A balloon filled with an ideal is taken from the surface of the sea deep to a depth of 100 m. What will be its volume in terms of its original volume ?

Explanation

To solve the problem of finding the volume of a balloon filled with an ideal gas when taken from the surface of the sea to a depth of 100 meters, we can use the principles of gas laws, particularly Boyle's Law. Boyle's Law states that for a given mass of gas at constant temperature, the pressure and volume of the gas are inversely related.

### Step-by-Step Solution:

**Step 1: Understand the Initial Conditions**
- At the surface of the sea, the pressure (P1) is equal to atmospheric pressure, which is approximately 1 atm (or 101.3 kPa).
- Let the initial volume of the balloon be V1.

**Step 2: Calculate the Pressure at 100 m Depth**
- The pressure increases with depth in a fluid. The pressure at a depth (h) in a fluid can be calculated using the formula:
\[ P = P_0 + \rho g h \]
where:
- \( P_0 \) = atmospheric pressure (1 atm)
- \( \rho \) = density of seawater (approximately 1025 kg/m³)
- \( g \) = acceleration due to gravity (approximately 9.81 m/s²)
- \( h \) = depth (100 m)

**Step 3: Calculate the Total Pressure at Depth**
- Substituting the values:
\[ P = 1 \, \text{atm} + (1025 \, \text{kg/m}^3)(9.81 \, \text{m/s}^2)(100 \, \text{m}) \]
- Convert 1 atm to Pascals: \( 1 \, \text{atm} = 101325 \, \text{Pa} \)
- Calculate the hydrostatic pressure:
\[ \text{Hydrostatic Pressure} = 1025 \times 9.81 \times 100 \approx 1000000 \, \text{Pa} = 10 \, \text{atm} \]
- Therefore, the total pressure at 100 m depth:
\[ P_2 = 1 \, \text{atm} + 10 \, \text{atm} = 11 \, \text{atm} \]

**Step 4: Apply Boyle's Law**
- According to Boyle's Law:
\[ P_1 V_1 = P_2 V_2 \]
Rearranging gives:
\[ V_2 = \frac{P_1 V_1}{P_2} \]
Substituting the values:
\[ V_2 = \frac{1 \, \text{atm} \cdot V_1}{11 \, \text{atm}} = \frac{V_1}{11} \]

**Step 5: Conclusion**
- The volume of the balloon at a depth of 100 m will be:
\[ V_2 = \frac{V_1}{11} \]
- Thus, the volume of the balloon decreases to one eleventh of its original volume.