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1 ME235 Thermodynamics I Winter 2021 April 8 th , 2021 Open Book Exam Duration: 1.5 hours (15 mins extra to wrap up, upload PDF file and SUBMIT the exam on Canvas) Short answer questions: 35 points Problem 5: 20 points Problem 6: 45 points Total: 100 points Instructions: (1) Two sections in this quiz a. Section 1: Short questions including multiple choice, true/false, short essay questions. b. Section 2: Honor code + Long numerical problems (2) Generally, no partial credit for the short answer questions (3) State all assumptions and the steps used in your analysis to obtain the final solutions for the long questions. You will not be given credit for implied work. (4) Remember to check unit conversions in your work, and to substitute T in K for ideal gas equations, when you calculate Carnot performance metrics, and in TdS equations (5) Throughout your analysis, you may choose to obtain the solution in terms of the given variables and substitute numerical values only in the last few steps Logistics instructions: (6) Please ensure that you submit the exam (quiz) on Canvas well before the scheduled time. You should plan on completing the exam within 90 mins and using the additional 15 mins to upload all files and CLICK the SUBMIT button. Canvas will shut and will not take your submissions afterward. (7) For any clarifications during the exam please go to the provided Zoom meeting room and wait until your turn. (8) Please only upload ONE PDF file for the signed honor code + solutions for the long numerical problems. Name your PDF file as “Last name, First name.pdf” (9) If you lose access to internet at any point of time during the exam, please call +1-612- 356-6066 to let Prof. Bala Chandran know about this issue.

2 Section I: Short Questions (35 points) 1. (6 points) Liquid water at 1.5 MPa and 120°C expands continuously through a throttling valve to 100 kPa. Select ALL applicable options for what happens at the outlet of the valve compared to the inlet a. all the liquid evaporates b. some of the liquid evaporates c. the temperature increases d. the temperature does not change e. temperature decreases f. enthalpy remains constant b, e and f are the correct answers. Some of the liquid will evaporate to form a 2-phase mixture as the pressure decreases when it flows through a valve and the evaporation will result in the decrease of temperature; enthalpy will remain constant in a throttling in a process. You will come to the exact same conclusions, even if you want to the thermodynamic tables and applied ℎ 𝑖𝑖𝑖𝑖 = ℎ 𝑜𝑜𝑜𝑜𝑜𝑜 and tried to get at the thermodynamic state conditions based on the provide inlet and outlet conditions. 