Example Problem with Complete Solution

4A-1 : Work for a Cycle Carried Out in a Closed System 6 pts
Air undergoes a three-process cycle.  Find the net work done for 2 kg of air if the processes are :
Process 1-2: constant pressure expansion
Process 2-3: constant volume    
Process 3-1: constant temperature compression      
               
Sketch the cycle path on a PV Diagram.
               
Data : T1 100 oC        
  T2 600 oC        
  P1 200 kPa        
               
Given:
T1
100 oC Find:
T2 600 oC Sketch cycle on a PV Diagram.
P1 200 kPa Wcycle = ??? kJ
m 2.0 kg
Assumptions: - The gas is a closed system
- Boundary work is the only form of work interaction
- Changes in kinetic and potential energies are negligible.
- Air behaves as an ideal gas.
  This must be verified at all three states.
Part a.) Figure 1
Part b.) Since Wcycle = W12 + W23 + W31, we will work our way around the cycle and calculate each work term along the way.
Step 1-2 is isobaric, therefore, the definition of boundary work becomes:
Eqn 1
We can simplify Eqn 1 using the fact that P2 = P1 and the Ideal Gas EOS :
Eqn 2
Eqn 3
We can determine the number of moles of air in the system from the given mass of air and its molecular weight.
Eqn 4
MWair 29 g/mole n 68.97 mole
Plug values into Eqn 3 : R 8.314 J/mole-K
W12 286.69 kJ
Because the volume is constant in step 2-3: W23 0 kJ
Step 3-1 is isothermal, therefore, the definition of boundary work becomes:
Eqn 5
The problem is that we don't know either P3 or V3.  Either one would be useful in evaluating W31 because we know P1 and we can determine V1 from the Ideal Gas EOS, Eqn 2.
We can evaluate V3 using the fact that V3 = V2.  Apply the the Ideal Gas EOS to state 2.
Eqn 6
V3 2.503 m3/mole
Next, we can apply Eqn 6 to state 1 : V1 1.070 m3/mole
Now, we can plug values into Eqn 4 to evaluate W13 :
W31 -181.89 kJ
Sum the work terms for the three steps to get Wcycle:
Wcycle = 104.8 kJ