Physics 231 Topic 14: Laws of Thermodynamics Alex Brown Dec 7-11 2015 MSU Physics 231 Fall 2015 1 8th 10 pm correction for 3rd exam 9th 10 pm attitude survey (1% for participation) 10th 10 pm concept test timed (50 min) (1% for performance) 11th 10 pm last homework set 17th 8-10 pm final (Thursday) VMC E100 MSU Physics 231 Fall 2015 2
Clicker Question! ce 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 100oC MSU Physics 231 Fall 2015 3 Key Concepts: Laws of Thermodynamics Laws of Thermodynamics 1st Law: U = Q + W 2nd 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 MSU Physics 231 Fall 2015 4 Engine based on a container of an idea gas where the P, V and T change (n is fixed) piston area A y P,V,T n 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 MSU Physics 231 Fall 2015 5 Process visualized with a P-V diagram for the gas inside P isobaric line: pressure is constant volume changes i iso-volumetric line: volume is constant pressure V changes
n fixed MSU Physics 231 Fall 2015 6 PV = n R T 12) (ideal gas equation from chapter P = n R T/V = c T/V P (c = constant) lines with constant T T1 T2 T3
iso-thermal lines T4 T1 < T2 < T3 < T4 V MSU Physics 231 Fall 2015 7 A Piston Engine piston area A y Pi Vi T i Piston is moved downward slowly so that the gas remains in thermal equilibrium: Volume decreases (obviously) Temperature increases
Work is done on the gas Pf Vf Tf Ti < Tf vin vout vout > vin (speeds) work is done on the gas and temperature increases MSU Physics 231 Fall 2015 8 piston Isobaric Compression area A The pressure does not change while pushing down the piston (isobaric compression).
y P i Vi T i P Pf Vf Tf f i Vf Vi W = work done on the gas by pushing down on the piston P V MSU Physics 231 Fall 2015 9
piston Isobaric Compression area A The pressure does not change while lowering the piston (isobaric compression). y P i Vi T i P f Pf Vf Tf W = work done on the gas W = F d = - P A y (P=F/A) W = - P V = - P (Vf-Vi) (in Joule) Sign of the work done on the gas: + if V < 0 - if V > 0
i P work is the area under the curve in a P-V diagram with V decreasing Vf Vi V MSU Physics 231 Fall 2015 10 piston Non-isobaric Compression area A In general, the pressure can change when lowering the piston. y
P i Vi T i P Pf 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. f i Pi Vf Vi V MSU Physics 231 Fall 2015 11 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 2015 12 Work Done on Gases: Getting the Signs Right! P i V If the arrow goes from left to right (volume becomes larger) W < 0 and Wg = -W > 0 positive work (Wg) is done by the gas on the piston. the internal energy of the gas goes down MSU Physics 231 Fall 2015
13 iso-volumetric process P v Work done on/by gas: W = Wg = - PV = 0 MSU Physics 231 Fall 2015 14 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? ork: area under PV diagram ork done by the gas: volume must become larger, which aves (a) or (c). Area is larger for (a). MSU Physics 231 Fall 2015 15
=50 kg A=100 cm2 = 0.010 m2 mass and area of the li Patm a) What is the pressure PA? PA 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 2015 16 =50 kg A=100 cm2 = 0.010 m2 mass and area of the li Patm a) What is the pressure PA?
