4A1 :  Work for a Cycle Carried Out in a Closed System  6 pts 

Ten kilograms of carbon dioxide (CO_{2}) is held in a pistonandcylinder device. The CO_{2} undergoes a thermodynamic cycle consisting of three processes. The processes are : Process 12: constant pressure expansion 

Process 23: constant volume Process 31: constant temperature compression a.) Sketch the cycle path on a PV Diagram b.) Calculate the net work done in kJ Data: T_{1} = 145^{o}C, T_{2} = 440^{o}C, P_{1} = 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 CO_{2} 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.  
T_{1}  145  ^{o}C  b.)  W_{cycle} =  ???  kJ  
T_{2}  440  ^{o}C  
P_{1}  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  CO_{2} behaves as an ideal gas. This must be verified at all three states.  
Equations / Data / Solve:  
Part a.) 


Part b.)  Since W_{cycle} = W_{12} + W_{23} + W_{31}, we will work our way around the cycle and calculate each work term along the way.  
Step 12 is isobaric, therefore, the definition of boundary work becomes:  

Eqn 1  
We can simplify Eqn 1 using the fact that P_{2} = P_{1} and the Ideal Gas EOS :  

Eqn 2  

Eqn 3  
We can determine the number of moles of CO_{2} in the system from the given mass of CO_{2} and its molecular weight.  

Eqn 4  
MW_{CO2}  44.01  g/mole  n  227.22  mole  
Plug values into Eqn 3 :  R  8.314  J/moleK  
W_{12}  557.29  kJ  
Because the volume is constant in step 23:  W_{23}  0  kJ  
Step 31 is isothermal, therefore, the definition of boundary work becomes:  

Eqn 5  
The problem is that we don't know either P_{3} or V_{3}. Either one would be useful in evaluating W_{31} because we know P_{1} and we can determine V_{1} from the Ideal Gas EOS, Eqn 2.  
We can evaluate V_{3} using the fact that V_{3} = V_{2}. Apply the the Ideal Gas EOS to state 2.  

Eqn 6  
V_{3}  8.981  m^{3}  
Next, we can apply Eqn 6 to state 1 :  V_{1}  5.266  m^{3}  
Now, we can plug values into Eqn 4 to evaluate W_{13} :  W_{31}  421.71  kJ  
Sum the work terms for the three steps to get W_{cycle }:  W_{cycle} =  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/molK  V_{1}  23.18  L/mol  
V_{2} = V_{3}  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.)  W_{cycle} =  135.6  kJ 