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A soft plastic bottle is filled with water (density 1 g/cc) and holds an inverted glass test-tube that traps some air (ideal gas), as shown. The test-tube has mass 5 g and is made of thick glass of density 2.5 g/cc. The bottle is initially sealed at atmospheric pressure p0 = 10⁵ Pa, with the trapped air volume v0 = 3.3 cc. When the bottle is squeezed at constant temperature, the internal pressure rises and the trapped-air volume shrinks. The test-tube just begins to sink (without changing orientation) when the pressure has risen to p0 + Delta p, where the trapped-air volume is v0 - Delta v. Write Delta v = X cc and Delta p = Y x 10³ Pa. Find X (the change in trapped-air volume at which the tube starts to sink).
- 0.3
- 0.5
- 0.7
- 1.0
Correct answer: 0.3
Solution
This is a Cartesian diver. Glass volume = 5/2.5 = 2 cc. The system (test-tube plus trapped air) floats while its average density is below water's. It just sinks when the total weight equals the buoyant force from the displaced water (glass volume + trapped air volume). Weight = mass of glass * g (the trapped air mass is negligible) = 5 g * g. Buoyancy = rho_water * (V_glass + v) * g = 1 * (2 + v) * g. Setting weight = buoyancy at the threshold: 5 = (2 + v), so v = 3 cc. Thus the air volume at sinking is 3 cc, and Delta v = v0 - v = 3.3 - 3 = 0.3 cc, so X = 0.3.
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