1. MSU Physics 231 Fall 2015 1 Physics 231 Topic 14: Laws of Thermodynamics Alex Brown Dec 7-11 2015

2. MSU Physics 231 Fall 2015 2 9 th 10 pm attitude survey (1% for participation) 11 th 10 pm last homework set 8 th 10 pm correction for 3 rd exam 10 th 10 pm concept test timed (50 min) (1% for performance) 17 th 8-10 pm final (Thursday) VMC E100

3. MSU Physics 231 Fall 2015 3 Clicker Question! Ice is heated steadily and becomes liquid and then vapor. During this process: a)the temperature rises continuously. b)when the ice turns into water, the temperature drops for a brief moment. c) the temperature is constant during the phase transformations d) the temperature cannot exceed 100 o C

4. MSU Physics 231 Fall 2015 4 Key Concepts: Laws of Thermodynamics Laws of Thermodynamics 1 st Law:  U = Q + W 2 nd Law: Heat flows from hotter  cooler Thermodynamic Processes Adiabatic (no heat flow) Work done in different processes Heat Engines & Refrigerators Carnot engine & efficiency Entropy Relationship to heat, energy. Statistical interpretation Covers chapter 14 in Rex & Wolfson

5. MSU Physics 231 Fall 2015 5 Engine based on a container of an idea gas where the P, V and T change (n is fixed) 1)Put in contact with a source of heat at high T during which heat energy flows in and piston is pushed up. 2)Put in contact with a source of heat at low T during which piston is pushed down and heat flows out. 3)Comes back to it original state (e.g. same value of P, V, and T) 4)End result is that we have turned heat energy into work piston P,V,T area A yy n fixed

6. MSU Physics 231 Fall 2015 6 Process visualized with a P-V diagram for the gas inside isobaric line: pressure is constant volume changes iso-volumetric line: volume is constant pressure changes n fixed V P i

7. MSU Physics 231 Fall 2015 7 P V lines with constant T iso-thermal lines PV = n R T (ideal gas equation from chapter 12) P = n R T/V = c T/V (c = constant) T1T1 T2T2 T3T3 T4T4 T 1 < T 2 < T 3 < T 4

8. MSU Physics 231 Fall 2015 8 A Piston Engine Piston is moved downward slowly so that the gas remains in thermal equilibrium: Volume decreases (obviously) Temperature increases Work is done on the gas v in v out v out > v in (speeds) work is done on the gas and temperature increases T i < T f P f V f T f piston P i V i T i area A yy

9. MSU Physics 231 Fall 2015 9 Isobaric Compression The pressure does not change while pushing down the piston (isobaric compression). W = work done on the gas by pushing down on the piston VV f V i f i P P piston P i V i T i area A yy P f V f T f

10. MSU Physics 231 Fall 2015 10 Isobaric Compression The pressure does not change while lowering the piston (isobaric compression). W = work done on the gas W = F d = - P A  y (P=F/A) W = - P  V = - P (V f -V i ) (in Joule) Sign of the work done on the gas: + if  V < 0 - if  V > 0 work is the area under the curve in a P-V diagram with V decreasing VV f V i f i P P piston P i V i T i area A yy P f V f T f

11. MSU Physics 231 Fall 2015 11 Non-isobaric Compression In general, the pressure can change when lowering the piston. The work (W) done by the piston on the gas when going from an initial state (i) to a final state (f) is the area under the line on the P-V diagram with V decreasing. VV f V i f PiPi P i PfPf piston P i V i T i area A yy

12. MSU Physics 231 Fall 2015 12 Work Done on Gases: Getting the Signs Right! If the arrow goes from right to left (volume becomes smaller) positive work is done by pushing the piston down on the gas (W > 0) the internal (kinetic) energy of the gas goes up V P i

13. MSU Physics 231 Fall 2015 13 Work Done on Gases: Getting the Signs Right! If the arrow goes from left to right (volume becomes larger) W 0 positive work (W g ) is done by the gas on the piston. the internal energy of the gas goes down V P i

14. MSU Physics 231 Fall 2015 14 iso-volumetric process v P Work done on/by gas: W = W g = - P  V = 0

15. MSU Physics 231 Fall 2015 15 Clicker Quiz! A gas is enclosed in a cylinder with a moveable piston. The figures show 4 different PV diagrams. In which case is the work done by the gas largest? Work: area under PV diagram Work done by the gas: volume must become larger, which leaves (a) or (c). Area is larger for (a).

