A fixed thermally conducting cylinder has a radius R and height L0. The cylinder is open at its bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is P0.The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then be
Question
A fixed thermally conducting cylinder has a radius R and height L0. The cylinder is open at its bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is P0.The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then be
Solution
The pressure in the cylinder can be calculated using the formula for pressure in a fluid column, which is P = P0 + ρgh, where P0 is the atmospheric pressure, ρ is the density of the fluid (in this case, the gas in the cylinder), g is the acceleration due to gravity, and h is the height of the fluid column.
In the initial state, the height of the gas column is L and the pressure is P0. When the piston is pulled out and held at a distance 2L from the top, the height of the gas column doubles. However, since the volume of the gas doubles as well, the density of the gas halves (since density is mass/volume and the mass of the gas remains constant).
Therefore, in the final state, the pressure P' in the cylinder is given by P' = P0 + (ρ/2)g(2L) = P0 + ρgL = P.
So, the pressure in the cylinder between its top and the piston remains the same as the initial pressure P0 when the piston is pulled out slowly and held at a distance 2L from the top.
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