Lecture27

Physics 101: Lecture 27 Thermodynamics Today’s lecture will cover Textbook Chapter 15.1-15.6 Check your grades in grade book!! HE 3 review Sun 8pm 141 Loomis HE 3 Mon Dec 6 7pm Final

First Law of Thermodynamics Energy Conservation  U = Q + W Heat flow into system Increase in internal energy of system Equivalent ways of writing 1st Law: Q =  U - W 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) Work done on system V P 1 2 3 V 1 V 2 P 1 P 3

Signs Example You are heating some soup in a pan on the stove. To keep it from burning, you also stir the soup. Apply the 1 st law of thermodynamics to the soup. What is the sign of ( A=Positive B= Zero C=Negative ) 1) Q 2) W 3)  U Positive, heat flows into soup Zero, is close to correct Positive, Soup gets warmer

Work Done on a System ACT W = -p  V : only for constant Pressure W < 0 if  V > 0 negative work required to expand system W > 0 if  V < 0 positive work required to contract system W = 0 if  V = 0 no work needed to keep system at const V W = - (work done BY system) = -F d cos  = -P A d = -P A  y = -P  V The work done on the gas as it contracts is A) Positive B) Zero C) Negative M M  y

Thermodynamic Systems and P-V Diagrams ideal gas law: PV = nRT for n fixed, P and V determine “state” of system T = PV/nR U = (3/2)nRT = (3/2)PV 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 V P A B C V 1 V 2 P 1 P 3

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 Pa , where V 1 =2m 3 and V 2 =3m 3 . Find T 1 , T 2 ,  U, W, Q. (R=8.31 J/k mole) 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 = 1500 + 1000 = 2500 J V P 1 2 V 1 V 2 P

First Law of Thermodynamics Isochoric 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. Q =  U - W 2.  U = (3/2) nR  T = 1500 J 3. W = -P  V = 0 J 4. Q =  U - W = 1500 + 0 = 1500 J V P 2 1 V P 2 P 1 requires less heat to raise T at const. volume than at const. pressure

Homework Problem: Thermo I W tot = ?? V P 1 2 3 4 V P W = -P  V (<0) 1 2 3 4  V > 0 V P W = -P  V = 0 1 2 3 4  V = 0 V P W = -P  V (>0) 1 2 3 4  V < 0 V W = -P  V = 0 1 2 3 4 P  V = 0 V P 1 2 3 4 W tot < 0

WORK ACT If we go the opposite direction for the cycle (4,3,2,1) the net work done on the system will be A) Positive B) Negative V P 1 2 3 4 If we go the other way then…

PV 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? A. Case 1 B. Case 2 C. Same A B 4 2 3 9 V(m 3 ) Case 1 A B 4 2 3 9 V(m 3 ) P(atm) Case 2 P(atm) Net Work = area under P-V curve Area the same in both cases! correct

PV 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? A. Case 1 B. Case 2 C. Same  U = 3/2 (p f V f – p i V i ) Case 1:  U = 3/2(4x9-2x3)=45 atm-m 3 Case 2:  U = 3/2(2x9-4x3)= 9 atm-m 3 A B 4 2 3 9 V(m 3 ) Case 1 A B 4 2 3 9 V(m 3 ) P(atm) Case 2 P(atm) correct

PV ACT3 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 C. Same Q =  U - W W is same for both  U is larger for Case 1 Therefore, Q is larger for Case 1 A B 4 2 3 9 V(m 3 ) Case 1 A B 4 2 3 9 V(m 3 ) P(atm) Case 2 P(atm) correct

First Law Questions Q =  U - W Which part of cycle has largest change in internal energy,  U ? 2  3 (since U = 3/2 pV ) Which part of cycle involves the least work W ? 3  1 (since W = -p  V) What is change in internal energy for full cycle?  U = 0 for closed cycle (since both p & V 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) Some questions: Heat flow into system Increase in internal energy of system Work done on system V P 1 2 3 V 1 V 2 P 1 P 3

Special PV Cases Constant Pressure (isobaric) Constant Volume Constant Temp  U = 0 Adiabatic Q=0 V P W = -P  V (<0) 1 2 3 4  V > 0 V P W = -P  V = 0 1 2 3 4  V = 0

Preflights 1-3 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 “ the change in U=Q+W ” -W (800) = Q hot (1000) - Q cold (200) Efficiency = -W/Q hot = 800/1000 = 80% 80% efficient 20% efficient 25% efficient correct

Reversible? Most “physics” processes are reversible: you could play movie backwards and still looks fine. (drop ball vs throw ball up) Exceptions: Non-conservative forces (friction) Heat Flow: Heat never flows spontaneously from cold to hot

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

Lecture27

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