2. (6 points) An inventor claims to have developed a resistance heater that supplies 2 kWh of heat energy to a room for each kWh of electricity that it consumes. Is this a reasonable claim? Explain briefly. No, this is NOT a reasonable claim. Even though there can be a 1:1 conversion of work to heat in a resistance heater, the heat energy produced CANNOT be larger than the electrical work input. On the other hand, a heat-pump operates differently and on a thermodynamic cycle. Here, 𝑄𝑄 𝐻𝐻 > 𝑊𝑊 𝑖𝑖𝑖𝑖 because | 𝑄𝑄 𝐻𝐻 | = | 𝑊𝑊 𝑖𝑖𝑖𝑖 | + 𝑄𝑄 𝐿𝐿  3 points for correct conclusion about the claim  3 points for justification 3. (6 points) Why are engineers interested in reversible processes and reversible thermodynamic cycles even though they can never be realized in practice? Explain briefly. Provide ONE example each for (a) a reversible process (what comes as a close approximation), and (b) an irreversible process. We are interested in reversible processes because they are idealized processes which can help establish performance limits of a process in any device. For example, a reversible turbine will produce more work output between the same inlet and outlet conditions as compared to a turbine with irreversibilities; a Carnot heat engine, which is a fully reversible thermodynamic cycle, produces the maximum amount of work output between any two given thermal reservoirs. Examples of: • a reversible process – slow, quasi-static expansion/compression of any substance in a frictionless and isothermal/adiabatic piston-cylinder; heat transfer across infinitesimally small temperature differences (e.g. with large surface areas involved in heat transfer); elastic expansion/compression of springs • an irreversible process – fast, unrestrained expansion/compression of any substance in a piston cylinder; heat-transfer across a finite temperature difference; inelastic/plastic deformation of solids -> 3 points for why reversible processes need to be considered ->1.5 points per correct example of reversible and irreversible process

3 4. (4 points) A pure substance is compressed reversibly to higher pressure while heat is added to the pure substance. An irreversible pathway then follows to return the substance back to its starting point. The entropy change of the pure substance for the irreversible pathway is _________. (a) positive (b) negative (c) zero (d) Not enough information (b) Entropy change will be negative because the entropy change from the initial (1) to the final state (2) in during the reversible process should be positive because heat is added, and then when we return the substance back to its initial state, the entropy change should be negative as ∆𝑆𝑆 𝑟𝑟𝑟𝑟𝑟𝑟 , 1→2 + ∆𝑆𝑆 𝑖𝑖𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 , 2→1 = 0 because entropy is a thermodynamic property and its difference b/n the same initial and final state should be 0. 