An upright cylinder, 1.00 m tall and closed at its lower end, is fitted with a l

Question

An upright cylinder, 1.00 m tall and closed at its lower end, is fitted with a light piston that is free to slide (see the figure). Initially the piston is in the center. A cuplike cavity is formed by the top of the piston and the upper cylinder walls. Water is poured into the cavity until it is full. Assume that the lower portion of the cylinder contains an ideal gas at constant temperature. Suppose instead of water that we use another liquid say Mercury which has a density of 13600 kg/m3. At what fraction of the total height of the cylinder will the piston be when the cavity is full? (The tolerance is 0.001).

Explanation / Answer

1. The initial force on the lower face of the piston was equal to the piston's weight plus (area of piston x atmospheric pressure). So the initial pressure in the gas was: atmospheric pressure + (piston weight per unit area) = 101.3 kPa + (piston weight per unit area) The final pressure in the gas must be: atmospheric pressure + (piston weight per unit area) + (water weight per unit area) = 101.3 kPa + (piston weight per unit area) + (1000 kg/m^3)(g)(1m - h) But also, the final pressure must be related to the final volume by (0.5m) (p-initial) = h (p-final), or (0.5m)(101.3 kPa) + (0.5m)(piston wt/area) = h (101.3 kPa) + h(piston wt/area) + (h-h^2) (1000 kg/m^3) g This can't be solve without knowing the piston weight per unit area, but the "light piston" phrase suggests that we neglect it, so that 9800 h^2 - 9800h + 50760 = 101300 h For convenience I rewrite this as 0 = 98 h^2 - 1111 h + 506.5 h (in meters) = (1111/196) plus or minus sqrt(1111^2-384*506.5)/(196) = 5.668 m - 5.203 m = 36.5 cm

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