial volume was `2mL`.

Options

A. 14 mL
B. 12.72 mL
C. 11.31 mL
D. 15 mL

Explanation

To solve the problem of finding the final volume of a bubble rising from the bottom of a lake to the surface, we can use the combined gas law, which relates pressure, volume, and temperature. Here are the steps to find the solution:

### Step 1: Convert Temperatures to Kelvin
The temperatures given in Celsius need to be converted to Kelvin since gas laws require absolute temperature.

- **Initial Temperature (T1)**:
\[ T1 = 8^\circ C + 273 = 281 \, K \]

- **Final Temperature (T2)**:
\[ T2 = 25^\circ C + 273 = 298 \, K \]

### Step 2: Identify Given Values
From the problem statement, we have:
- Initial Pressure (P1) = 6.0 atm
- Initial Volume (V1) = 2 mL
- Final Pressure (P2) = 1.0 atm
- Initial Temperature (T1) = 281 K
- Final Temperature (T2) = 298 K

### Step 3: Use the Combined Gas Law
The combined gas law can be expressed as:
\[ \frac{P1 \cdot V1}{T1} = \frac{P2 \cdot V2}{T2} \]
We need to rearrange this equation to solve for the final volume (V2):
\[ V2 = \frac{P1 \cdot V1 \cdot T2}{P2 \cdot T1} \]

### Step 4: Substitute the Values
Now, substitute the known values into the equation:
\[ V2 = \frac{(6.0 \, \text{atm}) \cdot (2 \, \text{mL}) \cdot (298 \, K)}{(1.0 \, \text{atm}) \cdot (281 \, K)} \]

### Step 5: Calculate V2
Now we can perform the calculation:
\[ V2 = \frac{(6.0) \cdot (2) \cdot (298)}{(1.0) \cdot (281)} \]
Calculating the numerator:
\[ 6.0 \cdot 2 \cdot 298 = 3588 \]
Calculating the denominator:
\[ 1.0 \cdot 281 = 281 \]
Now divide the two results:
\[ V2 = \frac{3588}{281} \approx 12.77 \, \text{mL} \]

### Step 6: Final Answer
The final volume of the bubble when it reaches the surface is approximately:
\[ V2 \approx 12.77 \, \text{mL} \]