5. (3 points) For an ideal gas, if the pressure decreases while keeping the specific entropy a constant (i.e., an isentropic process), its specific enthalpy __________. (a) increases (b) decreases (c) remains constant (d) not enough information (b) Enthalpy will decrease because temperature decreases (enthalpy is only a function of temperature for an ideal gas) from TdS equations for an ideal gas, 𝑅𝑅𝑅𝑅𝑅𝑅 � 𝑃𝑃 2 𝑃𝑃 1 � = 𝐶𝐶 𝑝𝑝0 ln � 𝑇𝑇 2 𝑇𝑇 1 � 7. (4 points) Two reversible cycles for heat engines are represented on a T-S diagram. Cycle 1 has 50% more work generated than cycle 2. What can you say about the areas inside the T-S diagrams? (a) The area is 50% larger for cycle 1. (b) The area is 50% larger for cycle 2. (c) The area is not related to work. (d) The area is not related to any physical property. (a) Area in a T-s plot for a reversible cycle is equal to the net work produced in the cycle. Therefore the area should be 50% larger for cycle 1 6. (6 points) Two piston-cylinders are at the same pressure. Cylinder 1 contains saturated liquid water and cylinder 2 contains the same mass of saturated water vapor. 1 kJ of heat is added very slowly and reversibly to each container at constant pressure. It is proposed that the entropy of cylinder 1 increases more than the entropy of cylinder 2. Is this proposal true/false? Explain your justification and include T-s sketches including the vapor dome for the two processes. Note: You can attach a figure on Canvas using the upload image functionality in the text box here and/or include it is part of the PDF being uploaded for the numerical problems This proposal is true. Because the heat is added slowly, lets treat both these processes to be reversible. Because for the same amount of heat added (at constant pressure), saturated liquid will change its entropy at a constant temperature as compared to the saturated vapor, which will increase in temperature for a const. pressure heat transfer process. Because 𝑞𝑞 12 = ∫ 𝑇𝑇𝑇𝑇𝑇𝑇 = 𝑇𝑇 ( 2 1 𝑇𝑇 2 − 𝑇𝑇 1 ) , for a constant temperature

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4 heat addition in cylinder 1, the magnitude of entropy change will be larger as compared to when both temperature and entropy can change in cylinder 2, 𝑞𝑞 34 = 𝑞𝑞 12 = ∫ 𝑇𝑇𝑇𝑇𝑇𝑇 4 3 . See figure below for the plot .  