16. MSU Physics 231 Fall 2015 16 a)What is the pressure P A ? b)If the inside temperature is raised the lid moves up by 5 cm. How much work is done by the gas? M=50 kg A=100 cm 2 = 0.010 m 2 mass and area of the lid PAPA P atm

17. MSU Physics 231 Fall 2015 17 a)What is the pressure P A ? b)If the inside temperature is raised the lid moves up by 5 cm. How much work is done by the gas? M=50 kg A=100 cm 2 = 0.010 m 2 mass and area of the lid PAPA P atm a)P A = P atm + Mg/A = 1.50 x 10 5 b) W g = P A  V = 75.0 J

18. MSU Physics 231 Fall 2015 18 One mole of an ideal gas initially at 0 ° C undergoes an expansion at constant pressure of one atmosphere to four times its original volume. a)What is the new temperature? b)What is the work done by the gas? For ideal gas PV=nRT

19. MSU Physics 231 Fall 2015 19 One mole of an ideal gas initially at 0 ° C undergoes an expansion at constant pressure of one atmosphere to four times its original volume. a)What is the new temperature? b)What is the work done by the gas? For ideal gas PV=nRT a) Use PV = nRT to get T f = (V f /V i ) T i = 1092 K b) W = -P  V – P(4V i -V i ) = -3PV i = -3P(nRT i /P) W g = -W = 3nRT i = 6806 J

20. MSU Physics 231 Fall 2015 20 First Law of Thermodynamics By performing work on an object the internal energy can increased By transferring heat to an object the internal energy can increased The change in internal energy depends on the work done on the object and the amount of heat transferred to the object. Internal energy (KE+PE) where KE is the kinetic energy associated with translational, rotational, vibrational motion of atoms

21. MSU Physics 231 Fall 2015 21 First Law of Thermodynamics  U = U f - U i = Q + W  U= change in internal energy Q= energy transfer through heat (+ if heat is transferred to the system) W= energy transfer through work (+ if work is done on the system) This law is a general rule for conservation of energy

22. MSU Physics 231 Fall 2015 22 Applications to ideal gas in a closed container (number of moles, n, is fixed) PV = n R T (chapter 12) U = (d/2) n R T (chapter 12) (d=3 monatomic) (d=5 diatomic) So U = (d/2) P V (useful for P-V diagram) (d/2) n R = constant So  U = (d/2) n R  T Example for P-V diagram (in class)

23. MSU Physics 231 Fall 2015 23 First Law: Isobaric Process A gas in a cylinder is kept at 1.0x10 5 Pa. The cylinder is brought in contact with a cold reservoir and 500 J of heat is extracted from the gas. Meanwhile the piston has sunk and the volume decreased by 100cm 3. What is the change in internal energy? Q = -500 J  V = -100 cm 3 = -1.0x10 -4 m 3 W = - P  V = 10 J  U = Q + W = - 500 + 10 = - 490

24. MSU Physics 231 Fall 2015 24 First Law: General Case V(m 3 ) P (Pa) 1 4 3 6 In ideal gas (d=3) is compressed A) What is the change in internal energy B) What is the work done on the gas? C) How much heat has been transferred to the gas? A)U = (3/2)PV  U = 3/2(P f V f - P i V i ) = 3/2[6x1 - 3x4] = -9 J i f B) Work: area under the P-V graph: (9 + 4.5) = 13.5 (positive since work is done on the gas) C)  U = Q+W so Q =  U-W = -9 - 13.5 = -22.5 J Heat has been extracted from the gas.

25. MSU Physics 231 Fall 2015 25 Types of Processes A: Iso-volumetric  V=0 B: Adiabatic Q=0 C: Isothermal  T=0 D: Isobaric  P=0 P

26. MSU Physics 231 Fall 2015 26 Iso-volumetric Process (  V = 0)  V = 0 W = 0 (area under the curve is zero) 4)  U = Q = (d/2) n R  T 5) P/T = constant When  P = + (like in the figure)  T = + (5)  U = + (4) Q = + (4) (heat added) When  P = -  T = -  U = - Q = - (heat extracted) 1) PV = n R T 2)  U = W + Q 3)  U = (d/2) n R  T

27. MSU Physics 231 Fall 2015 27 Isobaric Process (  P = 0) 1) PV = n R T 2)  U = W + Q 3)  U = (d/2) n R  T  P = 0 4) W = - P  V = - n R  T 5) Q =  U - W = [(d+2)/2] n R  T 6) V/T = constant When  V = - (like in the figure)  T = - (6) W = + (4) (work done on gas)  U = - (3) Q = - (5) (heat extracted) When  V = +  T = + W = - (work done by gas)  U = + Q = + (heat added)

