UB, Phy101: Chapter 15, Pg 1 Physics 101: Chapter 15 Thermodynamics, Part I l Textbook Sections 15.1 – 15.5.

Slides:



Advertisements
Similar presentations
QUICK QUIZ 22.1 (end of section 22.1)
Advertisements

IB Physics Topic 3 & 10 Mr. Jean May 7 th, The plan: Video clip of the day Thermodynamics Carnot Cycle Second Law of Thermodynamics Refrigeration.
Physics 101: Lecture 30, Pg 1 PHY101: Lecture 30 Thermodynamics l New material: Chapter 15 l The dynamics of thermal processes (equilibrium, work, energy)
Thermodynamics April 27, 2015April 27, 2015April 27, 2015.
UB, Phy101: Chapter 14, Pg 1 Okay, so I was watching this video of some U.S. warplane breaking the sound barrier and there was this, sort of, vertical.
Kinetic Theory and Thermodynamics
L 19 - Thermodynamics [4] Change of phase ice  water  steam
Entropy and the Second Law of Thermodynamics
Physics 101: Lecture 25, Pg 1 Work Done by a System W W yy W = p  V W > 0 if  V > 0 expanding system does positive work W < 0 if  V < 0 contracting.
Second Law of Thermodynamics Physics 202 Professor Lee Carkner Lecture 18.
Physics 101: Lecture 31, Pg 1 Physics 101: Lecture 31 Thermodynamics, part 2 l Review of 1st law of thermodynamics l 2nd Law of Thermodynamics l Engines.
Second Law of Thermodynamics Physics 202 Professor Lee Carkner Lecture 18.
The Second Law of Thermodynamics Physics 102 Professor Lee Carkner Lecture 6.
Gas Laws. Molecular Picture of Gas Gas is made up of many individual molecules Number density is number of molecules/volume: –N/V =  /m –  is the mass.
The Second Law of Thermodynamics Physics 102 Professor Lee Carkner Lecture 7.
Physics 101: Lecture 27, Pg 1 Physics 101: Lecture 27 Thermodynamics l Today’s lecture will cover Textbook Chapter l Check your grades in grade.
The Second Law of Thermodynamics Physics 102 Professor Lee Carkner Lecture 7.
Phy 202: General Physics II Ch 15: Thermodynamics.
1 Thermal Physics 13 - Temperature & Kinetic Energy 15 - Laws of Thermodynamics.
MHS Physics Department AP Unit II C 2 Laws of Thermodynamics Ref: Chapter 12.
Important Terms & Notes Conceptual Physics Mar. 12, 2014.
Physics Lecture Notes The Laws of Thermodynamics
Chapter 15. ThermodynamicsThermodynamics  The name we give to the study of processes in which energy is transferred as heat and as work  There are 4.
Physics 12 Giancoli Chapter 15
Results from kinetic theory, 1 1. Pressure is associated with collisions of gas particles with the walls. Dividing the total average force from all the.
Physics 213: Lecture 3, Pg 1 Packet 3.4 Thermodynamics Thermodynamics l Internal Energy l W = PΔV l 1 st Law of Thermodynamics: ΔU = Q – W l Define: Adiabatic,
Chapter 15: Thermodynamics
Thermodynamics. First Law of Thermodynamics Energy Conservation  U = Q - W Heat flow into system Increase in internal energy of system Work done by system.
Physics 101: Lecture 28, Pg 1 Physics 101: Lecture 28 Thermodynamics II l Today’s lecture will cover Textbook Chapter Final.
Thermodynamics … the study of how thermal energy can do work
Laws of Thermodynamics Thermal Physics, Lecture 4.
Thermodynamics The First Law of Thermodynamics Thermal Processes that Utilize an Ideal Gas The Second Law of Thermodynamics Heat Engines Carnot’s Principle.
Lecture Outline Chapter 12 College Physics, 7 th Edition Wilson / Buffa / Lou © 2010 Pearson Education, Inc.
Chapter 19. “A theory is the more impressive the greater the simplicity of its premises, the more different kinds of things it relates, and the more extended.
Physics 101: Lecture 28, Pg 1 Physics 101: Lecture 28 Thermodynamics II l Today’s lecture will cover Textbook Chapter Final Check Final Exam.
