Question

Two kilograms of CO₂ gas is contained in a piston-cylinder assembly at a pressure of 6.5 bar and a temperature of 300 K. The piston has a mass of 5000 kg and surface area of 1m². The friction of the piston on the walls is significant and cannot be ignored. The atmospheric pressure is 1.01325 bar. The latch holding the piston in position is suddenly removed and the gas is allowed to expand. The expansion is arrested when the volume is double the original volume. Determine the work appearing in the surroundings. Will it be the same as the work done by the gas?

Answer 1

Molecular weight of CO2 = (12 + 2*16) = 44 kg/kmol.

So, 2 kg CO2 = 2 kg * 1kmol/44 kg = 1/22 kmol of CO2 = n, number of moles in the system.

Initial system pressure, Pi = 6.5 bar = 6.5 bar * 100 kPa/1bar = 650 kPa.

Initial temperature of system = Ti = 300K.

By ideal gas law, P.V = nRT, where R = 8.314 kPa.m3/kmol.K

so, initial volume = Vi = n.R.Ti/Pi = 1/22 * 8.314 * 300 / 650 = 0.1744 m3.

Final volume after expansion = Vf = 2*Vi.

Now, work done by a gas system on the surrounding = surrounding pressure * (final volume - initial volume).

Surrounding pressure = atmospheric pressure + pressure applied by piston.

atmospheric pressure = 1.01325 bar = 101.325 kPa.

pressure by piston = mass of piston * gravity / surface area = 5000 kg* 9.81 m/sec2 / 1m2 = 49050 Pa = 49.05 kPa.

Surrounding pressure, Ps = 101.325 + 49.05 = 150.375 kPa.

So, work done by gas on the surrounding = Ps * (Vf - Vi) = Ps*(2Vi - Vi) = 150.375 kPa * 0.1744 m3 = 26.2254 kJ.

= the work appearing in the surrounding.

Gas has to done some work to overcome the frictional resistance. So, total work done by the gas = work done on the surrouonding + work done to overcome friction.