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Work Done by a System

y. W. W. Work Done by a System. 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. Homework Problem: Thermo I. 1. P. 2. 4. 3. P. P. W = P D V (>0).

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Work Done by a System

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  1. y W W Work Done by a System • W = p V • 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

  2. Homework Problem: Thermo I 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 1 P 2 If we gothe other way thenWtot < 0 4 4 3 3 4 3 V DV = 0 V DV < 0 V Wtot = ??

  3. Now try this: P 1 P1 P3 2 Area = (V2-V1)x(P1-P3)/2 3 V V1 V2

  4. P Work done by system 1 P1 P3 2 3 V V1 V2 First Law of ThermodynamicsEnergy Conservation DU = Q- W Increase in internalenergy of system Heat flow into system Equivalent ways of writing 1st Law: Q = DU + W Q = DU + p DV

  5. P 1 P 2 V V1 V2 First Law of ThermodynamicsExample 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. 1. pV1 = nRT1 T1 = pV1/nR = 120K 2. pV2 = nRT2 T2 = pV2/nR = 180K 3. DU = (3/2) nR DT = 1500 J DU = (3/2) p DV = 1500 J (has to be the same) 4. W = p DV = 1000 J 5. Q = DU + W = 1500 + 1000 = 2500 J

  6. First Law of ThermodynamicsExample 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. pV1 = nRT1 p1 = nRT1/V = 1000 Pa 2. pV2 = nRT1 p2 = nRT2/V = 1500 Pa 3. DU = (3/2) nR DT = 1500 J 4. W = p DV = 0 J 5. Q = DU + W = 0 + 1500 = 1500 J requires less heat to raise T at const. volume than at const. pressure 1 V V

  7. P(atm) P(atm) A B 4 4 A B 2 2 Case 1 Case 2 correct 3 9 3 9 V(m3) V(m3) Chapter 15,Problem 1 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 Net Work = area under P-V curve Area the same in both cases!

  8. P(atm) P(atm) A B 4 4 A B 2 2 Case 1 Case 2 correct 3 9 3 9 V(m3) V(m3) Chapter 15,Problem 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? 1. Case 1 2. Case 2 3. Same U = 3/2 (pV) Case 1: (pV) = 4x9-2x3=30 atm-m3 Case 2: (pV) = 2x9-4x3= 6 atm-m3

  9. P(atm) P(atm) A B 4 4 A B 2 2 Case 1 Case 2 correct 3 9 3 9 V(m3) V(m3) Chapter 15,Problem 2 (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 Q = U + W W is same for both U is larger for Case 1 Therefore, Q is larger for Case 1

  10. P Work done by system 1 P1 P3 2 Q= DU + W Increase in internalenergy of system 3 Heat flow into system V V1 V2 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? • U = 0  Q = W = area of triangle (>0)

  11. Work done by system Increase in internalenergy of system Heat flow into system Recap: • 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

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