A balloon contains 14.0 L of air at 760 torr. What will be the volume of the balloon when it is taken to a depth of 10 ft. in a swimming pool? Assume that the temperature of the air and water are equal. (density : Hg=13.6g/mL.)

Options

A. 11
B. 11.3
C. 10
D. 10.8

Explanation

To find the volume of the balloon at a depth of 10 feet in a swimming pool, we can use the principles of gas laws, specifically Boyle's Law, which states that for a given mass of gas at constant temperature, the product of pressure and volume is constant (P1V1 = P2V2).

### Step-by-Step Solution:

**Step 1: Identify the initial conditions.**
- Initial volume (V1) = 14.0 L
- Initial pressure (P1) = 760 torr

**Step 2: Convert the initial pressure to Pascals.**
- 1 torr = 133.322 Pa
- P1 = 760 torr × 133.322 Pa/torr = 101325 Pa

**Step 3: Calculate the pressure at a depth of 10 feet.**
- First, convert feet to meters:
- 10 ft = 10 × 0.3048 m = 3.048 m
- Calculate the pressure increase due to the water column:
- Pressure increase (hρg) = height × density of water × gravitational acceleration
- Density of water = 1000 kg/m³
- Gravitational acceleration (g) = 9.81 m/s²
- hρg = 3.048 m × 1000 kg/m³ × 9.81 m/s² = 29953.88 Pa
- Total pressure at depth (P2) = atmospheric pressure + pressure due to water
- P2 = 101325 Pa + 29953.88 Pa = 131278.88 Pa

**Step 4: Use Boyle's Law to find the final volume (V2).**
- According to Boyle's Law: P1V1 = P2V2
- Rearranging gives us: V2 = (P1V1) / P2
- Substitute the values:
- V2 = (101325 Pa × 14.0 L) / 131278.88 Pa
- V2 = 10.8 L (approximately)

### Final Answer:
The volume of the balloon when taken to a depth of 10 feet in a swimming pool is approximately **10.8 liters**.

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