A sealed vertical cylinder of radius R and height h = 0.636 m is initially fille

Question

A sealed vertical cylinder of radius R and height h = 0.636 m is initially filled halfway with water, and the upper half is filled with air. The air is initially at standard atmospheric pressure, p0 = 1.01·105 Pa. A small valve at the bottom of the cylinder is opened, and water flows out of the cylinder until the reduced pressure of the air in the upper part of the cylinder prevents any further water from escaping. By what distance is the depth of the water lowered? (Assume that the temperature of water and air do not change and that no air leaks into the cylinder.)

Explanation / Answer

Here ,

let the depth of water remaining is d

h = 0.636 m

Po = 1.01 *10^5 Pa

Using ideal gas equation of air

Pressure * volume = constant

Pf * (h - d) = Po * h/2

Pf = 1.01 *10^5 * (0.636/2)/(0.636 - d)

Now , for the water to be in equilibrium

Po = Pf + p * g * d

1.01 *10^5 = 1.01 *10^5 * (0.636/2)/(0.636 - d) + 1000 * 9.8 * d

solving for d

d = 0.308 m

depth of lowered = 0.636/2 - 0.308

depth of lowered = 0.01 m

the depth of water is lowered by 0.01 m

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