A cylinder that has a 50.0-cm radius and is 50.0 cm deep is filled with air at 2

Question

A cylinder that has a 50.0-cm radius and is 50.0 cm deep is filled with air at 20.0°C and 1.00 atm. A 25.0-kg piston is now lowered into the cylinder, compressing the air trapped
inside as it takes equilibrium height hi. Finally, a 30.0-kg dog stands on the piston,
further compressing the air, which remains at 20°C. (a) How far down (Dh ) does
the piston move when the dog steps onto it? (b) To what temperature should the gas be warmed
to raise the piston and dog back to hi ?

Explanation / Answer

As the temperature in the whole process remains the same...It is an iso-Thermal process So, PV = constant Initially for the equlibrium of the piston of 25.0 kg weight...its weight must be equal to pressure acting on the piston... ==> mg = Force acting = Pressure difference x Aread of the piston = P x p(.5)2 ==> 25 x 9.8 = P x p x .25 ==> P = 311.944 N/m2 Initail volume before keeping the piston = ( p(.5)^2)(.5) = .39269 m3 Initial pressure = 1 atm = 101010.1 N/m2 So final pressure = 101010.1 + 311.944 = 101322.044 As height after placing the piston = hi ===> Final volume = ( p(.5)^2)hi = 0.7854hi Sumbstituting in P1V1 = P2V2 ==> 101010.1 x .39269 = 101322.044 x 0.7854hi ==> hi = 0.49845 m Similarly after a dog of 30 kg weight sits Pressure difference = ( 25 +30 ) x 9.8/(p x .25) = 686.2761 N/m2 Final pressure = 101010.1 + 686.2761 = 101696.3761 N/m2 Sumbstituting in P1V1 = P2V2 ==> 101010.1 x .39269 = 101696.3761 x 0.7854hf ==> hf = 0.49661 m So the piston moves by (0.49845 -0.49661) = 0.0018368 m = 1.8368 mm after the dog gets on the piston. Now for the second part....The dog will be staying on the piston...So the pressure will be constant...Hence iso-baric process......So, V/T = const initial volume = 0.7854hf = 0.7854 x 0.49661 = 0.39004 m3 Initail temp = 20 C = 273 + 20 k = 293 K Final volume = full volume = ( p(.5)2)(.5) = .39269 m3 Applying V1/T1 = V2/T2 ==> 0.39004/293 = .39269/T ==> T = 294.9926 K = 21.9926 C

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