Welcome back to Physics 211

1. Welcome back to Physics 211 Today’s agenda: work, heat, 1st law of thermodynamics applications 2nd law of thermodynamics

2. Review of work • Gas is example of non-rigid body • Many components whose positions may vary independently • Careful how we define work • Ex. spring could have Fnet=0 but Wnet not 0 • Same is true with gas – no change in K.E of CM gas but work is done …

3. How ? Wnet=SFi. Dxi if rigid Dxi=Ds independent of I  Wnet=Ds.SFi = Fnet.Ds in this case Wnet depends only on Fnet Otherwise need to sum up work done on each component separately Wnet=SFi. Dxi

4. Work in gases • W=-PDV if P constant • W= -PdV if P varies with V • As usual this can be interpreted as area under curve in PV-diagram P V1 V2 V

5. Two moles of a gas are enclosed in a cylinder with a movable piston that can be locked. During the process from point X to point Y in the PV-diagram below, is the work done on the gas by the piston… 1. positive, 2. negative, or 3. zero? 4. Can’t tell.

6. Two moles of a gas are enclosed in a cylinder with a movable piston. Is the magnitude of the work done on the gas in the process labeled “TOP” the magnitude of the work done on the gas in the process labeled “BOTTOM”? 1. greater than, 2. less than, 3. equal to, or 4. impossible to compare to

7. Work does what ? • For ideal gas work done W=-PDV goes into increasing K.E of individual molecules • If gas is real there will be intermolecular forces also. Then external work may also change associated potential energies • Sum of all these molecular energies is called internal energy U

8. Internal energy of an ideal gas For a monatomic gas (e.g., He, Ne, Ar) the internal energy U is the sum of the translational kinetic energies of the particles: U=N(1/2mv2)=3/2NkBT=3/2nRT

9. How else can we change U? • Suppose you do no work on gas but allow heat to flow in • Again, in thermal equilibrium mean energy of gas molecules goes up • Equivalent to increasing internal energy U

10. What is heat ? • A way of injecting energy into a system at a microscopic level • Heat flows between 2 objects at different T’s (mean K.E’s) • Collisions between more energetic molecules transfer energy to less energetic molecules – Q tells me how much energy is transferred this way

11. How to change the internal energy of a gas system? • By placing it in thermal contact with some other system that is at a different temperature, (i.e., transferring heat). • By using mechanical means to transfer energy to the system (i.e., performing work).

12. First Law of Thermodynamics • Always possible to define an internal energy function U • may contain potential energies of interaction, rotational and vibrational K’s • Change in U can be effected by work W or transfer of heat Q DU=Q+W W – Macroscopic way to change energy Q -- Microscopic way to change energy

13. In an isothermal compression of an ideal gas, the temperature of the gas does not change. Is the work done on the gas by the piston… 1. positive, 2. negative, or 3. zero? 4. Not sure.

14. In an isothermal compression of an ideal gas, the temperature of the gas does not change. Is the change in internal energy of the gas… 1. positive, 2. negative, or 3. zero? 4. Not sure.

15. In an isothermal compression of an ideal gas, the temperature of the gas does not change. Is the heat transferred to the gas… 1. positive, 2. negative, or 3. zero? 4. Not sure.

16. Implications of first law: Heat transfer can be non-zero even if temperature does not change Temperature can change even if there is no heat transfer

17. Heat engines DU=0 over cycle so, Q=-Won gas=Wby gas Gas process that runs in a cycle can absorb heat Q from a heat source and perform work W (e.g. on a mechanical device).

18. Stirling engine demo Runs on hot water and ice

19. Stirling engine demo Does not run on hot water alone

20. First law for cyclic process implies • Qnet=Wnet • i.e Qin-Qout=Wnet • Can Qout be zero ? i.e can we convert heat to work with 100% efficiency ? • 2nd law addresses this question

21. The Second Law of Thermodynamics It is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the absorption of energy from a reservoir and the performance of an equal amount of work. From: Serway/Beichner

22. Implications of the second law Some of the energy absorbed from the “heat source” is not used to do work but instead is given off (also as heat) to the cooler surroundings. This “loss of energy” is not due to friction or other imperfections of the mechanical device used (such as bad insulation).