Question

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 × 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 × 105 Pa).

Answer 1

concept :-
The excess pressure inside the soap bubble is given by ∆P = 4S/R
where, S is surface tension of the soap bubble . and R is radius of soap bubble .
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Given ,
surface tension of the soap bubble (S) = 2.5 × 10^-2 N/m
radius of soap bubble (R) = 5mm = 5× 10^-3 m
Excess pressure inside the soap bubble = 4S/R
= 4 × 2.5 × 10^-2 /5 × 10^-3
= 20 Pa

Excess pressure inside the air bubble = 2S/R
= 2 × 2.5 × 10^-2 /5 × 10^-3
= 10 Pa

pressure inside the air bubble = atmospheric pressure + pressure due to soap solution + Excess pressure inside the bubble
= 1.01 × 10^5 Pa + dgh + 10 Pa
= 1.013 × 10^5 + (1.2 × 10³×9.8 × 0.4) + 10
= 1.05714 × 10^5 Pa

Answer 2

Answer:

Explanation:

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