1. Physics 101: Lecture 27, Pg 1 Physics 101: Lecture 27 Thermodynamics l Today’s lecture will cover Textbook Chapter 15.1-15.6

2. Physics 101: Lecture 27, Pg 2 First Law of Thermodynamics Energy Conservation  U = Q - W 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) V P 1 2 3 V 1 V 2 P1P3P1P3 ideal gas

3. Physics 101: Lecture 27, Pg 3 Signs Example l 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

4. Physics 101: Lecture 27, Pg 4 Work Done by a Gas ACT M M yy The work done by a gas as it contracts is A) PositiveB) ZeroC) Negative

5. Physics 101: Lecture 27, Pg 5 Thermodynamic Systems and P-V Diagrams l ideal gas law: PV = nRT (nR = Nk B ) l for n fixed, P and V determine “state” of system l Examples: è which point has highest T? è which point has lowest U? V P A B C V 1 V 2 P1P3P1P3

6. Physics 101: Lecture 27, Pg 6 First Law of Thermodynamics Isobaric Example V P 1 2 V 1 V 2 P 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)

7. Physics 101: Lecture 27, Pg 7 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. V P 2 1 V P2P1P2P1

8. Physics 101: Lecture 27, Pg 8 WORK ACT If we go the opposite direction for the cycle (4,3,2,1) the net work done by the system will be A) PositiveB) Negative

9. Physics 101: Lecture 27, Pg 9 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 by the system bigger? A. Case 1 B. Case 2 C. Same A B 4 2 39 V(m 3 ) Case 1 A B 4 2 39 V(m 3 ) P(atm) Case 2 P(atm)

10. Physics 101: Lecture 27, Pg 10 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 bigger? A. Case 1 B. Case 2 C. Same A B 4 2 39 V(m 3 ) Case 1 A B 4 2 39 V(m 3 ) P(atm) Case 2 P(atm)

11. Physics 101: Lecture 27, Pg 11 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 bigger? A. Case 1 B. Case 2 C. Same A B 4 2 39 V(m 3 ) Case 1 A B 4 2 39 V(m 3 ) P(atm) Case 2 P(atm)

12. Physics 101: Lecture 27, Pg 12 Q =  U + W V P 1 2 3 V 1 V 2 P1P3P1P3 Which part of cycle has largest change in internal energy,  U ? l Which part of cycle involves the least work W ? l What is change in internal energy for full cycle? l What is net heat into system for full cycle (positive or negative)? Some questions: First Law Questions

13. Physics 101: Lecture 27, Pg 13 Special PV Cases l Constant Pressure (isobaric) l Constant Volume Constant Temp  U = 0 l Adiabatic Q=0 V P W = P  V (>0) 1 2 3 4  V > 0V P W = P  V = 0 1 2 3 4  V = 0

14. Physics 101: Lecture 27, Pg 14 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

15. Physics 101: Lecture 27, Pg 15Reversible? l Most “physics” processes are reversible, you could play movie backwards and still looks fine. (drop ball vs throw ball up) l Exceptions: è Non-conservative forces (friction) è Heat Flow: »Heat never flows spontaneously from cold to hot

16. Physics 101: Lecture 27, Pg 16Summary: è 1st Law of Thermodynamics: Energy Conservation Q =  U + W Heat flow into system Increase in internal energy of system Work done by system V P l point on p-V plot completely specifies state of system (pV = nRT) l work done is area under curve l U depends only on T (U = 3nRT/2 = 3pV/2) for a complete cycle  U=0  Q=W