28. MSU Physics 231 Fall 2015 28 molar heat capacities Constant volume Q = (d/2) n R  T = C v n  T where C v = (d/2) R molar heat capacity at constant volume Constant pressure Q = [(d+2)/2] n R  T = C P n  T where C P = [(d+2)/2] R molar heat capacity at constant pressure For all  U = (d/2) n R  T = C v n  T

29. MSU Physics 231 Fall 2015 29 1) PV = n R T 2)  U = W + Q 3)  U = (d/2) n R  T Isothermal Process (  T = 0)  T = 0 work done on gas is the  U = 0 area under the curve: Q = -W PV = constant When  V = - (like in the figure)  P = + (like in the figure) W = + (work done on gas, from area) Q = - (heat extracted, Q = -W) When  V = +  P = - W = - (work done by gas) Q = + (heat added)

30. MSU Physics 231 Fall 2015 30 1) PV = n R T 2)  U = W + Q 3)  U = (d/2) n R  T Adiabtic Process (Q = 0) Q = 0 (system is isolated) W =  U (work goes into internal energy) P (V)  = constant  = C p /C v = (d+2)/d  > 1 When  V = - (like in the figure)  P = + (like in the figure)  T = + (see figure)  U = + (3) W = + (work done on gas, area) When  V = +  P = -  T = -  U = - W = - (work done by gas) dashed lines are isotherms

31. MSU Physics 231 Fall 2015 31 Ideal gas (monatomic d = 3) (diatomic d = 5) C v = ( d/2) R C p = [( d+2)/2] R Process UUQW Isobaric nC v  TnC p  T-P  V Adiabatic nC v  T0 UU Isovolumetric nC v  T UU0 Isothermal nC v  T=0-W -nRTln(V f /V i ) General nC v  T  U-W (PV Area) negative if V expands

32. MSU Physics 231 Fall 2015 32 First Law: Adiabatic process A piston is pushed down rapidly. Because the transfer of heat through the walls takes a long time, no heat can escape. During the moving of the piston, the temperature has risen 100 0 C. If the container contained 10 mol of an ideal gas, how much work has been done during the compression? (d=3) U = (3/2) nRT Q = 0 and  U = Q + W piston P,V,T area A yy W =  U = (3/2) nR  T = (3/2)(10)(8.31)(100) = 1.25x10 4 J

33. MSU Physics 231 Fall 2015 33 Clicker Quiz! A vertical cylinder with a movable cap is cooled. The process corresponding to this is: a)CB b)AB c)AC d)CA e)Not shown After the cooling of the gas and the lid has come to rest, the pressure is the same as before the cooling process.

34. MSU Physics 231 Fall 2015 34 Adiabatic process An molecular hydrogen gas goes from P 1 = 9.26 atm and V 1 = 0.0118 m 3 to P 2 and V 2 via an adiabatic process. If P 2 = 2.66 atm, what is V 2 ? H 2 (d=5) and adiabatic: PV  = Constant with  = C p /C v = (d+2)/d= 7/5 P 1 (V 1 ) 1.4 = P 2 (V 2 ) 1.4 (V 2 ) 1.4 = (P 1 /P 2 )(V 1 ) 1.4 = 0.0069 V 2 = 0.0069 0.714 = 0.029 (1/1.4) = 0.714)

35. MSU Physics 231 Fall 2015 35 Cyclic processes (monatomic with d=3) In a cyclic process, The system returns to its original state. Therefore, the internal energy must be the same after completion of the cycle [U = (3/2) PV and  U=0] P (Pa) V (m 3 ) 10 50 5.0 1.0 A B C

36. MSU Physics 231 Fall 2015 36 P (Pa) V (m 3 ) 10 50 5.0 1.0 Cyclic Process: Step by Step (1) Process A to B Negative work is done on the gas: (the gas is doing positive work). W= - Area under P-V diagram = - [ (50-10)  (1.0-0.0) +½(50-10)  (5.0-1.0) ] = - 40 - 80 = - 120 J (work done on gas) W g = 120 J (work done by gas)  U= 3/2 (P B V B - P A V A ) = 1.5  [(1)(50) - (5)(10)] = 0 The internal energy has not changed  U=Q+W so Q =  U-W = 120 J Heat that was added to the system was used to do the work! A B C

37. MSU Physics 231 Fall 2015 37 P (Pa) V (m 3 ) 10 50 5.0 1.0 Process B-C W = Area under P-V diagram = - [(50-10)  (1.0-0.0)] W=40 J Work was done on the gas  U= 3/2(P c V c -P b V b ) = 1.5  [(1)(10) - (1)(50)] = - 60 J The internal energy has decreased by 60 J  U=Q+W so Q =  U-W = - 60 - 40 J = - 100 J 100 J of energy has been transferred out of the system. A B C Cyclic Process: Step by Step (2)

