Example Problem with Complete Solution

4A-1 : Work for a Cycle Carried Out in a Closed System 6 pts
Ten kilograms of carbon dioxide (CO2) is held in a piston-and-cylinder device. The CO2 undergoes a thermodynamic cycle consisting of three processes. The processes are :
Process 1-2: constant pressure expansion
Process 2-3: constant volume
Process 3-1: constant temperature compression
a.) Sketch the cycle path on a PV Diagram
b.) Calculate the net work done in kJ
Data: T1 = 145oC, T2 = 440oC, P1 = 150 kPa
Read : Work your way around the cycle, step by step. The work for the cycle is the sum of the work for each step.
Assume the CO2 behaves as an ideal gas throughout all three process steps.
Apply the definition of
boundary work or PV work to each step in the cycle.
Diagram: See the solution to part (a).
Given: m 10 kg Find: a.) Sketch cycle on a PV Diagram.
T1 145 oC b.) Wcycle = ??? kJ
T2 440 oC
P1 150 kPa
Assumptions: 1 - The gas is held in a closed system.
2 - Boundary work is the only form of work interaction
3 - Changes in kinetic and potential energies are negligible.
4 - CO2 behaves as an ideal gas. This must be verified at all three states.
Equations / Data / Solve:
Part a.)
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 CO2 in the system from the given mass of CO2 and its molecular weight.
Eqn 4
MWCO2 44.01 g/mole n 227.22 mole
Plug values into Eqn 3 : R 8.314 J/mole-K
W12 557.29 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 8.981 m3
Next, we can apply Eqn 6 to state 1 : V1 5.266 m3
Now, we can plug values into Eqn 4 to evaluate W13 : W31 -421.71 kJ
Sum the work terms for the three steps to get Wcycle : Wcycle = 135.6 kJ
Verify : Only the ideal gas assumption can be verified.
We need to determine the specific volume and check if :
Eqn 7
R 8.314 J/mol-K V1 23.18 L/mol
V2 = V3 39.53 L/mol
The ideal gas assumption is valid because V > 20 L/mole in all three states.
Answers : a.) See the sketch, above. b.) Wcycle = 135.6 kJ