PA b) If the inside temperature is raised the lid moves up by 5 cm. How much work is done by the gas? a) PA = Patm + Mg/A = 1.50 x 105 b) Wg = PA V = 75.0 J MSU Physics 231 Fall 2015 17 For ideal gas PV=nRT ne mole of an ideal gas initially at 0 C undergoes an xpansion at constant pressure of one atmosphere to our times its original volume. ) What is the new temperature? ) What is the work done by the gas? MSU Physics 231 Fall 2015 18
For ideal gas PV=nRT ne mole of an ideal gas initially at 0 C undergoes an xpansion at constant pressure of one atmosphere to our times its original volume. ) What is the new temperature? ) What is the work done by the gas? a) Use PV = nRT to get Tf = (Vf/Vi) Ti = 1092 K b) W = -PV P(4Vi-Vi) = -3PVi = -3P(nRTi/P) Wg = -W = 3nRTi = 6806 J MSU Physics 231 Fall 2015 19 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 MSU Physics 231 Fall 2015 20 motion of atoms First Law of Thermodynamics U = Uf - Ui = 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 2015 21
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 monatomic) (chapter 12) (d=3 (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 2015 22 First Law: Isobaric Process A gas in a cylinder is kept at 1.0x105 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 100cm3. What is the change in internal energy? Q = -500 J V = -100 cm3 = -1.0x10-4 m3 W = - P V = 10 J U = Q + W = - 500 + 10 = - 490 MSU Physics 231 Fall 2015 23 First Law: General Case P (Pa) 6 In ideal gas (d=3) is compressed f i 3 1
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? 4 V(m3) A) U = (3/2)PV U = 3/2(PfVf - PiVi) = 3/2[6x1 - 3x4] = -9 J 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. MSU Physics 231 Fall 2015 24 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 2015 25 1) PV = n R T Iso-volumetric Process (V = 0) 2) U = W + Q 3) U = (d/2) n R T 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 2015
26 Isobaric Process (P = 0) 1) PV = n R T 2) U = W + Q 3) U = (d/2) n R T P = 0 4) W = - PV = - 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 2015 27
molar heat capacities Constant volume Q = (d/2) n R T = Cv n T where Cv = (d/2) R Constant pressure molar heat capacity at constant volume Q = [(d+2)/2] n R T = CP n T where CP = [(d+2)/2] R For all molar heat capacity at constant pressure U = (d/2) n R T = Cv n T MSU Physics 231 Fall 2015 28
Isothermal Process (T = 0) 1) PV = n R T 2) U = W + Q 3) U = (d/2) n R T T = 0 the U = 0 Q = -W PV = constant work done on gas is area under the curve: 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 2015 29 Adiabtic Process (Q = 0) 1) PV = n R T 2) U = W + Q Q = 0 (system is isolated) W = U (work goes into internal energy) 3) U = (d/2) n R T P (V) = constant (d+2)/d = Cp/Cv = >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 2015 30 Process Isobaric U nCv T Q nCp T W -P V Adiabatic
nCv T 0 U Isovolumet ric nCv T U 0 Isothermal nCv T=0 -W -nRTln(Vf/ Vi) nCv T
U-W General Ideal gas (monatomic d = 3) Cv = (d/2) R Cp = [(d+2)/2]MSU RPhysics 231 Fall 2015 (PV Area) negative if V expands (diatomic d = 5) 31 piston area A y P,V,T First Law: Adiabatic process
A piston is pushed down rapidly. Because th transfer of heat through the walls takes a long time, no heat can escape. During the moving of the piston, the temperature has risen 1000C. 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) = 1.25x104 J MSU Physics 231 Fall 2015 32 Clicker Quiz! A vertical cylinder with a movable cap is cooled. The process corresponding to this is: a) b)
c) d) e) CB AB AC CA 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 2015 33 Adiabatic process An molecular hydrogen gas goes from P1 = 9.26 atm and V1 = 0.0118 m3 to P2 and V2 via an adiabatic process. If P2 = 2.66 atm, what is V2 ? H2 (d=5) and adiabatic:
PV =Constant with = Cp/Cv = (d+2)/d= 7/5 P1 (V1)1.4 = P2 (V2)1.4 (V2)1.4 = (P1 /P2)(V1)1.4 = 0.0069 MSU Physics 231 Fall 2015 34 Cyclic processes (monatomic with d=3) P (Pa) A 5.0 1.0 B C 10 50 V (m3) 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 2015 35 Cyclic Process: Step by Step (1) P (Pa) B C 10 Negative work is done on the (the- gas doingP-V positive work). W= Areaisunder diagram A 5.0
1.0 Process A to B gas: 50 V (m3) = - [ (50-10)(1.0-0.0) +(50-10)(5.0-1.0) ] = - 40 - 80 = - 120 J (work done on gas) Wg = 120 J (work done by gas) U = 3/2 (PBVB - PAVA) = 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! MSU Physics 231 Fall 2015 36
Cyclic Process: Step by Step (2) P (Pa) Process B-C W = Area under P-V diagram A 5.0 1.0 = - [(50-10)(1.0-0.0)] W=40 J Work was done on the gas B C 10 50 V (m3) U = 3/2(PcVc-PbVb) = 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. MSU Physics 231 Fall 2015 37 Cyclic Process: Step by Step (3) P (Pa) Process C-A W=-Area under P-V diagram W=0 J No work was done on/by the gas. A 5.0 1.0 B
C 10 50 V (m3) U = 3/2(PcVc-PbVb)= = 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 2015 38 Summary of the process Quantity Process P (Pa) A 5.0
1.0 B C 10 50 A-B Work on gas (W) Heat(Q) U A-B -120 J 120 J 0J
B-C MSU Physics 231 Fall 2015 C-A 39 P (Pa) What did we do? A 5.0 1.0 B C 10 50 Quantity Process
Work on gas (W) Heat(Q) A-B -120 J 120 J 0J B-C 40 J -100 J -60 J C-A 0J
60 J 60 J -80 J 80 J 0 SUM (net) V (m3) U The gas performed net work (80 J) (Wg = 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 Wg = (5-1)(50-10)/2 = 80 MSU Physics 231 Fall 2015
40 P (Pa) Quantity Process A 5.0 1.0 B C 10 50 Work on gas (W) Heat(Q) U
80 J 0 SUM (net) V (m3) Qh = 180 heat input from hot source Qc = 100 heat output to cold source (wasted heat) Wg = -W = Qh Qc = 80 work output by gas (engine) MSU Physics 231 Fall 2015 41 Generalized Heat Engine Water turned to steam Wg = Qh - Qc Heat reservoir Th efficiency: Wg/Qh
e = 1 - Qc/Qh Qh (heat input) Work is done engine The steam moves a piston Wg Work Qc (heat output) The steam is condensed Cold reservoir Tc The efficiency is determined by how much of the heat you supply to the engine is turned into work instead of being los as waste. MSU Physics 231 Fall 2015
42 Reverse Direction: The Fridge MSU Physics 231 Fall 2015 43 Heat Pump (fridge) heat is expelled to outside heat reservoir Th Qh work is done a piston compresses the coolant work engine Qc the fridge is cooled cold reservoir Tc W
Coefficient of performance COP = |Qc|/W Qc: amount of heat removed W: work input W= Qh - Qc MSU Physics 231 Fall 2015 44 P (Pa) A 5.0 1.0 C 10 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
B amount of work done on the gas to remove 50 V (m3)heat from the cold source. MSU Physics 231 Fall 2015 45 ) ) ) ) ) Clicker Quiz! P (Pa) 3x105 Consider this clockwise cyclic process. Which of the following is true? 1x105
1 This This This This This is is is is is a a a a a 3 V (m3) heat engine and the work done by the gas is +4x1 heat engine and the work done by the gas is +6x1 heat engine and the work done by the gas is 4x1
fridge and the work done on the gas is +4x105 J fridge and the work done on the gas is +6x105 J Clockwise: work done by the gas, so heat engine Work by gas=area enclosed = (3-1) x (3x105-1x105) = 4x105 MSU Physics 231 Fall 2015 46 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 2015 47 AB isothermal expansion W-, Q+ BC adiabatic expansion Carnot engine
W-, T- Q=0 T=0 Th T=0 Q=0 W+, T+compression DA adiabatic W+, QTc CD isothermal compression MSU Physics 231 Fall 2015 48 inverse Carnot cycle Carnot cycle Work done by engine: Weng Weng = Qh - Qc
Efficiency: ecarnot = 1-(Tc/Th) A heat pump or a fridge! By doing work we can transport heat e = 1-(Qc/Qh) also holds since MSU Physics 231 Fall 2015 49 Carnot engine e =1 - (Qc/Qh ) always ecarnot =1 - (Tc/Th ) carnot only!! In general: e < ecarnot 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 buil MSU Physics 231 Fall 2015