2 points for stating that proposal is true  2 points for justification  2 points for correct T-s plot (not required to show shaded area for heat transfer) Section II: Honor Code and Numerical Problems Please upload this as one PDF file named as “Last name, First name.pdf”. Please include the signed Honor Code and if possible, also include your full name in all the sheets in which you are writing the answers. Full Name: I have observed the Honor Code and neither given nor received aid on this exam SOLUTIONS Signature 8. (20 points) A heat pump removes heat from a lake that has an average wintertime temperature of 6 °C and supplies heat into a house having an average temperature of 27 °C. (a) (10 points) If the house steadily loses heat to the atmosphere at the rate of 64,000 kJ/h, determine the minimum electrical power that needs to be supplied to the heat pump, in kW. Explain why what you are calculating is the minimum power input. (b) (5 points) A heat exchanger is used to transfer energy from the lake water to the heat pump. If the lake water temperature decreases by 5 °C as it flows through the heat exchanger, determine the

5 minimum mass flow rate of the lake water in kg/s. Use a constant specific heat capacity of water to be 4200 J/kg/K. (c) (5 points) Find the rate of entropy change (kW/K) based in the lake water as it flows through the heat exchanger using the mass flow rate computed in (b). (a) The heat supply of the house should compensate the heat lost to the atmosphere to keep the average temperature: 𝑄𝑄 𝐻𝐻 ̇ = 𝑄𝑄 𝑙𝑙𝑜𝑜𝑙𝑙𝑜𝑜 ̇ = 64,000 𝑘𝑘𝑘𝑘 / ℎ 3600 𝑇𝑇 / ℎ = 17.78 𝑘𝑘𝑊𝑊  2 points for statement and unit conversion To minimize the electrical power, the best possible efficient method is using the Carnot cycle. The coefficient of performance of such a cycle: 𝐶𝐶𝐶𝐶𝐶𝐶 = 𝑇𝑇 𝐻𝐻 𝑇𝑇 𝐻𝐻 − 𝑇𝑇 𝐿𝐿 = 273.15 + 27 27 − 6 = 14.3 = 𝑄𝑄 𝐻𝐻 ̇ 𝑊𝑊 ̇ 𝑊𝑊 ̇ = 𝑄𝑄 𝐻𝐻 ̇ 𝐶𝐶𝐶𝐶𝐶𝐶 = 17.78 𝑘𝑘𝑊𝑊 14.3 = 𝟏𝟏 . 𝟐𝟐𝟐𝟐 𝒌𝒌𝒌𝒌  3 points for correct inference of minimizing power input  3 points for correct equation about the Carnot cycle COP  2 points for correct result (b) The rate of heat transfer in the heat exchanger: 𝑄𝑄 𝐿𝐿 ̇ = 𝑄𝑄 𝐻𝐻 ̇ − 𝑊𝑊 ̇ = 17.78 − 1.24 = 16.54 𝑘𝑘𝑊𝑊 Consider the heat exchanger as a control volume system, we find the energy equation: 𝑄𝑄 𝐿𝐿 ̇ = 𝑚𝑚̇ ∆ℎ 𝑤𝑤𝑤𝑤𝑜𝑜𝑟𝑟𝑟𝑟 = 𝑚𝑚̇ 𝐶𝐶∆𝑇𝑇 𝑤𝑤𝑤𝑤𝑜𝑜𝑟𝑟𝑟𝑟 𝑚𝑚̇ = 𝑄𝑄 𝐿𝐿 ̇ 𝐶𝐶∆𝑇𝑇 𝑤𝑤𝑤𝑤𝑜𝑜𝑟𝑟𝑟𝑟 = 16.54 𝑘𝑘𝑊𝑊 4.2 𝑘𝑘𝑘𝑘 / 𝑘𝑘𝑘𝑘𝑘𝑘 × 5 𝑘𝑘 = 𝟎𝟎 . 𝟕𝟕𝟕𝟕𝟕𝟕 𝒌𝒌𝒌𝒌 / 𝒔𝒔  3 points for correct equation  2 points for correct result (c) For pure liquids of lake water, the entropy change rate can be computed:

