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Topic 10: Thermal physics 10.2 Processes

Topic 10: Thermal physics 10.2 Processes. The first law of thermodynamics 10.2.1 Deduce an expression for the work involved in a volume change of a gas at constant pressure. 10.2.2 State the first law of thermodynamics.

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Topic 10: Thermal physics 10.2 Processes

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  1. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics 10.2.1 Deduce an expression for the work involved in a volume change of a gas at constant pressure. 10.2.2 State the first law of thermodynamics. 10.2.3 Identify the first law of thermodynamics as a statement of energy conservation. 10.2.4 Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas. 10.2.5 Draw and annotate thermodynamic processes and cycles on P-V diagrams. 10.2.6 Calculate from a P-V diagram the work done in a thermodynamic cycle. 10.2.7 Solve problems involving state changes of a gas.

  2. x A Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Deduce an expression for the work involved in a volume change of a gas at constant pressure. Suppose we take a beaker that is filled with an ideal gas, and stopper it with a gas-tight cork and a weight, as shown. The weight F causes a pressure in the gas having a value given by P = F/A, where A is the area of the cork in contact with the gas. If we now heat up the gas it will expand against the cork, pushing it upward: The dashed red box shows the change in volume ∆V. Note that ∆V = Ax. ∆V P is constant. Why?

  3. F x ∆V A Work done by expanding gas (constant P) W = P∆V Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Deduce an expression for the work involved in a volume change of a gas at constant pressure. From the previous slide: P = F/A F = PA and ∆V = Ax. The work W done by the gas is just the force F it exerts on the weighted cork times the distance x it moves the cork. Thus W = Fx = PAx= P∆V. FYI If ∆V > 0 (gas expands) then W > 0. If ∆V < 0 (gas contracts) then W < 0.

  4. first law of thermodynamics Q = ∆U + W Topic 10: Thermal physics10.2 Processes The first law of thermodynamics State the first law of thermodynamics. Consider the previous “system” containing gas, and heat, and providing mechanical work. Just as we have done before, we will use Q to represent a quantity of heat. If heat is added to our system (as it was) then Q > 0. If heat is removed from our system then Q < 0. If the gas expands then W > 0. If the gas contracts then W < 0. Finally, we define the change in internal energy of the system as ∆U. The first law of thermodynamics relates Q, W and ∆U as follows:

  5. first law of thermodynamics Q = ∆U + W Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Identify the first law of thermodynamics as a statement of energy conservation. What the first law shows is that when heat energy Q is added to a system, some (or all of it) may be used to change the internal energy ∆U of the system, and some (or all of it) may be used to provide mechanical work W. FYI Sometimes the first law is expressed ∆U = Q – W. In this form we see that there are two ways to change the internal energy of a system: (1)By passing heat Q through the system boundary. (2)By doing work W through the system boundary.

  6. first law of thermodynamics Q = ∆U + W Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Identify the first law of thermodynamics as a statement of energy conservation. In words, the first law says “Heat added to a closed system can change its internal energy and cause it to do work on its environment.” This is a statement of the conservation of energy. All of the energy added to the system is accounted for. FYI Recall that ∆Uconsists of both ∆EP (manifested in phase change) and ∆EK (temperature change) of the substance. Lack of internal forces in an ideal gas means that ∆EP = 0.

  7. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas. For an ideal gas, the equation of state (PV = nRT) tells us all of the important things about a gas through the state variables P,V, n, and T. When we add (or remove) heat Q from a closed system containing an ideal gas, or have it do work W on the external world, its state variables may change. We callchanging the state of a system a process. A process in which the state of a gas is changed without changing its volume is calledisochoric. FYI Don’t confuse “state” with “phase” in Topic 10.

  8. 20 10 30 0 Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas. A process in which the state of a gas is changed without changing its volume is calledisochoric. We have already seen an isochoric process when we studied the concept of absolute zero: p During an isochoric process the temperature and the pressure change. NOT the volume. T (°C) 200 0 300 100 -200 -100 -300

  9. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas. A process in which the state of a gas is changed without changing its volume is calledisochoric. PRACTICE: Show that the first law of thermodynamics reduces to Q = ∆U for an isochoric process. SOLUTION: Recall that the work done by a gas is given by W = P∆V. Isochoric means ∆V = 0. Thus W = 0. The first law, Q = ∆U + W, thus reduces to Q = ∆U.

