Since the question concerns the entire apparatus, the system is taken as the gas, piston, and cylinder. No work is done during the process, because no force external to the system moves, and no heat is transferred through the vacuum surrounding the apparatus. (101.3)(0.2-0.1)kPaHence Q and W are zero, and the total energy the system does not change, Without further information we can say nothing about the distribution of energy among the parts of system. This may well be different than the initial distribution.
Example2.3
If the process described in Ex. 2.2 is repeated, not in a vacuum but in air at atmospheric pressure of 101.3kPa, what is the energy change of the apparatus? Assume the rate of heat exchange between the apparatus and the surrounding air is slow compared with the rate at which the process occurs.
Solution 2.3
The system is chosen exactly as before, but in this case work is done by the system in pushing back the atmosphere. This work is given by the product of the force exerted by atmospheric pressure on the back side of the piston and the displacement of the piston. If the area of piston is A, the force is F=PatmA. The displacement of the piston is equal to the volume change of the gas divided by the area of the piston, or△l=△Vt/A. Work is done by the system on the surroundings. By Eq.(1.1),
Work done by system=F △l=Patm△Vt kN
Since W is work done on the system, it is the negative of this result:
W=-10.13kN m=-10.13kJ
Heat transfer between the system and surrounding is also possible in this case, but the problem is worked for the instant after the process has occurred and before appreciable heat transfer has had time to take place. Thus Q is assumed to be zero in Eq. (2.2), giving △(Energy of the system)=Q + W=0-10.13=-10.13 kJ
The total energy of the system has decreased by an amount equal to work done on the surroundings.