Lecture 15

• Clarifications about lecture 14

• heat engines and refrigerators using ideal gas as working substance

Clarification about lecture 14 (I)• For both gases being

monoatomic or both being diatomic, use

• ...but not for one monoatomic and the other diatomic (as in example)

E1f

N1= E2f

N2= Etot

N1+N2!

E1f = N1N1+N2

Etot; E2f = N2N1+N2

Etot

Clarification about lecture 14 (II)• collision is reversible:

transfer energy from system 2 (faster atom) to system 1 (slower atom)

• ...but less likely to find faster atoms in system 2 than in system 1

• on the average, collisions transfer energy from system 1 to system 2

T1 > T2

Clarification about lecture 14 (III)• entropy is measure of disorder...and never decreases...

• Initial state more ordered (less entropy): less energetic atoms on one side, more energetic on the other....than equilibrium: same energy for atoms on both sides

• ...but initial state less ordered than if heat from cold to hot (more energetic side becomes even more energetic)

Refrigerators• closed cycle uses external work to

remove heat from cold reservoir and exhaust heat to hot reservoir (2nd law does not allow spontaneous): e.g. air-conditioner or kitchen...make air that is cooler than environment even colder

• exhaust more heat than removed from inside (cool room by leaving refrigerator door open?)

• coefficient of performance:

• perfect refrigerator ( ) forbidden by 2nd law (informal statement # 3): real refrigerator uses work ( )

!Eth = 0 (cyclical) : QH = QC + Win

Win = 0; K =!K <!

Example

• 1.0 L of 20 degree Celsius water is placed in a refrigerator. The refrigerator’s motor must supply an extra 8.0 W power to chill the water to 5 degree Celsius in 1.0 hr. What is the refrigerator’s coefficient of performance?

No perfect Heat Engine ( )• connect perfect engine to refrigerator: no net work for 2

combined, but heat transferred from cold to hot (not allowed by 2nd law)

• informal statement # 4: no perfect heat engine, must waste heat...

(! = WoutQH

>! 1 from energy conservation)

! = WoutQH

< 1

! = 1

Example of Proof by Contradiction

• want to prove statement “A” (e.g., there exists a perfect engine) is not true

• assume A is true, find a violation of basic law (e.g. ,2nd law of thermodynamics) assumption is incorrect, A is not true

• Using only energy conservation and heat not transferred from cold to hot, deduce heat engines and refrigerators exist; must use closed-cycle processes; no perfect...

• upper limit on ?!, K

Unanswered questions

Summary

Ideal gas Heat Engines

Ideal gas summary I

• closed cycle trajectory: clockwise for Wout > 0Wout = Wexpand ! |Wcompress| = area inside closed curve

Ideal gas summary II• depends only on T

• identify each process, draw pV diagram

• use ideal gas law to know n, p, V, T at one point

• use ideal gas law and equations for specific processes for p, V, T at beginning/end of each process

• calculate for each process

• by adding `s: confirm by area within curve

• add positive values of Q to find

• check:

Strategy for heat engine problems

(!Eth)net = 0, ! < 1, signs of Ws and Q...

QH

WsWout

Q, Ws and !Eth

Eth