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Please fill in on-line course evaluation. Final Exam May 3, 10:30 am-12:30 pm ELLT 116 Lecture 26 Purdue University, Physics 220

Purdue University, Physics 220 Lecture 26 Thermodynamics I Lecture 26 Purdue University, Physics 220

Purdue University, Physics 220 Lecture 26 Purdue University, Physics 220

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) minus the work done by the system (W) DU = Q - W Work done by system Increase in internal energy of system Heat flow into system Equivalent ways of writing 1st Law: Q = DU + W Lecture 26 Purdue University, Physics 220

Purdue University, Physics 220 Lecture 26 Purdue University, Physics 220

Purdue University, Physics 220 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 1st law of thermodynamics to the soup. What is the sign of A) Q B) W C) DU Positive, heat flows into soup Negative, stirring does work on soup Positive, soup gets warmer Lecture 26 Purdue University, Physics 220

Purdue University, Physics 220 Work Done by a System M M y W = F d cosq= P A d = P A Dy = P DV 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 Lecture 26 Purdue University, Physics 220

Purdue University, Physics 220 PV Diagram 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? 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 V1 V2 P1 P3 Lecture 26 Purdue University, Physics 220

Purdue University, Physics 220 Example (Isobaric) 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. V P 1 2 V1 V2 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) D pV = (3/2) (3000-2000)J =1500 J (has to be the same) 4. W = p DV = 1000 J 5. Q = DU + W = 1500 + 1000 = 2500 J Lecture 26 Purdue University, Physics 220

Purdue University, Physics 220 Example (Isochoric) 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 2 P2 P1 1 1. pV1 = nRT1  p1 = nRT1/V = 1000 Pa 2. pV2 = nRT2  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 CV=CP-R V V Lecture 26 Purdue University, Physics 220

Purdue University, Physics 220 Question 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 A B 4 2 3 9 V(m3) Case 1 P(atm) Case 2 Net Work = area under P-V curve Area the same in both cases! Lecture 26 Purdue University, Physics 220

Purdue University, Physics 220 Question 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 A B 4 2 3 9 V(m3) Case 1 P(atm) Case 2 U = 3/2 (PV) Case 1: (PV) = 4x9-2x3=30 atm*m3 Case 2: (PV) = 2x9-4x3= 6 atm*m3 Lecture 26 Purdue University, Physics 220

Purdue University, Physics 220 iClicker 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 A B 4 2 3 9 V(m3) Case 1 P(atm) Case 2 Q = U + W W is same for both U is larger for Case 1 Therefore, Q is larger for Case 1 Lecture 26 Purdue University, Physics 220

Purdue University, Physics 220 Example (Isothermal) W=nRTln(Vf/Vi) for isothermal expansion Lecture 26 Purdue University, Physics 220

Purdue University, Physics 220 Special PV Cases V P W = PDV (>0) 1 2 3 4 DV > 0 Constant Pressure isobaric Constant Volume isochoric Constant Temp DU = 0 isothermal Adiabatic Q=0 (no heat is transferred) V P W = PDV = 0 1 2 3 4 DV = 0 Lecture 26 Purdue University, Physics 220

Purdue University, Physics 220 First Law Questions Q = DU + W Work done by system V P 1 2 3 V1 V2 P1 P3 Increase in internal energy of system Heat flow into system Some questions: 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) Lecture 26 Purdue University, Physics 220

Purdue University, Physics 220 1st Law Summary 1st Law of Thermodynamics: Energy Conservation Q = DU + W Heat flow into system Increase in internal energy of system Work done by system 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 Lecture 26 Purdue University, Physics 220

Purdue University, Physics 220 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 a colder body to a hotter body. Lecture 26 Purdue University, Physics 220

Example: 19 19

Example: 20 20

Example: 21 21

Example: 22 22

Purdue University, Physics 220 iClicker Path A and path B both take system from initial (i) to final state (f). Which of the following is true: A) WA=WB and QA=QB B) WA=WB and QAQB C) WAWB and QA=QB D) WAWB and QAQB WA=Pi(Vf-Vi) WB=Pf(Vf-Vi) UA=QA-WA UB=QB-WB Lecture 26 Purdue University, Physics 220