  10. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas. A process in which the state of a gas is changed without changing its volume is calledisochoric. EXAMPLE: Show that for an isolated ideal gas P Tduring an isochoric process. SOLUTION: Use PV = nRT. Then P = (nR/V)T Isolated means n is constant (no gas is added to or lost from the system). Then n and V are constant (as is R). Thus P = (nR/V)T = (CONST)T P T. (Isochoric)

  11. x A Work done by expanding gas (constant P) W = P∆V Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas. A process in which the state of a gas is changed without changing its pressure is calledisobaric. We have already seen such a process, when we derived the formula for the work done by a gas: For an isobaric process the first law can therefore be written Q = ∆U + P∆V. ∆V

  12. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas. A process in which the state of a gas is changed without changing its pressure is calledisobaric. EXAMPLE: Show that for an isolated ideal gas V Tduring an isobaric process. SOLUTION: Use PV = nRT. Then V = (nR/P)T Isolated means n is constant (no gas is added to or lost from the system). Then n and P are constant (as is R). Thus V = (nR/P)T = (CONST)T V T. (Isobaric)

  13. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas. A process in which the state of a gas is changed without changing its pressure is calledisobaric. PRACTICE: Show that for an isolated ideal gas W = nR∆T during an isobaric process. SOLUTION: From PV = nRT we can write (if n and P are constant) P∆V = nR∆T. Recall W = P∆V. Thus W = nR∆T. (Isobaric)

  14. 20 10 30 0 Why do we wait before recording our values? Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas. If the state of a gas is changed without changing its temperature the process is calledisothermal. EXAMPLE: A graduated syringe which is filled with air is placed in an ice bath and allowed to reach the temperature of the water. Demonstrate that P1V1 = P2V2. SOLUTION: Record initial states after a wait: P1= 15, V1 = 10, and T1 = 0ºC. Record final states after a wait: P2= 30, V2 = 5, and T2 = 0ºC. P1V1 = 15(10) = 150 = 30(5) = P2V2. How do we know that the process is isothermal?

  15. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas. If the state of a gas is changed without changing its temperature the process is calledisothermal. PRACTICE: Show that for an isolated ideal gas P1V1 = P2V2during an isothermal process. SOLUTION: From PV = nRT we can write (if n and T are constant) P1V1 = nRT P2V2 = nRT. Thus P1V1 = nRT = P2V2. (Isothermal)

  16. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas. If the state of a gas is changed without adding or losing heat the process is calledadiabatic. PRACTICE: Show that for an isolated ideal gas W = -∆Uduring an adiabatic process. SOLUTION: From Q = ∆U + W we can write (if n is constant and Q is zero) Q = ∆U + W 0 = ∆U + W W = -∆U. FYI We accomplish this by insulating the container.

  17. z y x Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Draw and annotate thermodynamic processes and cycles on P-V diagrams. Perhaps you have enjoyed the pleasures of analytic geometry and the graphing of surfaces in 3D. The three variables of a surface are x, y, and z, and we can describe any surface using the "state" variables x, y, and z: The equation "of state" of a sphere is x2 + y2 + z2 = r2, where r is the radius of the sphere: FYI We “built” the 3D sphere with layers of 2D circles. We have transformed a 3D surface into a stack of 2D surfaces.

  18. P P V V T isotherms T4 T3 Topic 10: Thermal physics10.2 Processes T2 The first law of thermodynamics Draw and annotate thermodynamic processes and cycles on P-V diagrams. The three state variables (if n is kept constant) of a gas are analogous. We can plot the three variables P, V, and T on mutually perpen- dicular axes like this: We have made layers in T. Thus each layer has a single temperature. T1 FYI Each layer is an isotherm. The 3D graph (below) can then be redrawn in its simpler 2D form (above) without loss of information.

  19. P V Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Draw and annotate thermodynamic processes and cycles on P-V diagrams. A thermodynamic process involves moving from one state to another state. This could involve changing any or even all of the state variables (P, V, or T). EXAMPLE: In the P-V graph shown, identify each process type as ISOBARIC, ISOTHERMAL, OR ISOVOLUMETRIC (isochoric). SOLUTION: AB is isothermal (constant T). BC is isobaric (constant P). CA is isochoric (constant V). A isotherm B C isotherm FYI The purple line could be an adiabatic process.

