NINE kg of AIR is initially contained in a rigid tank at 1000 kPa and ONE kg of
ID: 1861407 • Letter: N
Question
NINE kg of AIR is initially contained in a rigid tank at 1000 kPa and ONE
kg of AIR occupies space in a piston-cylinder device. The valve in the Air Tank pipe connecting the tank to the cylinder is accidentally bumped slightly open causing air to slowly leak from the tank to the cylinder. The heavy piston moves freely (without friction) as shown. The process occurs slowly enough that there is sufficient heat transfer to maintain a uniform and constant temperature of 300 K. Neglect changes in kinetic & potential energy of the air. The surrounding atmospheric pressure is 100 kPa. Given: piston weight = 120 kN, area = 0.60 m2.
FIND: The mass and volume of air in the cylinder when equilibrium is finally achieved and the total heat transfer required, QIN (kJ). NOTE: Complete a p-V diagram, fill out a property table and clearly show your analysis.
Explanation / Answer
procedure can be as follows
The molar mass of dry air is 28.97 grams/mole,
so 1 kg = 34.52 moles and 10 kg = 345.2 moles.
The pressure in the cylinder at all times will be
100 kPa plus (120 kN)/(0.60 m^2) = 300 kPa
If the 9 kg of air in the tank were initially at 1000 kPa and 300K,
the volume of the tank is nRT/p
= (310.7 mol)(8.314 J/mol K)(300 K)/(1000000 Pa) = 0.775 m^3
The final volume of the whole 10 kg of air will be
nRT/p = (345.2 mol)(8.314 J/mol K)(300 K)/(300000 Pa) = 2.87 m^3
The 1 kg of air that was originally in the cylinder occupied a volume of 0.287 m^3.
The mass of air in the cylinder when equilibrium is achieved is
(10 kg)(2.87 m^3 - 0.775 m^3)/(2.87 m^3) = 7.3 kg
And the final volume of the cylinder is 2.87 m^3 - 0.775 m^3 = 2.1 m^3
The height of the part of the cylinder below the piston
changed by (2.095 m^3 - 0.287 m^3)/(0.60 m^2) = 3.0 meters,
so the 120 kN piston was lifted through 3 meters,
which means 360 kJ of work were done by the gas
and 360 kJ of heat must've been supplied to keep
the temperature from falling.
What at first bothered me about this answer is that
it didn't seem to match W = p delta-V, but then I
realized I should probably use gauge pressure in the
calculation of p delta-V, so then it does work out:
(200 kPa)(1.8 m^3) = 360 kJ.
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