38. MSU Physics 231 Fall 2015 38 P (Pa) V (m 3 ) 10 50 5.0 1.0 Process C-A W=-Area under P-V diagram W=0 J No work was done on/by the gas.  U = 3/2(P c V c -P b V b )= = 1.5  [ (5)(10) - (1)(10) ] = 60 J The internal energy has increased by 60 J  U=Q+W so Q =  U-W = 60-0 J = 60 J 60 J of energy has been transferred into the system. A B C Cyclic Process: Step by Step (3)

39. MSU Physics 231 Fall 2015 39 Summary of the process Quantity Process Work on gas (W) Heat(Q) UU A-B-120 J120 J0 J B-C40 J-100 J-60 J C-A0 J60 J SUM (net) -80 J80 J0 A-B B-C C-A P (Pa) V (m 3 ) 10 50 5.0 1.0 A B C

40. MSU Physics 231 Fall 2015 40 What did we do? The gas performed net work (80 J) (W g = -W) while net heat was supplied (80 J): We have built an engine that converts heat energy into work! P (Pa) V (m 3 ) 10 50 5.0 1.0 A B C When the path on the P-V diagram is clockwise work is done by the gas (engine) – heat engine The work done by the gas is equal to the area of the loop W g = (5-1)(50-10)/2 = 80 Quantity Process Work on gas (W) Heat(Q) UU A-B-120 J120 J0 J B-C40 J-100 J-60 J C-A0 J60 J SUM (net) -80 J80 J0

41. MSU Physics 231 Fall 2015 41 P (Pa) V (m 3 ) 10 50 5.0 1.0 A B C Quantity Process Work on gas (W) Heat(Q) UU A-B-120 J120 J0 J B-C40 J-100 J-60 J C-A0 J60 J SUM (net) -80 J80 J0 Q h = 180 heat input from hot source Q c = 100 heat output to cold source (wasted heat) W g = -W = Q h – Q c = 80 work output by gas (engine) efficiency e = W g /Q h = 80/180 = 0.4444

42. MSU Physics 231 Fall 2015 42 Generalized Heat Engine Heat reservoir T h Cold reservoir T c engine Work Q h (heat input) Q c (heat output) W g = Q h - Q c efficiency: W g /Q h e = 1 - Q c /Q h WgWg The efficiency is determined by how much of the heat you supply to the engine is turned into work instead of being lost as waste. Water turned to steam The steam moves a piston Work is done The steam is condensed

43. MSU Physics 231 Fall 2015 43 Reverse Direction: The Fridge

44. MSU Physics 231 Fall 2015 44 heat reservoir T h cold reservoir T c engine work QhQh QcQc W heat is expelled to outside a piston compresses the coolant work is done the fridge is cooled Coefficient of performance COP = |Q c |/W Q c : amount of heat removed W: work input W= Q h - Q c Heat Pump (fridge)

45. MSU Physics 231 Fall 2015 45 On the P-V diagram the heat pump (fridge) is given by a path that goes counter clockwise. The area inside the loop is the amount of work done on the gas to remove heat from the cold source. P (Pa) V (m 3 ) 10 50 5.0 1.0 A B C

46. MSU Physics 231 Fall 2015 46 Clicker Quiz! V (m 3 ) P (Pa) 13 3x10 5 1x10 5 Consider this clockwise cyclic process. Which of the following is true? a)This is a heat engine and the work done by the gas is +4x10 5 b)This is a heat engine and the work done by the gas is +6x10 5 c)This is a heat engine and the work done by the gas is –4x10 5 d)This is a fridge and the work done on the gas is +4x10 5 J e)This is a fridge and the work done on the gas is +6x10 5 J Clockwise: work done by the gas, so heat engine Work by gas=area enclosed = (3-1) x (3x10 5 -1x10 5 ) = 4x10 5 J

47. MSU Physics 231 Fall 2015 47 What is the most efficient engine we can make given a hot and a cold reservoir? What is the best path to take on the P-V diagram?

48. MSU Physics 231 Fall 2015 48 Carnot engine A  B isothermal expansion D  A adiabatic compression W+,  T+ ThTh C  D isothermal compression TcTc W+, Q- B  C adiabatic expansion W-,  T- W-, Q+ Q=0  T=0 Q=0  T=0

49. MSU Physics 231 Fall 2015 49 Carnot cycle Work done by engine: W eng W eng = Q h - Q c Efficiency: e carnot = 1-(T c /T h ) e = 1-(Q c /Q h ) also holds since this holds for any engine inverse Carnot cycle A heat pump or a fridge! By doing work we can transport heat

50. MSU Physics 231 Fall 2015 50 Carnot engine e =1 - (Q c /Q h ) always e carnot =1 - (T c /T h ) carnot only!! In general: e < e carnot The Carnot engine is the most efficient way to operate an engine based on hot/cold reservoirs because the process is reversible: it can be reversed without loss or dissipation of energy Unfortunately, a perfect Carnot engine cannot be built.

