Physics 231. Topic 14: Laws of Thermodynamics. Alex Brown Dec MSU Physics 231 Fall

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1 Physics 231 Topic 14: Laws of Thermodynamics Alex Brown Dec MSU Physics 231 Fall

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

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 MSU Physics 231 Fall

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 Entropy Carnot engine & efficiency Relationship to heat, energy. Statistical interpretation Covers chapter 14 in Rex & Wolfson MSU Physics 231 Fall

5 piston Engine based on a container of an idea gas where the P, V and T change (n is fixed) area A 1) Put in contact with a source of heat at high T during which heat energy flows in and piston is pushed up. P,V,T y 2) Put in contact with a source of heat at low T during which piston is pushed down and heat flows out. n fixed 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 MSU Physics 231 Fall

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

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

8 piston area A y 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 P i V i T i P f V f T f T i < T f v out v in v out > v in (speeds) work is done on the gas and temperature increases MSU Physics 231 Fall

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

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

11 piston area A Non-isobaric Compression In general, the pressure can change when lowering the piston. y P P f P i V i T i f 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. P i i V f V i V MSU Physics 231 Fall

12 Work Done on Gases: Getting the Signs Right! P i V 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 MSU Physics 231 Fall

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

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

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). MSU Physics 231 Fall

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

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

18 For ideal gas PV=nRT 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? MSU Physics 231 Fall

19 For ideal gas PV=nRT 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? 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 MSU Physics 231 Fall

20 First Law of Thermodynamics By transferring heat to an object the internal energy can increased By performing work on 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 MSU Physics 231 Fall

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 MSU Physics 231 Fall

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) MSU Physics 231 Fall

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 = = MSU Physics 231 Fall

24 P (Pa) 6 3 f First Law: General Case i 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? 1 4 V(m 3 ) A) U = (3/2)PV U = 3/2(P f V f -P i V i ) = 3/2[6x1-3x4] = -9 J B) Work: area under the P-V graph: ( ) = 13.5 (positive since work is done on the gas) C) U = Q+W so Q = U-W = = J Heat has been extracted from the gas. MSU Physics 231 Fall

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

26 1) PV = n R T 2) U = W + Q 3) U = (d/2) n R T 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) MSU Physics 231 Fall

27 1) PV = n R T 2) U = W + Q 3) U = (d/2) n R T Isobaric Process ( P = 0) 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) MSU Physics 231 Fall

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 MSU Physics 231 Fall

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 W nrt ln 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) MSU Physics 231 Fall

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) dashed lines are isotherms When V = + P = - T = - U = - W = - (work done by gas) MSU Physics 231 Fall

31 Process U Q W Isobaric nc v T nc p T -P V Adiabatic nc v T 0 U Isovolumetric nc v T U 0 Isothermal nc v T=0 -W -nrtln(v f /V i ) General nc v T U-W (PV Area) negative if V expands Ideal gas (monatomic d = 3) (diatomic d = 5) C v = (d/2) R C p = [(d+2)/2] R MSU Physics 231 Fall

32 piston area A y P,V,T 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 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 W = U = (3/2) nr T = (3/2)(10)(8.31)(100) = 1.25x10 4 J MSU Physics 231 Fall

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. MSU Physics 231 Fall

34 Adiabatic process An molecular hydrogen gas goes from P 1 = 9.26 atm and V 1 = 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 = V 2 = = (1/1.4) = 0.714) MSU Physics 231 Fall

35 Cyclic processes (monatomic with d=3) P (Pa) 5.0 A 1.0 C B V (m 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] MSU Physics 231 Fall

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

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

38 Cyclic Process: Step by Step (3) 5.0 P (Pa) A Process C-A W=-Area under P-V diagram W=0 J No work was done on/by the gas. 1.0 C B V (m 3 ) 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. MSU Physics 231 Fall

39 Summary of the process 5.0 P (Pa) A Quantity Process Work on gas (W) Heat(Q) U A-B -120 J 120 J 0 J B-C 40 J -100 J -60 J 1.0 C B C-A 0 J 60 J 60 J SUM (net) -80 J 80 J 0 V (m 3 ) A-B B-C C-A MSU Physics 231 Fall

40 5.0 P (Pa) A What did we do? Quantity Process Work on gas (W) Heat(Q) U A-B -120 J 120 J 0 J B-C 40 J -100 J -60 J 1.0 C B C-A 0 J 60 J 60 J SUM -80 J 80 J 0 (net) V (m 3 ) 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! 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 MSU Physics 231 Fall

41 5.0 P (Pa) A Quantity Process Work on gas (W) Heat(Q) U A-B -120 J 120 J 0 J B-C 40 J -100 J -60 J 1.0 C B V (m 3 ) C-A 0 J 60 J 60 J SUM (net) -80 J 80 J 0 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 = MSU Physics 231 Fall

42 Generalized Heat Engine Water turned to steam Heat reservoir T h W g = Q h -Q c efficiency: W g /Q h The steam moves a piston The steam is condensed Q h engine Q c Cold reservoir T c (heat input) W g (heat output) Work is done Work e = 1 - Q c /Q h 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. MSU Physics 231 Fall

43 Reverse Direction: The Fridge MSU Physics 231 Fall

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

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

46 P (Pa) 3x10 5 1x10 5 Clicker Quiz! Consider this clockwise cyclic process. Which of the following is true? 1 3 V (m 3 ) 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 MSU Physics 231 Fall

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? MSU Physics 231 Fall

48 A B isothermal expansion B C adiabatic expansion W-, Q+ Carnot engine W-, T- T=0 Q=0 T h Q=0 T=0 W+, T+ D A adiabatic compression W+, Q- T c C D isothermal compression MSU Physics 231 Fall

49 Carnot cycle inverse 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 A heat pump or a fridge! By doing work we can transport heat MSU Physics 231 Fall

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. MSU Physics 231 Fall

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 = = 240 J MSU Physics 231 Fall

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 MSU Physics 231 Fall

53 Entropy Lower Entropy Higher Entropy MSU Physics 231 Fall

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

55 Entropy The CHANGE in entropy (S): Adiabatic process Q=0 and S= 0 (J/K unit) 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 MSU Physics 231 Fall

56 Entropy and Work S hot = -Q hot /T hot = J / 400K = -60 J/K S cold = Q cold /T cold = J / 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. Entropy represents an inefficiency wherein energy is lost and cannot be used to do work. Cold mass gained less potential to do work than host mass lost. Net loss in the ability to do work. MSU Physics 231 Fall

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 MSU Physics 231 Fall

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

59 Heat transfer via conduction T h T c Conduction occurs if there is a temperature difference between two parts of a conducting medium Rate of energy transfer P A P = Q/ t (unit Watt = J/s) P = k A (T h -T c )/ x = k A T/ x 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) MSU Physics 231 Fall

60 Multiple Layers T h k 1 k 2 T c A P Q t A( T i h ( L T i / c k i ) ) L 1 L 2 ( x) MSU Physics 231 Fall

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 04 ) = Power radiated Power absorbed where T: temperature of object (K) T 0 : temperature of surroundings (K) = x10-8 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) MSU Physics 231 Fall

62 Wavelength where the radiant energy is maximum where b= m K Wiens displacement constant MSU Physics 231 Fall