50 Example The efficiency of a Carnot engine is 30%. The engine absor 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 (Qc/Qh) Qc = Qh(1-e) = 800(1-0.3) = 560 J b) for a Carnot engine: efficiency: e = 1 - (Tc/Th ) Tc = Th(1-e) = 500(1-0.3) = 350 c) W = Qh Qc = 800 560 = 240 J MSU Physics 231 Fall 2015 51 The 2nd law of thermodynamics 1st 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 2nd 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 2015 52 Entropy Lower Entropy Higher Entropy MSU Physics 231 Fall 2015 53 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 MSU Physics 231 Fall 2015 54 Entropy The CHANGE in entropy (S): (J/K unit) 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 MSU Physics 231 Fall 2015 55 Entropy and Work Entropy represents an inefficiency wherein energy is lost and cannot be used to do
work. Shot = -Qhot/Thot = -24000J / 400K = -60 J/K Scold = Qcold/Tcold = +24000J / 300K = +80 J/K Shot + Scold = -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. MSU Physics 231 Fall 2015 56 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 mc cc ( Tf - Tc) = mh ch (Th - Tf) the final temperature is: mc cc Tc + mh ch Th Tf =
mc c c + mh ch MSU Physics 231 Fall 2015 57 Phase Change GAS(high T) Q=cgasmT Gas liquid Q=mLv Q=csolidmTSolid (low T) liquid solid liquid (medium T) Q=cliquidmT MSU Physics 231 Fall 2015
Q=mLf 58 Heat transfer via conduction Tc Th 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 (Th-Tc)/x = k A T/x k: thermal conductivity Unit: J/(m s oC) x Metals k~300 J/(m s oC) Gases k~0.1
J/(m s oC) Nonmetals k~1 J/(m s oC) MSU Physics 231 Fall 2015 59 Multiple Layers Th k 1 Tc k2 A L1 L2 Q A(Th Tc ) P t ( Li / ki ) i (x)
MSU Physics 231 Fall 2015 60 Net Power Radiated (photons) An object emits AND receives radiation, energy radiated per second = net power radiated (J/s) PNET = A e (T4-T04) = Power radiated Power absorbed where T: temperature of object (K) T0: temperature of surroundings (K) = 5.6696x10-8 W/m2K4 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 2015 61 Wavelength where the radiant energy is maximum where b=2.90103 m K Wiens displacement constant
Distributed Systems Brief Overview CNT 5517-5564 ... Service-oriented computing Service registration/discovery Service composition Cloud Computing The Emerging Computing Model: Cloud, Edge & Beneath Cheaper to provide services to a thin client than to maintain a fat client in a changing...
The tax, interest, penalty or late fee for which the Nodal Officer has issued a notice, in relation to any proceedings under the Relevant Act; The tax, interest, penalty or late fee as determined to be payable by the assessee...
May 2018 - DOT&E efforts with Chief Management Officer Test Reform Management Group on T&E Cost Fidelity (TE-7) April 2019 - DOT&E established Cost of T&E Working Group. Involves DD,DTE&P and DTRMC. Support requested from Services and Agencies. In Early...
Kelvin K. Shen, Ph.D. Dr. Kelvin Shen is a technical and marketing consultant for fire retardant chemicals. His last industrial position was Sr. Global Market Development Manager of Fire Retardant Industry at Rio Tinto Minerals (former U.S. Borax/ Luzenac). He...
The Reference List. A list . at the end. of your essay of every source you have referenced (paraphrased or quoted) in your essay. Provides the information necessary for a reader to locate and retrieve any sources cited in your...
In the North the Danes and Vikings made their own laws! But by the 10th century, as England was once again close to becoming a single kingdom - it was the King (Anglo-Saxon) who was responsible for making laws.
A Digitally Programmable Polyphase Filter for Bluetooth By ... 57dB image band rejection CMOS gm-C polyphase filter with automatic frequency tuning for Bluetooth," Proc. Int. Symp. Circuits and Systems, ISCAS' 2002., vol. 5, pp. V-169 - II-172, 2002. A. Emira,...
Ready to download the document? Go ahead and hit continue!