An air bubble formed under water at temperature `17^oC` and the pressure 1.8 atm started to move upwards and reaches the surface where temperature is `27^oC` and pressure in 1 atm. The volume of the bubble at the surface will become

Options

A. 2.32 times of the initial volume
B. 1.86 times of the initial volume
C. 0.5 times of the initial volume
D. 1.51 times of the initial volume

Answer

To solve the problem of finding the volume of an air bubble at the surface of the water, we can use the combined gas law derived from the ideal gas law. The combined gas law states that:

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

Where:
- \(P_1\) = initial pressure
- \(V_1\) = initial volume
- \(T_1\) = initial temperature (in Kelvin)
- \(P_2\) = final pressure
- \(V_2\) = final volume
- \(T_2\) = final temperature (in Kelvin)

### Step 1: Convert temperatures from Celsius to Kelvin
To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]

- For \(T_1 = 17°C\):
\[ T_1 = 17 + 273.15 = 290.15 \, K \approx 290 \, K \]

- For \(T_2 = 27°C\):
\[ T_2 = 27 + 273.15 = 300.15 \, K \approx 300 \, K \]

### Step 2: Substitute the known values into the combined gas law
We know:
- \(P_1 = 1.8 \, atm\)
- \(P_2 = 1 \, atm\)
- \(T_1 = 290 \, K\)
- \(T_2 = 300 \, K\)

Substituting these values into the equation:
\[ \frac{1.8 \, atm \cdot V_1}{290 \, K} = \frac{1 \, atm \cdot V_2}{300 \, K} \]

### Step 3: Rearrange the equation to solve for \(V_2\)
Cross-multiplying gives:
\[ 1.8 \, atm \cdot V_1 \cdot 300 \, K = 1 \, atm \cdot V_2 \cdot 290 \, K \]

Rearranging to isolate \(V_2\):
\[ V_2 = \frac{1.8 \cdot 300}{290} \cdot V_1 \]

### Step 4: Calculate \(V_2\)
Now, calculate the numerical value:
\[ V_2 = \frac{540}{290} \cdot V_1 \approx 1.86 \cdot V_1 \]

### Conclusion
Thus, the volume of the bubble at the surface will become approximately \(1.86\) times the initial volume \(V_1\).

### Final Answer
The volume of the bubble at the surface will become \(1.86 \, V_1\).

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