Thermodynamics How Energy Is Transferred As Heat and Work Animation Courtesy of Louis Moore.
© 2005 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their.
Thermodynamics. Announcements – 1/21 Next Monday, 1/26 – Readiness Quiz 1 –Chapter 19, sections 1 – 4 –Chapter 20, sections 1 – 4 Next Wednesday, 1/28.
Chapter 13: Thermodynamics
Thermodynamics Internal energy of a system can be increased either by adding energy to the system or by doing work on the system Remember internal energy.
Physics 207: Lecture 27, Pg 1 Physics 207, Lecture 27, Dec. 6 l Agenda: Ch. 20, 1 st Law of Thermodynamics, Ch. 21  1 st Law of thermodynamics (  U=
Chapter 11 Laws of Thermodynamics. Chapter 11 Objectives Internal energy vs heat Work done on or by a system Adiabatic process 1 st Law of Thermodynamics.
B2 Thermodynamics Ideal gas Law Review PV=nRT P = pressure in Pa V = volume in m3 n = # of moles T= temperature in Kelvin R = 8.31 J K -1 mol -1 m = mass.
Physics 207: Lecture 29, Pg 1 Physics 207, Lecture 29, Dec. 13 l Agenda: Finish Ch. 22, Start review, Evaluations  Heat engines and Second Law of thermodynamics.
Thermodynamics Davidson College APSI Ideal Gas Equations P 1 V 1 / T 1 = P 2 V 2 / T 2 PV = n R T (using moles) P V = N k B T (using molecules)  P:
203/4c18:1 Chapter 18: The Second Law of Thermodynamics Directions of a thermodynamic process Reversible processes: Thermodynamic processes which can be.
Physics 101: Lecture 28, Pg 1 Physics 101: Lecture 28 Thermodynamics II l Today’s lecture will cover Textbook Chapter Final Check Final Exam.
Thermodynamic Processes
Physics 101: Lecture 28, Pg 1 Physics 101: Lecture 28 Thermodynamics II l Today’s lecture will cover Textbook Chapter
Physics 101: Lecture 27, Pg 1 Physics 101: Lecture 27 Thermodynamics l Today’s lecture will cover Textbook Chapter Check your grades in grade.
Physics 101: Lecture 26, Pg 1 Physics 101: Lecture 26 Thermodynamics II Final.
Thermodynamics II Thermodynamics II. THTH TCTC QHQH QCQC W HEAT ENGINE THTH TCTC QHQH QCQC W REFRIGERATOR system l system taken in closed cycle   U.
Lecture 26: Thermodynamics II l Heat Engines l Refrigerators l Entropy l 2 nd Law of Thermodynamics l Carnot Engines.
Chapter 12 Laws of Thermodynamics. Chapter 12 Objectives Internal energy vs heat Work done on or by a system Adiabatic process 1 st Law of Thermodynamics.
Chapter 20 - Thermodynamics A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007.
THE SECOND LAW OF THERMODYNAMICS Entropy. Entropy and the direction of time Microscopically the eqs. of physics are time reversible ie you can turn the.
THERMODYNAMICS THE NEXT STEP. THERMAL PROPERTIES OF MATTER STATE VARIABLES – DESCRIBE THE SUBSTANCE –PRESSURE –TEMPERATURE –VOLUME –QUANITY OF SUBSTANCE.
Lecture Outline Chapter 12 College Physics, 7 th Edition Wilson / Buffa / Lou © 2010 Pearson Education, Inc.
Lecture 26Purdue University, Physics 2201 Lecture 26 Thermodynamics I PHYSICS 220.
Thermodynamics AP B. ‘its hot enough to fry an egg on the sidewalk’
Lecture 27Purdue University, Physics 2201 Lecture 27 Thermodynamics II Physics 220.
Physics 101: Lecture 26, Pg 1 Chapter 15, Problem 3 Consider a hypothetical device that takes 1000 J of heat from a hot reservoir at 300K, ejects 200 J.
Physics 101: Lecture 27, Pg 1 Physics 101: Lecture 27 Thermodynamics l Today’s lecture will cover Textbook Chapter
Chapter 11 Super Review. 1. A two mole sample of a gas has a temperature of 1000 K and a volume of 6 m 3. What is the pressure?
Physics 101: Lecture 25 Thermodynamics
Physics 101: Lecture 28 Thermodynamics II
Please fill in on-line course evaluation.
Physics 101: Lecture 27 Thermodynamics
Physics 101: Lecture 31 Thermodynamics, part 2
Presentation transcript:

UB, Phy101: Chapter 15, Pg 1 Physics 101: Chapter 15 Thermodynamics, Part I l Textbook Sections 15.1 – 15.5

UB, Phy101: Chapter 15, Pg 2 The Ideal Gas Law (review) l pV = Nk B T è N = number of molecules »N = number of moles (n) x N A molecules/mole è k B = Boltzmann’s constant = 1.38 x J/K l pV = nRT è R = ideal gas constant = N A k B = 8.31 J/mol/K =.0823 atm-l/mol/K

UB, Phy101: Chapter 15, Pg 3 Summary of Kinetic Theory: The relationship between energy and temperature (for monatomic ideal gas) l Internal Energy U = number of molecules x ave KE/molecule = N (3/2)k B T = (3/2)nRT = (3/2)pV Careful with units: R = 8.31 J/mol/K m = molar mass in kg v rms = speed in m/s

UB, Phy101: Chapter 15, Pg 4 Thermodynamic Systems and p-V Diagrams l ideal gas law: pV = nRT l for n fixed, p and V determine “state” of system è T = pV/nR è U = (3/2)nRT = (3/2)pV l examples: è which point has highest T? » 2 è which point has lowest U? » 3 è to change the system from 3 to 2, energy must be added to system V P V 1 V 2 P1P3P1P3

UB, Phy101: Chapter 15, Pg 5 Work Done by a System W W yy W = p  V W > 0 if  V > 0 expanding system does positive work W < 0 if  V < 0 contracting system does negative work W = 0 if  V = 0 system with constant volume does no work

UB, Phy101: Chapter 15, Pg 6 Problem: Thermo I V P V P W = P  V (>0)  V > 0V P W = P  V =  V = 0 V P W = P  V (<0)  V < 0 V W = P  V = P  V = 0 V P If we go the other way then W tot < 0 V P W tot > 0 W tot = ??

UB, Phy101: Chapter 15, Pg 7 V P Now try this: V 1 V 2 P1P3P1P3 Area = (V 2 -V 1 )x(P 1 -P 3 )/2

UB, Phy101: Chapter 15, Pg 8 First Law of Thermodynamics Energy Conservation  U = Q - W Heat flow into system Increase in internal energy of system Work done by system V P V 1 V 2 P1P3P1P3 Equivalent ways of writing 1st Law: Q =  U + W Q =  U + p  V

UB, Phy101: Chapter 15, Pg 9 First Law of Thermodynamics Example V P 1 2 V 1 V 2 P 2 moles of monatomic ideal gas is taken from state 1 to state 2 at constant pressure p=1000 Pa, where V 1 =2m 3 and V 2 =3m 3. Find T 1, T 2,  U, W, Q. 1. pV 1 = nRT 1  T 1 = pV 1 /nR = 120K 2. pV 2 = nRT 2  T 2 = pV 2 /nR = 180K 3.  U = (3/2) nR  T = 1500 J  U = (3/2) p  V = 1500 J (has to be the same) 4. W = p  V = 1000 J 5. Q =  U + W = = 2500 J