  20. P V Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Draw and annotate thermodynamic processes and cycles on P-V diagrams. A thermodynamic cycle is a set of processes which ultimately return a gas to its original state. EXAMPLE: A fixed quantity of a gas undergoes a cycle by changing between the following three states: State A: (P = 2 Pa, V = 10 m3) State B: (P = 8 Pa, V = 10 m3) State C: (P = 8 Pa, V = 25 m3) Each process is a straight line, and the cycle goes like this: ABCA. Sketch the complete cycle on a P-V diagram. SOLUTION: Scale your axes and plot your points… B C 8 2 A 10 25

  21. P V B C 8 2 A 10 25 Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle. A thermodynamic cycle is a set of processes which ultimately return a gas to its original state. EXAMPLE: A fixed quantity of a gas undergoes the cycle shown here (from the last example): (a) Find the work done during the process AB. (b) Find the work done during the process BC. SOLUTION: Use W = P∆V. (a) From A to B: ∆V = 0. Thus the W = 0. (b) From B to C: ∆V = 25 – 10 = 15; P = 8. Thus W = P∆V = 8(15) = 120 J.

  22. P V B C 8 2 A 10 25 Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle. A thermodynamic cycle is a set of processes which ultimately return a gas to its original state. EXAMPLE: A fixed quantity of a gas undergoes the cycle shown here (from the last example): (c) Find the work done during the process CA. SOLUTION: (c) Use W = Area under the P-V diagram. Observe that ∆V is negative when going from C (V = 25) to A (V = 10). Observe that P is NOT constant so W  P∆V. W = Area = -[ (2)(15) + (1/2)(6)(15) ] = -75 J.

  23. P V B C 8 2 A 10 25 Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle. A thermodynamic cycle is a set of processes which ultimately return a gas to its original state. EXAMPLE: A fixed quantity of a gas undergoes the cycle shown here (from the last example): (d) Find the work done during the cycle ABCA. SOLUTION: (d) Just total up the work done in each process. WAB= 0 J. WBC= +120 J. WCA= -75 J. Wcycle= 0 + 120 – 75 = +45 J. FYI Work is done on the external environment during each cycle.

  24. P V B C 8 2 A 10 25 Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle. A thermodynamic cycle is a set of processes which ultimately return a gas to its original state. PRACTICE: Find the total work done if the previous cycle is reversed. SOLUTION: We want ACBA. WAC= Area =+[ (2)(15) + (1/2)(6)(15) ] = +75 J. WCB= P∆V= 8(10–25) = -120 J. WBA= 0 J (since ∆V = 0). Wcycle= 75 - 120 + 0 = -45 J. FYI Reversing the cycle reverses the sign of the work.

  25. hot reservoir at TH QH W engine QL cold reservoir at TL Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle. A heat engine is any device which converts heat energy into mechanical work. To make a heat engine we need a source of heat energy (coal, gas, etc.) and a working fluid which undergoes thermodynamic change of state causing work do be done on the external environment. Common working fluids are water (made into steam) and petrol-air mixtures (ignited). The energy flow diagram of a heat engine is shown here:

  26. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle. EXAMPLE: An internal combustion engine is an example of a heat engine that does work on the environment. A four-stroke engine is animated here. http://www.animatedengines.com/otto.html http://chemcollective.org/activities/simulations/engine

  27. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle. EXAMPLE: From the animation of the second link on the previous slide, show the direction of each process with an arrow. SOLUTION: View the animation. Or reason that POSITIVE work must be done by the cycle. FYI Remember “work done BY system” is positive, and “work done ON system” is negative.

