Question

A bubble having a diameter of 1.00 cm is released from the bottom of a swimming pool where the depth is 5.00 m. What will the diameter of the bubble be when it reaches the surface? The temperature of the water at the surface is 20.0^oC, whereas it is 15.0^oC at the bottom. (The density of water is 1.00 x 10^3 kg/m3.)

Answer 1

The diameter of the bubble be when it reaches the surface is 1.15 cm.

The given parameters;

• diameter of the bubble at bottom, d = 1.00 cm
• radius of the bubble at the bottom, r = 0.5 cm = 0.005 m
• depth of the pool, h = 5 m
• temperature of the water at bottom, T₁ = 15 ⁰C = 15 + 273 = 288 K
• temperature of the water at top, T₂ = 20 ⁰C = 20 + 273 = 293 K

The total pressure at bottom of the pool is calculated as follows;

$P_1 = \rho gh \ + P_{atm}\\\\P_1 = (1000 \times 9.8 \times 5) \ + \ (1.01 \times 10^5)\\\\P_1 = 150,000 \ Pa$

The pressure of water at the surface (top);

$P_2 = 1 \ atm = 1.01 \times 10^{5} \ Pa$

The volume of the bubble at the bottom is calculated as follows;

$V_1 = \frac{4}{3} \pi r^3\\\\V_1 = \frac{4}{3}\pi \times (0.005)^3\\\\V_1 = 5.24 \times 10^{-7} \ m^3$

Apply general gas law to determine the volume of the bubble at the top;

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \\\\V_2 = \frac{P_1V_1T_2}{T_1 P_2} \\\\V_2 = \frac{(150,000) \times (5.24 \times 10^{-7}) \times (293)}{(288)\times (1 \times 10^{5})} \\\\V_2 = 7.996 \times 10^{-7} \ m^3$

The radius of the bubble at the surface is calculated as follows;

$V_2 = \frac{4}{3} \pi r^3\\\\4\pi r^3 = 3V_2\\\\r^3 = \frac{3V_2}{4\pi} \\\\r^3 = \frac{3 \times 7.996 \times 10^{-7} }{4\pi } \\\\r^3 = 1.909 \times 10^{-7} \\\\r = \sqrt[3]{(1.909 \times 10^{-7})} \\\\r = 0.00576 \ m$

The diameter of the bubble is calculated as follows;

d = 2r = 2(0.00576 m)

d = 0.0115 m

d = 1.15 cm

Thus, the diameter of the bubble be when it reaches the surface is 1.15 cm.

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