UB, Phy101: Chapter 15, Pg 10 First Law of Thermodynamics Example 2 moles of monatomic ideal gas is taken from state 1 to state 2 at constant volume V=2m 3, where T 1 =120K and T 2 =180K. Find Q. 1. pV 1 = nRT 1  p 1 = nRT 1 /V = 1000 Pa 2. pV 2 = nRT 1  p 2 = nRT 2 /V = 1500 Pa 3.  U = (3/2) nR  T = 1500 J 4. W = p  V = 0 J 5. Q =  U + W = = 1500 J requires less heat to raise T at const. volume than at const. pressure V P 2 1 V P2P1P2P1

UB, Phy101: Chapter 15, Pg 11 Preflight Shown in the picture below are the pressure versus volume graphs for two thermal processes, in each case moving a system from state A to state B along the straight line shown. In which case is the work done by the system the biggest? 1. Case 1 2. Case 2 3. Same A B V(m 3 ) Case 1 A B V(m 3 ) P(atm) Case 2 P(atm) correct Net Work = area under P-V curve Area the same in both cases!

UB, Phy101: Chapter 15, Pg 12 Preflight Shown in the picture below are the pressure versus volume graphs for two thermal processes, in each case moving a system from state A to state B along the straight line shown. In which case is the change in internal energy of the system the biggest? 1. Case 1 2. Case 2 3. Same A B V(m 3 ) Case 1 A B V(m 3 ) P(atm) Case 2 P(atm) correct  U = 3/2  (pV) Case 1:  (pV) = 4x9-2x3=30 atm-m 3 Case 2:  (pV) = 2x9-4x3= 6 atm-m 3

UB, Phy101: Chapter 15, Pg 13 Preflight (followup) Shown in the picture below are the pressure versus volume graphs for two thermal processes, in each case moving a system from state A to state B along the straight line shown. In which case is the heat added to the system the biggest? 1. Case 1 2. Case 2 3. Same A B V(m 3 ) Case 1 A B V(m 3 ) P(atm) Case 2 P(atm) correct Q =  U + W W is same for both  U is larger for Case 1 Therefore, Q is larger for Case 1

UB, Phy101: Chapter 15, Pg 14 Q =  U + W Heat flow into system Increase in internal energy of system Work done by system V P V 1 V 2 P1P3P1P3 Which part of cycle has largest change in internal energy,  U ? 2  3 (since U = 3/2 pV) l Which part of cycle involves the least work W ? 3  1 (since W = p  V) l What is change in internal energy for full cycle?  U = 0 for closed cycle (since both p & V are back where they started) l What is net heat into system for full cycle?  U = 0  Q = W = area of triangle (>0) Some questions: DEMO

UB, Phy101: Chapter 15, Pg 15Recap: è 1st Law of Thermodynamics: Energy Conservation Q =  U + W Heat flow into system Increase in internal energy of system Work done by system V P l point on p-V plot completely specifies state of system (pV = nRT) l work done is area under curve l U depends only on T (U = 3nRT/2 = 3pV/2) for a complete cycle  U=0  Q=W

UB, Phy101: Chapter 15, Pg 16 Thermodynamics, part 2 l engines and refrigerators l 2nd law of thermodynamics

UB, Phy101: Chapter 15, Pg 17 THTH TCTC QHQH QCQC W HEAT ENGINE THTH TCTC QHQH QCQC W REFRIGERATOR system Engines and Refrigerators l system taken in closed cycle   U system = 0 l therefore, net heat absorbed = work done Q H - Q C = W (engine) Q C - Q H = -W (refrigerator) energy into green blob = energy leaving green blob

UB, Phy101: Chapter 15, Pg 18 THTH TCTC QHQH QCQC W HEAT ENGINE The objective: turn heat from hot reservoir into work The cost: “waste heat” 1st Law: Q H -Q C = W efficiency e  W/Q H =W/Q H = (Q H -Q C )/Q H = 1-Q C /Q H