  28. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle. EXAMPLE: From the same animation, label the COMPRESSION, POWER, EXHAUST, AND INTAKE STROKES. Which stroke has a positive area under it? Which has a negative area? SOLUTION: A STROKE involves a change in volume. Since ∆V > 0 for the power stroke it has a positive area. The compression stroke has a negative area. COMPRESSION STROKE POWER STROKE EXHAUST STROKE INTAKE STROKE

  29. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle. EXAMPLE: From the same animation, show that the work done in a cycle is positive. SOLUTION: The work done during the power stroke is positive (sketched in red). The work done during the compression stroke is negative (sketched in purple). There is more positive than negative. Thus W > 0. W = area between graphs. COMPRESSION STROKE POWER STROKE EXHAUST STROKE INTAKE STROKE

  30. hot reservoir at TH QH W pump QL cold reservoir at TL Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Calculate from a P-V diagram the work done in a thermodynamic cycle. A heat pump is any device which can move heat from a low temperature reservoir to a high temperature reservoir. A refrigerator is an example of a heat pump. A common working fluid is Freon. The Freon vaporizes inside the fridge removing QL = LV. The Freon condenses outside the fridge releasing QH = Lf. FYI The compressor does the work ON the system.

  31. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. ”Thermally insulated” means that heat can not enter or leave the system. Thus Q = 0. This is the definition of an adiabatic process.

  32. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. W = P∆V = 1105(-3) = -3106 J. Or you can just find the area under the process.

  33. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. Adiabatic (and insulated) both mean that Q = 0. ∆V < 0 means W < 0 (-). The first law says Q = ∆U + W. Then 0 = ∆U + (-)  ∆U > 0.

  34. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. At A: PV = 25 = 10. At B: PV = 52 = 10. At C: PV = 6.82 = 13.6. Isothermal means T is constant. PV = nRT then becomes PV = CONST. Thus process AB is the isothermal one.

  35. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. The difference in work is the area between the graphs.

  36. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. 18 This can only be an estimate. 16 14 12 12 A tally of small rectangles should suffice. 10 9 9 8 8 7 7 6 5 5 4 4 19 There are about 170 of them. Each rectangle has a work value given by W = PV = (0.1105)(0.110-3) = 1 J. Thus the difference in work is 170 J.

  37. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. AC is the adiabatic compression. The first law says Q = ∆U + W. Adiabatic means Q = 0. ∆U = -W. Since W < 0 during the compression of a gas, ∆U = -W > 0. Thus the temperature increases.

  38. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. The gas only does external work when it expands. QR and SP are isochoric (isovolumetric) so W = 0. PQ is a compression so W < 0. RS is an expansion so W > 0.

  39. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. NOT adiabatic since Q  0 (Q = 8103 J). From PV = nRT we see that PiVi = 1.2105(0.05) = 6000 J = nRTi. PfVf = 1.2105(0.10) = 12000 J = nRTf. Thus T changes and the process is NOT isothermal. So the process is NEITHER.

  40. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. From W = P∆V we see that W = 1.2105(0.10 - 0.05) = 6.0103 J.

  41. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. From the first law: Q = ∆U + W, or 8.0103 = ∆U + 6.0103 ∆U = 2.0103 J.

  42. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. Work done ON a gas is a compression. Only process PQ has a decreasing volume.

  43. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. For an ideal gas, ISOTHERMAL means ∆U = 0. From the first law of thermodynamics we have Q = ∆U + W Q = 0 + W Q = W = 2500 J.

  44. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. From the first law we have q = ∆U + w. Keep T constant with ICE BATH. Keep V constant with FIXED CONTAINER. Keep q zero with INSULATION.

  45. QH W QL Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. A heat pump transfers heat from a cold reservoir to a hot one. Since this is not the natural direction of flow, work must be done ON the system to make this happen.

  46. It is clear from the diagramthat both AB and CD are isobaric (P = CONST). Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. From CD evaporation occurs. During phase changes ∆T = CONST (isothermal). From AB compression occurs. During phase changes ∆T = CONST (isothermal).

  47. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. Wherever W > 0 heat is removed from the cold reservoir. Thus BC and CD. Wherever W < 0 heat is forced into the hot reservoir. Thus DA and AB.

  48. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. From the diagram Ein= W + Qc. Eout= QH. From Ein= Eoutwe have W + QC = QH.

  49. Wherever W > 0 heat is removed from the cold reservoir. Thus where ∆V > 0. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. A B A refrigerator is a heat pump. Work must be done ON it. Thus W < 0. B A Area inside curves must be NEGATIVE. P = CONST

  50. Topic 10: Thermal physics10.2 Processes The first law of thermodynamics Solve problems involving state changes of a gas. For each square: W = P∆V W = (1105)(0.110-3) W = 10 J. There are about 47 squares. Thus W = 47(10 J) = 470 J (done ON substance).

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