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 after it has moved from a depth in water, we can use the Ideal Gas Law and the concept of combined gas laws. Here's a step-by-step solution:

### Step 1: Understand the Initial and Final Conditions
We have the following conditions:
- Initial temperature (T1) = 17°C = 290 K (convert to Kelvin by adding 273)
- Initial pressure (P1) = 1.8 atm
- Final temperature (T2) = 27°C = 300 K
- Final pressure (P2) = 1 atm

### Step 2: Use the Combined Gas Law
The combined gas law states that:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Where:
- \(P\) = pressure
- \(V\) = volume
- \(T\) = temperature in Kelvin

### Step 3: Rearranging the Equation
We need to find the ratio of the final volume (V2) to the initial volume (V1). Rearranging the combined gas law gives us:
\[ \frac{V_2}{V_1} = \frac{P_1}{P_2} \times \frac{T_2}{T_1} \]

### Step 4: Substitute the Values
Now, substituting the known values into the equation:
\[ \frac{V_2}{V_1} = \frac{1.8 \, \text{atm}}{1 \, \text{atm}} \times \frac{300 \, \text{K}}{290 \, \text{K}} \]

### Step 5: Calculate the Ratio
Calculating the right-hand side:
\[ \frac{V_2}{V_1} = 1.8 \times \frac{300}{290} \]

Calculating \(\frac{300}{290}\):
\[ \frac{300}{290} \approx 1.0345 \]

Now, multiply:
\[ \frac{V_2}{V_1} = 1.8 \times 1.0345 \approx 1.86 \]

### Step 6: Conclusion
Thus, the final volume \(V_2\) is approximately 1.86 times the initial volume \(V_1\). Therefore, the volume of the bubble at the surface will become:
\[ V_2 \approx 1.86 \, V_1 \]