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)