Question

Sketch a PV diagram and find the work done by the gas during the following stages. (Draw the PV diagram on paper. Your instructor may ask you to turn in this work.)

(a) A gas is expanded from a volume of 1.0 L to 2.6 L at a constant pressure of 3.1 atm.
J

(b) The gas is then cooled at constant volume until the pressure falls to 1.8 atm.
J

(c) The gas is then compressed at a constant pressure of 1.8 atm from a volume of 2.6 L to 1.0 L. [Note: Be careful of the signs.]
J

(d) The gas is heated until its pressure increases from 1.8 atm to 3.1 atm at a constant volume.
J

(e) Find the net work done during the complete cycle.
J

Answer 1

Observe my image:
http://img682.imageshack.us/img682/1225/pvrectangle.gif

I indicate Piston/cylinder assemblies of each state, plus how each transition is accomplished in in-between piston-cylinder assemblies. I've also included a P-V sketch of the cycle, although your instructor may want a nicely drawn graph which has proper scale on the axes and good lableing.

Identify each type of process:
1 to 2: isobaric heating/expansion (indicated by piston lifting a lot of sand due to heat of the flame)
2 to 3: isochoric cooling/de-pressurization (indicated by stops constraining piston and air cooling gas)
3 to 4: isobaric cooling/compression (indicated by piston falling under sand load and cooled by air)
4 to 1: isochoric heating (indicated by stops and flame)

We are interested in finding work done by gas (sign convention of energy is correct) in each stage.

Recall how work is found by integrating pressure relative to volume. In common cases, the integration has already been done for you. In the isobaric case, it turns in to a simple multiplication and subtraction, because of the integral of a constant. Think area under the P-V trajectory. What's the area of a rectangle? For the isochoric cases, it is trivially equal to zero because the piston moves nowhere.

In first process:
W12 = P1*(V2 - V1)

In second process:
W23 = 0

In third process:
W34 = P3*(V4 - V3)

In final process
W41 = 0

Data:
P1:=3.1 atm; P3:=1.8 atm; V1:=1.0 L; V2:=2.6 L;

Equating of pressures and volumes
P2:=P1; P4:=P3; V4:=V1; V3:=V2;

Results:
Part A: W12 = 7.68 Liter-atmospheres
Part B: W23 = 0
Part C: W34 = -4.32 Liter-atmospheres
Part D: W41 = 0

Convert to Joules (1 L-atm = 101.325 Joules):
Part A: W12 = 778.176 Joules
Part B: W23 = 0
Part C: W34 = -437.724 Joules
Part D: W41 = 0

Negative sign on W34 indicates work being done on gas.

Net work: W_net = W12 + W23 + W34 + W41

W_net = 340.452 Joules

Should you want an expression:
W_net = P1*(V2 - V1) + P3*(V4 - V3)

Simplified with equating of pressures and volumes:
W_net = (P1 - P3)*(V2 - V1), which is the area enclosed by the cycle trajectory on the P-V diagram