An air bubble of radius r is formed at a depth h below the surface of water. If P0 is the atmospheric pressure, ρ is the density of water, and T is the surface tension of water, then the pressure inside the bubble is

Engineering

Physics

Liquid Pressure

Surface Tension and Surface Energy

Question

An air bubble of radius r is formed at a depth h below the surface of water. If P₀ is the atmospheric pressure, ρ is the density of water, and T is the surface tension of water, then the pressure inside the bubble is

$P_{o} + hρg + \frac{4 T}{r}$

$P_{o} + \frac{2 T}{r}$

$P_{0} + \frac{4 T}{r}$

$P_{o} + hρg + \frac{2 T}{r}$

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An air bubble inside water has two pressure contributions: hydrostatic pressure from depth h, and excess pressure due to surface tension. The hydrostatic pressure outside the bubble is P₀ + ρgh. For an air bubble (liquid-gas interface), the excess pressure is 2T/r. Since this excess pressure acts inward, the pressure inside the bubble must be greater than outside to prevent collapse.

Therefore, the total pressure inside is: $P_{o} + ρ g h + \frac{2 T}{r}$

Final answer: $P_{o} + ρ g h + \frac{2 T}{r}$