UB, Phy101: Chapter 15, Pg 19 THTH TCTC QHQH QCQC W REFRIGERATOR The objective: remove heat from cold reservoir The cost: work 1st Law: Q H = W + Q C coeff of performance CP r  Q C /W = Q C /W = Q C /(Q H - Q C )

UB, Phy101: Chapter 15, Pg 20 Preflights Preflights Consider a hypothetical device that takes 1000 J of heat from a hot reservoir at 300K, ejects 200 J of heat to a cold reservoir at 100K, and produces 800 J of work. Does this device violate the first law of thermodynamics ? 1. Yes 2. No This device doesn't violate the first law of thermodynamics because no energy is being created nor destroyed. All the energy is conserved. correct l W (800) = Q hot (1000) - Q cold (200) l Efficiency = W/Q hot = 800/1000 = 80% 80% efficient

UB, Phy101: Chapter 15, Pg 21 New concept: Entropy (S) l A measure of “disorder” l A property of a system (just like p, V, T, U) è related to number of different “states” of system l Examples of increasing entropy: è ice cube melts è gases expand into vacuum l Change in entropy: è  S = Q/T »>0 if heat flows into system (Q>0) »<0 if heat flows out of system (Q<0)

UB, Phy101: Chapter 15, Pg 22 Second Law of Thermodynamics l The entropy change (Q/T) of the system+environment  0 è never < 0 è order to disorder l Consequences è A “disordered” state cannot spontaneously transform into an “ordered” state è No engine operating between two reservoirs can be more efficient than one that produces 0 change in entropy. This is called a “Carnot engine”

UB, Phy101: Chapter 15, Pg 23 THTH TCTC QHQH QCQC W HEAT ENGINE The objective: turn heat from hot reservoir into work The cost: “waste heat” 1st Law: Q H -Q C = W efficiency e  W/Q H =W/Q H = 1-Q C /Q H  S = Q C /T C - Q H /T H  0  S = 0 for Carnot Therefore, Q C /Q H  T C / T H Q C /Q H = T C / T H for Carnot Therefore e = 1 - Q C /Q H  1 - T C / T H e = 1 - T C / T H for Carnot e = 1 is forbidden! e largest if T C << T H Engines and the 2nd Law

UB, Phy101: Chapter 15, Pg 24 Preflight Consider a hypothetical device that takes 1000 J of heat from a hot reservoir at 300K, ejects 200 J of heat to a cold reservoir at 100K, and produces 800 J of work. Does this device violate the second law of thermodynamics ? 1. Yes 2. No. correct l W (800) = Q hot (1000) - Q cold (200) l Efficiency = W/Q hot = 800/1000 = 80% l Max eff = /300 = 67%

UB, Phy101: Chapter 15, Pg 25 Second Law of Thermodynamics l The entropy change (Q/T) of the system+environment  0 è never < 0 è order to disorder l Consequences è A “disordered” state cannot spontaneously transform into an “ordered” state è No engine operating between two reservoirs is more efficient than one that produces 0 change in entropy (“Carnot engine”) è Heat cannot be transferred spontaneously from cold to hot

UB, Phy101: Chapter 15, Pg 26 Preflight Preflight Consider a hypothetical refrigerator that takes 1000 J of heat from a cold reservoir at 100K and ejects 1200 J of heat to a hot reservoir at 300K. 1. How much work does the refrigerator do? 2. What happens to the entropy of the universe? 3. Does this violate the 2nd law of thermodynamics? Answers: 200 J Decreases yes THTH TCTC QHQH QCQC W Q C = 1000 J Q H = 1200 J Since Q C + W = Q H, W = 200 J  S H = Q H /T H = (1200 J) / (300 K) = 4 J/K  S C = -Q C /T C = (-1000 J) / (100 K) = -10 J/K  S TOTAL =  S H +  S C = -6 J/K  decreases (violates 2 nd law)