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A gas bubble (n moles, ideal gas) rises in a liquid column. Taking R as the universal gas constant, with the given symbols for densities, pressures, heights, and the depth variable y, the buoyancy force on the bubble is:

1. rho1 nRgT0 ((P0 + rho0 gH)/(P0 + rho0 gy))^(2/5) / ((P0 + rho0 gH)^(7/5))
2. rho1 nRgT0 / ((P0 + rho0 gH)^(2/5) [P0 + rho0 g(H - y)]^(3/5))
3. rho1 nRgT0 ((P0 + rho0 gH)/(P0 + rho0 gy))^(3/5) / ((P0 + rho0 gH)^(8/5))
4. rho1 nRgT0 / ((P0 + rho0 gH)^(3/5) [P0 + rho0 g(H - y)]^(2/5))

Correct answer: rho1 nRgT0 / ((P0 + rho0 gH)^(2/5) [P0 + rho0 g(H - y)]^(3/5))

Solution

The buoyant force is rho1 * V * g where V is the bubble volume at depth corresponding to y. The bubble expands adiabatically (gamma = 7/5) as the surrounding pressure drops from (P0 + rho0 gH) at the bottom to [P0 + rho0 g(H - y)]. Combining the adiabatic relation with the ideal-gas law gives V proportional to T0/[(P0+rho0 gH)^(2/5) (P0+rho0 g(H-y))^(3/5)], producing the matching expression.

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