Question

An air bubble of volume 20 cm3 is at the bottom of a lake 40 m deep, where the temperature is 4.00C. The bubble rises to the surface, which is at a temperature of 200C. Take the temperature of the bubble’s air to be the same as that of the surrounding water. Just as the bubble reaches the surface, what is its volume?

Answer 1

The volume of the bubble at the surface is 34.15 cm³

Data obtained from the question

•Initial volume (V₁) = 20 cm³

•Initial temperature (T₁) = 4 °C = 4 + 273 = 277 K

•Final temperature (T₂) = 200 °C = 200 + 273 = 473 K

•Final volume (V₂) =?

The final volume of the bubble can be obtained by using the Charles' law equation as shown below:

V₁ / T₁ = V₂ / T₂

20 / 277 = V₂ / 473

Cross multiply

277 × V₂ = 20 × 473

277 × V₂ = 9640

Divide both side by 277

V₂ = 9640 / 277

V₂ = 34.15 cm³

Thus, the volume of the bubble at the surface is 34.15 cm³

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