6 𝑇𝑇𝑆𝑆 = 𝑚𝑚𝐶𝐶 𝑇𝑇𝑇𝑇 𝑇𝑇 Integrate from 𝑇𝑇 𝑖𝑖 = 6 ℃ to 𝑇𝑇 𝑟𝑟 = 1 ℃ 𝑆𝑆 ̇ = 𝑚𝑚̇ 𝐶𝐶𝑅𝑅𝑅𝑅 � 𝑇𝑇 𝑟𝑟 𝑇𝑇 𝑖𝑖 � = 0.788 𝑘𝑘𝑘𝑘 / 𝑇𝑇 × 4.2 𝑘𝑘𝑘𝑘 / 𝑘𝑘𝑘𝑘𝑘𝑘 × ln 273.15 + 1 273.15 + 6 = −𝟎𝟎 . 𝟎𝟎𝟎𝟎𝟎𝟎𝟕𝟕 𝒌𝒌𝒌𝒌 / 𝑲𝑲  3 points for correct equation  2 points for correct result 9. (45 points) A compressor inputs a stream of air at 300 K, 100 kPa, and outputs to a line at 350 K, 200 kPa. The output line is used to recharge a partially deflated balloon by opening a valve. The balloon initially contains 0.003 kg of air at 300 K, 105 kPa and the valve is kept open until its pressure increases to a final value of 115 kPa. The air pressure in the balloon can be assumed to vary linearly with volume, P = 100 kPa + 𝛼𝛼 V, where 𝛼𝛼 is a constant term. Because the charging process occurs rapidly, there is not enough time for heat-transfer and therefore adiabatic conditions can be assumed. Assume ideal-gas behavior for air and use constant specific heats, C vo = 0.717 kJ/kg/K; C po = 1.004 kJ/kg/K; R = 0.287 kJ/kg/K (a) (2 points) What is the nature of the charging process in the balloon – steady or transient? (b) (10 points) Determine the final volume of air (in m 3 ) in the balloon and the moving boundary work-transfer (in kJ) for the air inside the balloon. Hint: Make use of the linear P-V relationship (c) ( 15 points) Find the final temperature of the air inside the balloon at the end of this charging process. Hint: Apply conservation of mass and energy by considering the balloon as your control volume (d) (8 points) What is the work input to the compressor (in kJ) during this charging process for the balloon? (e) (10 points) Determine the entropy change (in kJ/K) between the inlet and the outlet states of air in the compressor (i.e., states 1 and 2). Mark these inlet and outlet states on P-v and T-s plots. No need to show the process line on the plot, just show the states.

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7 (a) Transient process because the thermodynamic state is changing as a function of time during the charging process  2 points (b) To get the initial volume of the balloon, apply the ideal gas law: 𝐶𝐶 3 𝑉𝑉 3 = 𝑚𝑚 3 𝑅𝑅𝑇𝑇 3 𝑉𝑉 3 = 𝑚𝑚 3 𝑅𝑅𝑇𝑇 3 𝐶𝐶 3 = 0.003 𝑘𝑘𝑘𝑘 × 0.287 𝑘𝑘𝑘𝑘 / 𝑘𝑘𝑘𝑘𝑘𝑘 × 300 𝑘𝑘 105 𝑘𝑘𝐶𝐶𝑘𝑘 = 2.46 × 10 −3 𝑚𝑚 3  2 points for using ideal gas law correctly Apply the linear pressure-volume relation in the balloon: 𝐶𝐶 3 = 100 𝑘𝑘𝐶𝐶𝑘𝑘 + 𝛼𝛼𝑉𝑉 1 = 105 𝑘𝑘𝐶𝐶𝑘𝑘 𝛼𝛼 = 105 𝑘𝑘𝐶𝐶𝑘𝑘 − 100 𝑘𝑘𝐶𝐶𝑘𝑘 2.46 × 10 −3 𝑚𝑚 3 = 2.03 × 10 3 𝑘𝑘𝐶𝐶𝑘𝑘 / 𝑚𝑚 3  2 points for finding correct α value Now apply the linear P-V relation to the final state: 𝐶𝐶 4 = 100 𝑘𝑘𝐶𝐶𝑘𝑘 + 𝛼𝛼𝑉𝑉 4 = 115 𝑘𝑘𝐶𝐶𝑘𝑘 𝑉𝑉 4 = 115 𝑘𝑘𝐶𝐶𝑘𝑘 − 100 𝑘𝑘𝐶𝐶𝑘𝑘 2.03 × 10 3 𝑘𝑘𝐶𝐶𝑘𝑘 / 𝑚𝑚 3 = 𝟕𝟕 . 𝟑𝟑𝟎𝟎 × 𝟏𝟏𝟎𝟎 −𝟑𝟑 𝒎𝒎 𝟑𝟑  2 points for finding correct final volume For the work transfer of air inside the balloon: 𝑊𝑊 34 = 1 2 ( 𝐶𝐶 3 + 𝐶𝐶 4 )( 𝑉𝑉 4 − 𝑉𝑉 3 ) = 0.5 × (105 𝑘𝑘𝐶𝐶𝑘𝑘 + 115 𝑘𝑘𝐶𝐶𝑘𝑘 ) × (7.39 − 2.46) × 10 −3 𝑚𝑚 3 = 𝟎𝟎 . 