| 4A-1 : | Work for a Cycle Carried Out in a Closed System | 6 pts |
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| 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 |
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| 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 |
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| 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. |
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| 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.) | 
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| 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: | |||||||||||||||
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Eqn 1 | ||||||||||||||
| We can simplify Eqn 1 using the fact that P2 = P1 and the Ideal Gas EOS : | |||||||||||||||
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Eqn 2 | ||||||||||||||
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Eqn 3 | ||||||||||||||
| We can determine the number of moles of CO2 in the system from the given mass of CO2 and its molecular weight. | |||||||||||||||
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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: | |||||||||||||||
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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. | |||||||||||||||
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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 : | 
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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 | |||||||||
