Question 18 of 18advanced🔧 ApplyNumerical5 marks

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 2.50×102.50 \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 1.01×101.01 \times 10⁵ Pa).

Correct Answer

(i) The excess pressure inside the soap bubble is 20 Pa.

(ii) The pressure inside the air bubble at 40.0 cm depth is 1.06×101.06 \times 10⁵ Pa.

Exercise: EXERCISES | Q: 9.20 | (Page 22)
For More Understanding

Explanation

This question tests understanding of excess pressure in bubbles. A soap bubble has TWO liquid surfaces (inner and outer), so excess pressure = 4S/r. An air bubble inside a liquid has only ONE liquid surface, so excess pressure = 2S/r. For the air bubble, the outside pressure equals atmospheric pressure plus hydrostatic pressure due to the liquid column above it.

Solution Steps

  1. Step 1: For soap bubble, excess pressure = 4S/r = 4×2.504 \times 2.50 × 10⁻² / (5.00×105.00 \times 10⁻³) = 20 Pa

  2. Step 2: For air bubble, excess pressure = 2S/r = 2×2.502 \times 2.50 × 10⁻² / (5.00×105.00 \times 10⁻³) = 10 Pa

  3. Step 3: Calculate density of soap solution = 1.20×10001.20 \times 1000 = 1200 kg m⁻³

  4. Step 4: Calculate hydrostatic pressure = hρg = 0.40×12000.40 \times 1200 × 9.80 = 4704 Pa

  5. Step 5: Calculate outside pressure P₀ = 1.01×101.01 \times 10⁵ + 4704 = 1.057×101.057 \times 10⁵ Pa

  6. Step 6: Calculate pressure inside bubble Pᵢ = P₀ + excess pressure = 1.057×101.057 \times 10⁵ + 10 = 1.06×101.06 \times 10⁵ Pa

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