A sealed vertical cylinder of radius R and height h = 0.507 m is initially fille

Question

A sealed vertical cylinder of radius R and height h = 0.507 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

   Initial height of water column   h1 = h/2   =   0.2535   m    Area of corss section of cylinder   =   A    Initial pressure of air   P1 = 1.01 * 105 Pa    Initial volume of air   V1   = A * h/2   =   0.2535 * A         m3    Let the water falls by depth H, then final height of water column    h2 =   0.2535 - H    Final volume of air   V2   = (0.2535 + H) * A    And final pressure of air   =   P2    Since the process is isothermal   P1 * V1 = P2 * V2    1.01 * 105 * 0.2535 * A   =   P2 * (0.2535 + H) * A    P2   =   1.01 * 105 * 0.2535 / (0.2535 + H)    Outside the pressure is equal to atmospheric pressure, hence the pressure difference    P   =   1.01 * 105 -   1.01 * 105 * 0.2535 / (0.2535 + H)             =   1.01 * 105 * (1 - 0.2535 / (0.2535 + H))             =   1.01 * 105 * H / (0.2535 + H)    This pressure difference will create a force that balances the weight of water, then    P * A   =   m * g                   =   volume of water * * g                   =   A * (0.2535 - H) * * g    =>   P   =   (0.2535 - H) * * g    1.01 * 105 * H / (0.2535 + H)   =   (0.2535 - H) * * g    1.01 * 105 * H   =  (0.2535 + H) * (0.2535 - H) * 1000 * 9.8    101 * H   =   (0.25352 - H2) * 9.8    H2 + 10.306 * H - 0.0643   =   0    On solving above quadratic equation    H   =   6.23 * 10-3   m          =   6.23   mm        

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