Physics 101: Lecture 27 Thermodynamics

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1 Final hysics 101: Lecture 27 Thermodynamics Today s lecture will cover Textbook Chapter Check your grades in grade book!! hysics 101: Lecture 27, g 1

2 0 N=81 Mean=76.5 Number of students Scaled Score hysics 101: Lecture 27, g 2

3 First Law of Thermodynamics Energy Conservation The change in internal energy of a system ( U) is equal to the heat flow into the system (Q) plus the work done on the system (W) U = Q + W Increase in internal energy of system Heat flow into system Work done on system hysics 101: Lecture 27, g

4 Work Done on a System ACT M M The work done on the gas as it contracts is A) ositive B) Zero C) Negative W = work done ON system=- (work done BY system) = - W = -p : true for constant ressure W < 0 if > 0 negative work required to expand system W > 0 if < 0 positive work required to contract system W = 0 if = 0 no work needed to keep system at const hysics 101: Lecture 27, g

5 Thermodynamic Systems and - Diagrams ideal gas law: = nrt for n fixed, and determine state of system T = /nr U = (/2)nRT = (/2) for monatomic gas Examples (ACT): which point has highest T?» B which point has lowest U?» C to change the system from C to B, energy must be added to system A B 1 C hysics 101: Lecture 27, g 5

6 First Law of Thermodynamics Isobaric Example 2 moles of monatomic ideal gas is taken from state 1 to state 2 at constant pressure p=1000 a, where 1 =2m and 2 =m. Find T 1, T 2, U, W, Q. (R=8.1 J/k mole) 1. 1 = nrt 1 T 1 = 1 /nr = 120K = nrt 2 T 2 = 2 /nr = 180K. U = (/2) nr T = 1500 J U = (/2) p = 1500 J (has to be the same). W = -p = J 5. Q = U - W = = 2500 J hysics 101: Lecture 27, g 6

7 First Law of Thermodynamics Isochoric Example 2 moles of monatomic ideal gas is taken from state 1 to state 2 at constant volume =2m, where T 1 =120K and T 2 =180K. Find Q. 1. Q = U - W U = (/2) nr T = 1500 J. W = - = 0 J. Q = U - W = = 1500 J requires less heat to raise T at const. volume than at const. pressure hysics 101: Lecture 27, g 7

8 Homework roblem: Thermo I W tot =?? W = - (<0) W = - = 0 W tot < 0 > 0 = 0 W = - (>0) W = - = 0 < 0 = 0 hysics 101: Lecture 27, g 8

9 ACTs 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 on the system the biggest? (atm) (atm) A. Case 1 B. Case 2 B A C. Same correct 2 A Case 1 2 Case 2 B 9 (m ) 9 (m ) Net Work = area under - curve Area the same in both cases! hysics 101: Lecture 27, g 9

10 ACT 2 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? (atm) (atm) A. Case 1 correct B. Case 2 B A C. Same A 2 Case 1 2 Case 2 B 9 (m ) 9 (m ) U = /2 (p f f p i i ) Case 1: U = /2(x9-2x)=5 atm-m Case 2: U = /2(2x9-x)= 9 atm-m hysics 101: Lecture 27, g 10

11 ACT 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? A. Case 1 B. Case 2 correct (atm) (atm) C. Same B A 2 A Case 1 2 Case 2 B Q = U - W 9 (m ) 9 (m ) W is same for both U is larger for Case 1 Therefore, Q is larger for Case 1 hysics 101: Lecture 27, g 11

12 First Law Questions Q = U - W Work done on system Heat flow into system Some questions: Increase in internal energy of system Which part of cycle has largest change in internal energy, U? 2 (since U = /2 p) Which part of cycle involves the least work W? 1 (since W = -p ) What is change in internal energy for full cycle? U = 0 for closed cycle (since both p & are back where they started) What is net heat into system for full cycle (positive or negative)? U = 0 Q = -W = area of triangle (>0) 1 hysics 101: Lecture 27, g 12

13 Special Cases Constant ressure (isobaric) W = - (<0) Constant olume W = - = 0 > 0 Constant Temp U = 0 = 0 hysics 101: Lecture 27, g 1

14 reflights 1-1 Consider a hypothetical device that takes 1000 J of heat from a hot reservoir at 00K, 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 the change in U=Q+W 2. No correct -W (800) = Q hot (1000) - Q cold (200) Efficiency = -W/Q hot = 800/1000 = 80% 26% 7% 80% efficient 20% efficient 25% efficient 12% 21% 68% 0% 20% 0% 60% 80% 0% 20% 0% 60% 80% hysics 101: Lecture 27, g 1

15 Summary: 1st Law of Thermodynamics: Energy Conservation Q = U - W Work done on system Heat flow into system Increase in internal energy of system point on p- plot completely specifies state of system (p = nrt) work done is area under curve U depends only on T (U = nrt/2 = p/2) for a complete cycle U=0 Q=-W hysics 101: Lecture 27, g 15

PHYS 1101 Practice problem set 6, Chapter 19: 7, 12, 19, 30, 37, 44, 53, 61, 69

PHYS 1101 Practice problem set 6, Chapter 19: 7, 12, 19, 30, 37, 44, 53, 61, 69 PYS 0 Practice problem set 6, hapter 9: 7,, 9, 0, 7, 44,, 6, 69 9.7. Solve: (a) he heat extracted from the cold reservoir is calculated as follows: (b) he heat exhausted to the hot reservoir is K 4.0 00