𝟎𝟎𝟐𝟐𝟐𝟐𝟑𝟑 𝒌𝒌𝒌𝒌  2 points for correct equation  2 points for correct result (c) Take the volume inside the balloon as a control volume system, we can apply the mass and energy equation: 𝑀𝑀𝑘𝑘𝑇𝑇𝑇𝑇 𝑐𝑐𝑐𝑐𝑅𝑅𝑇𝑇𝑐𝑐𝑐𝑐𝑐𝑐𝑘𝑘𝑐𝑐𝑐𝑐𝑐𝑐𝑅𝑅 : 𝑚𝑚 𝑓𝑓 − 𝑚𝑚 𝑖𝑖 = � 𝑚𝑚 𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖 − � 𝑚𝑚 𝑜𝑜𝑜𝑜𝑜𝑜 𝑜𝑜𝑜𝑜𝑜𝑜 Energy conservation: 𝑚𝑚 𝑓𝑓 𝑢𝑢 𝑓𝑓 − 𝑚𝑚 𝑖𝑖 𝑢𝑢 𝑖𝑖 = 𝑄𝑄 − 𝑊𝑊 + ∑ 𝑚𝑚 𝑖𝑖𝑖𝑖 ℎ 𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖 − ∑ 𝑚𝑚 𝑜𝑜𝑜𝑜𝑜𝑜 ℎ 𝑜𝑜𝑜𝑜𝑜𝑜 𝑜𝑜𝑜𝑜𝑜𝑜  2 points for correct mass conservation equation  2 points for correct energy equation Mass and energy conservation equations simplify to: 𝑚𝑚 4 = 𝑚𝑚 2 + 𝑚𝑚 3 𝑚𝑚 4 𝑢𝑢 4 − 𝑚𝑚 3 𝑢𝑢 3 = −𝑊𝑊 34 + 𝑚𝑚 2 ℎ 2

8  2 points for correct simplifications of both equations Substituting for 𝑚𝑚 2 = 𝑚𝑚 4 − 𝑚𝑚 3 in energy conservation, 𝑚𝑚 4 𝑢𝑢 4 − 𝑚𝑚 3 𝑢𝑢 3 = −𝑊𝑊 34 + ( 𝑚𝑚 4 − 𝑚𝑚 3 ) ℎ 2 => 𝑚𝑚 4 ( 𝑢𝑢 4 − ℎ 2 ) + 𝑚𝑚 3 ( ℎ 2 − 𝑢𝑢 3 ) = −𝑊𝑊 34 => 𝑚𝑚 4 ( 𝑢𝑢 4 − 𝑢𝑢 2 − 𝐶𝐶 2 𝑐𝑐 2 ) + 𝑚𝑚 3 ( 𝑢𝑢 2 − 𝑢𝑢 3 + 𝐶𝐶 2 𝑐𝑐 2 ) = −𝑊𝑊 34 𝑚𝑚 4 𝐶𝐶 𝑟𝑟0 ( 𝑇𝑇 4 − 𝑇𝑇 2 ) − 𝑚𝑚 4 𝑅𝑅𝑇𝑇 2 + 𝑚𝑚 3 𝐶𝐶 𝑟𝑟0 ( 𝑇𝑇 2 − 𝑇𝑇 3 ) + 𝑚𝑚 3 𝐶𝐶 2 𝑐𝑐 2 = −𝑊𝑊 34 Applying, 𝑚𝑚 4 𝑇𝑇 4 = 𝑃𝑃 4 𝑉𝑉 4 𝑅𝑅 , we get: 𝐶𝐶 𝑟𝑟0 � 𝐶𝐶 4 𝑉𝑉 4 𝑅𝑅 � − 𝑚𝑚 4 𝑇𝑇 2 �𝐶𝐶 𝑟𝑟0 + 𝑅𝑅 �� 𝐶𝐶 𝑝𝑝0 � + 𝑚𝑚 3 𝐶𝐶 𝑟𝑟0 ( 𝑇𝑇 2 − 𝑇𝑇 3 ) + 𝑚𝑚 3 𝐶𝐶 2 𝑐𝑐 2 = −𝑊𝑊 34 All quantities are known in above equation except for 𝑚𝑚 4 𝑚𝑚 4 = �𝑊𝑊 34 + 𝐶𝐶 𝑟𝑟0 � 𝐶𝐶 4 𝑉𝑉 4 𝑅𝑅 � + 𝑚𝑚 3 𝐶𝐶 𝑟𝑟0 ( 𝑇𝑇 2 − 𝑇𝑇 3 ) + 𝑚𝑚 3 𝐶𝐶 2 𝑐𝑐 2 � 𝐶𝐶 𝑝𝑝0 𝑇𝑇 2  5 points for correct overall approach (includes 2 points for invoking ideal gas equation to transform T4 in terms of P4,V4) 𝑆𝑆𝑐𝑐𝑘𝑘𝑐𝑐𝑐𝑐 2: 𝐶𝐶 2 = 200 𝑘𝑘𝐶𝐶𝑘𝑘 , 𝑇𝑇 2 = 350 𝑘𝑘 ; 𝑆𝑆𝑐𝑐𝑘𝑘𝑐𝑐𝑐𝑐 3: 𝐶𝐶 3 = 105 𝑘𝑘𝐶𝐶𝑘𝑘 ; 𝑇𝑇 3 = 300 𝑘𝑘 ; 𝑚𝑚 3 = 0.003 𝑘𝑘𝑘𝑘 𝑆𝑆𝑐𝑐𝑘𝑘𝑐𝑐𝑐𝑐 4: 𝐶𝐶 4 = 115 𝑘𝑘𝐶𝐶𝑘𝑘 , 𝑉𝑉 4 = 7.39 ∗ 10 −3 𝑚𝑚 3 => 𝑚𝑚 4 = 𝟎𝟎 . 𝟎𝟎𝟎𝟎𝟕𝟕𝟕𝟕𝟎𝟎 𝒌𝒌𝒌𝒌 𝑇𝑇 4 = 𝐶𝐶 4 𝑉𝑉 4 𝑅𝑅𝑚𝑚 4 = 𝟑𝟑𝟑𝟑𝟕𝟕 𝑲𝑲  4 points for correct T4 and m4 values (d) The mass of air flows through the compressor: 𝑚𝑚 2 = 𝑚𝑚 4 − 𝑚𝑚 3 = 0.00875 − 0.003 = 0.00575 𝑘𝑘𝑘𝑘 Now take the compressor as a control volume system, apply energy conservation 𝑊𝑊 12 = 𝑚𝑚 2 ( ℎ 2 − ℎ 1 ) = 𝑚𝑚 2 𝐶𝐶 𝑝𝑝0 ( 𝑇𝑇 2 − 𝑇𝑇 1 ) = 0.00575 𝑘𝑘𝑘𝑘 × 1.004 𝑘𝑘𝑘𝑘 / 𝑘𝑘𝑘𝑘𝑘𝑘 × (350 𝑘𝑘 − 300 𝑘𝑘 ) = 𝟎𝟎 . 𝟐𝟐𝟕𝟕𝟕𝟕𝟕𝟕 𝒌𝒌𝒌𝒌  2 points for correct approach to calculated m2  3 points for correct equation for compressor  3 points for correct final answer

9 (e) For ideal gas, the entropy change rate can be computed: 𝑇𝑇 2 − 𝑇𝑇 1 = 𝐶𝐶 𝑝𝑝0 ln 𝑇𝑇 2 𝑇𝑇 1 − 𝑅𝑅 ln 𝐶𝐶 2 𝐶𝐶 1  3 points for correct entropy equation for ideal gas ∆𝑆𝑆 = 𝑚𝑚 2 ( 𝑇𝑇 2 − 𝑇𝑇 1 ) = 0.00575 [ 𝑘𝑘𝑘𝑘 ] ∗ � 1.004 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘𝑘𝑘 × ln 350 [ 𝑘𝑘 ] 300 [ 𝑘𝑘 ] − 0.287 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘𝑘𝑘 × ln 200 𝑘𝑘𝐶𝐶𝑘𝑘 100 𝑘𝑘𝐶𝐶𝑘𝑘 � = −𝟐𝟐 . 𝟎𝟎𝟐𝟐 × 𝟏𝟏𝟎𝟎 −𝟐𝟐 𝒌𝒌𝒌𝒌 / 𝑲𝑲  3 points for correct result Entropy is decreasing because of the larger increase in pressure despite an increase of T  4 points for showing the correct states in both plot(P,v,s)

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