51. MSU Physics 231 Fall 2015 51 Example The efficiency of a Carnot engine is 30%. The engine absorbs 800 J of heat energy per cycle from a hot reservoir at 500 K. Determine a) the energy expelled per cycle and b) the temperature of the cold reservoir c) how much work does the engine do per cycle? a)Generally for an engine: efficiency: e = 1 – (Q c /Q h ) Q c = Q h (1-e) = 800(1-0.3) = 560 J b) for a Carnot engine: efficiency: e = 1 - (T c /T h ) T c = T h (1-e) = 500(1-0.3) = 350 c) W = Q h – Q c = 800 – 560 = 240 J

52. MSU Physics 231 Fall 2015 52 The 2 nd law of thermodynamics 1 st law:  U=Q+W In a cyclic process (  U=0) Q=-W: we cannot do more work than the amount of energy (heat) that we put inside 2 nd law in equivalent forms: -Heat flows spontaneously ONLY from hot to cold masses -Heat flow is accompanied by an increase in the entropy (disorder) of the universe -Natural processes evolve toward a state of maximum entropy

53. MSU Physics 231 Fall 2015 53 Entropy Lower Entropy Higher Entropy

54. MSU Physics 231 Fall 2015 54 Reversing Entropy We can only reverse the increase in entropy if we do work on the system Do work to compress the gas back to a smaller volume

55. MSU Physics 231 Fall 2015 55 Entropy The CHANGE in entropy (S): Adiabatic process Q=0 and  S = 0 If heat flows out (Q < 0) then  S < 0 entropy decreases If heat flows in (Q > 0) then  S > 0 entropy increases For a Carnot engine, there is no change in entropy over one complete cycle

56. MSU Physics 231 Fall 2015 56 Entropy and Work Entropy represents an inefficiency wherein energy is “lost” and cannot be used to do work.  S hot = -Q hot /T hot = -24000J / 400K = -60 J/K  S cold = Q cold /T cold = +24000J / 300K = +80 J/K  S hot +  S cold = -60 J/K + 80 J/K = +20 J/K Entropy increases! Cold mass: Gained heat, can do more work. Hot mass: Lost heat, can do less work. Cold mass gained less potential to do work than host mass lost. Net loss in the ability to do work.

57. MSU Physics 231 Fall 2015 57 Review: calorimetry If we connect two objects with different temperature energy will transferred from the hotter to the cooler one until their temperatures are the same. If the system is isolated: Energy flow into cold part = Energy flow out of hot part m c c c ( T f - T c ) = m h c h (T h - T f ) the final temperature is: T f = m c c c T c + m h c h T h m c c c + m h c h

58. MSU Physics 231 Fall 2015 58 Phase Change GAS(high T) liquid (medium T) Solid (low T) Q=c gas m  T Q=c liquid m  T Q=c solid m  T Gas  liquid liquid  solid Q=mL v Q=mL f

59. MSU Physics 231 Fall 2015 59 Heat transfer via conduction Conduction occurs if there is a temperature difference between two parts of a conducting medium Rate of energy transfer P P = Q/  t (unit Watt = J/s) P = k A (T h -T c )/  x = k A  T/  x k: thermal conductivity Unit: J/(m s o C) Metals k~300 J/(m s o C) Gases k~0.1 J/(m s o C) Nonmetals k~1 J/(m s o C) ThTh TcTc xx A

60. MSU Physics 231 Fall 2015 60 Multiple Layers ThTh TcTc A L 1 L 2 (  x) k 1 k 2

61. MSU Physics 231 Fall 2015 61 Net Power Radiated (photons) An object emits AND receives radiation, energy radiated per second = net power radiated (J/s) P NET =  A e (T 4 -T 0 4 ) = Power radiated – Power absorbed where T: temperature of object (K) T 0 : temperature of surroundings (K)  = 5.6696x10 -8 W/m 2 K 4 A = surface area e = object dependent constant emissivity (0-1) for a black body e=1 (all incident radiation is absorbed)

62. MSU Physics 231 Fall 2015 62 Wavelength where the radiant energy is maximum where b=2.90×10 −3 m K Wiens displacement constant