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Q10.20:
What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is $2.50 \times 10^{-2} N m^{-1}$ ? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble ? (1 atmospheric pressure is $1.01 \times 10^5 Pa$).

Solution :

Part 1: Excess Pressure Inside a Soap Bubble

Given Variables:

  • Radius of the bubble, $r = 5.00 \text{ mm} = 5.00 \times 10^{-3} \text{ m}$
  • Surface tension of soap solution, $S = 2.50 \times 10^{-2} \text{ N m}^{-1}$

Step 1: Formula Selection

A soap bubble possesses two interfaces with air (inner and outer). Thus, the excess pressure ($P_e$) is given by [Per the Young-Laplace equation for a thin film]:

$P_e = \frac{4S}{r}$

Step 2: Numerical Computation

$P_e = \frac{4 \times (2.50 \times 10^{-2} \text{ N m}^{-1})}{5.00 \times 10^{-3} \text{ m}}$

$P_e = \frac{10 \times 10^{-2}}{5.00 \times 10^{-3}} = 2 \times 10^{1} \text{ Pa} = 20.0 \text{ Pa}$

r Soap Bubble (2 surfaces)

Part 2: Pressure Inside an Air Bubble at Depth

Given Variables:

  • Depth, $h = 40.0 \text{ cm} = 0.40 \text{ m}$
  • Relative density of solution, $\rho_{rel} = 1.20 \implies \rho = 1.20 \times 10^3 \text{ kg m}^{-3}$
  • Atmospheric pressure, $P_a = 1.01 \times 10^5 \text{ Pa}$
  • Acceleration due to gravity, $g \approx 9.8 \text{ m s}^{-2}$

Step 1: Formula Selection

For an air bubble inside a liquid, there is only one surface in contact with the liquid. The pressure inside the bubble ($P_{in}$) is the sum of atmospheric pressure, hydrostatic pressure, and the excess pressure [Per Pascal's Law and the Young-Laplace Equation]:

$P_{in} = P_a + h\rho g + \frac{2S}{r}$

Step 2: Numerical Computation

1. Hydrostatic pressure: $P_h = h\rho g = 0.40 \times 1200 \times 9.8 = 4704 \text{ Pa}$

2. Excess pressure for air bubble: $P_{ea} = \frac{2S}{r} = \frac{2 \times 2.50 \times 10^{-2}}{5.00 \times 10^{-3}} = \frac{0.05}{0.005} = 10 \text{ Pa}$

3. Total internal pressure: $P_{in} = 1.01 \times 10^5 + 4704 + 10$

$P_{in} = 101000 + 4714 = 105714 \text{ Pa}$

Rounding to significant figures: $P_{in} \approx 1.06 \times 10^5 \text{ Pa}$

Final Solution: The excess pressure in the soap bubble is $20.0 \text{ Pa}$, and the pressure inside the air bubble at the specified depth is $1.06 \times 10^5 \text{ Pa}$.


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