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Physics 101: Lecture 25 Thermodynamics

Final. Physics 101: Lecture 25 Thermodynamics. Check your grades in grade book Final is Cumulative. P. Work done by system. 1. P 1 P 3. 2. 3. V. V 1 V 2. First Law of Thermodynamics Energy Conservation.

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Physics 101: Lecture 25 Thermodynamics

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  1. Final Physics 101: Lecture 25 Thermodynamics • Check your grades in grade book • Final is Cumulative

  2. P Work done by system 1 P1 P3 2 3 V V1 V2 First Law of ThermodynamicsEnergy Conservation The change in internal energy of a system (DU) is equal to the heat flow into the system (Q) minus the work done by the system (W) DU = Q- W Increase in internalenergy of system Heat flow into system Equivalent ways of writing 1st Law: Q = DU + W 07

  3. y M M Work Done by a System ACT The work done by the gas as it contracts is A) Positive B) Zero C) Negative W = F d cosq =P A d = P A Dy = P DV W = p V :For constant Pressure W > 0 if V > 0 expanding system doespositivework W < 0 if V < 0 contracting system doesnegativework W = 0 if V = 0 system with constant volume doesnowork 13

  4. 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: • which point has highest T? • which point has lowest U? • to change the system from C to B, energy must be added to system P A B P1 P3 C V V1 V2 17

  5. P P W = PDV (>0) W = PDV = 0 1 1 2 2 4 4 3 3 DV > 0 V DV = 0 V Special PV Cases • Constant Pressure (isobaric) • Constant Volume (isochoric) • Constant Temp DU = 0 • Adiabatic Q=0 42

  6. P 1 P 2 V V1 V2 First Law of ThermodynamicsIsobaric Example 2 moles of monatomic ideal gas is taken from state 1 to state 2 at constant pressure p=1000 Pa, where V1 =2m3 and V2 =3m3. Find T1, T2, DU, W, Q. (R=8.31 J/k mole) 21

  7. First Law of ThermodynamicsIsochoric Example 2 moles of monatomic ideal gas is taken from state 1 to state 2 at constant volume V=2m3, where T1=120K and T2 =180K. Find Q. P P2 P1 2 1 V V 24

  8. Example: complete cycle 1 P 2 4 3 P P W = PDV (>0) W = PDV = 0 V 1 1 2 2 1 P 2 4 4 3 3 Wtot > 0 4 3 DV > 0 V DV = 0 V V P P W = PDV (<0) W = PDV = 0 1 1 2 2 4 4 3 3 V DV = 0 V DV < 0 Wtot = ?? 27

  9. 1 P 2 If we gothe other way then 4 3 V WORK Question If we go the opposite direction for the cycle (4,3,2,1) the net work done by the system will be A) Positive B) Negative 30

  10. PVQuestion 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? A. Case 1 B. Case 2 C. Same P(atm) P(atm) A B 4 4 A B 2 2 Case 1 Case 2 3 9 3 9 V(m3) V(m3) Net Work = area under P-V curve Area the same in both cases! 29

  11. P(atm) P(atm) A B 4 4 A B 2 2 Case 1 Case 2 3 9 3 9 V(m3) V(m3) PV Question 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 (pV) Case 1: (pV) = 4x9-2x3=30 atm-m3 Case 2: (pV) = 2x9-4x3= 6 atm-m3 31

  12. P(atm) P(atm) A B 4 4 A B 2 2 Case 1 Case 2 3 9 3 9 V(m3) V(m3) PV Question 3 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 34

  13. P Work done by system 1 P1 P3 2 Increase in internalenergy of system 3 Heat flow into system V V1 V2 First Law Questions Q= DU + W Some questions: • Which part of cycle has largest change in internal energy, DU ? 2  3 (since U = 3/2 pV) • Which part of cycle involves the least work W ? 3  1 (since W = pDV) • 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) 37

  14. Question 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 • W (800) = Qhot (1000) - Qcold (200) • Efficiency = W/Qhot = 800/1000 = 80% 45

  15. 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 47

  16. Work done by system Increase in internalenergy of system Heat flow into system Summary: • 1st Law of Thermodynamics: Energy Conservation Q= DU + W 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 DU=0  Q=W V 50

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