Summarizing, Key Point: An irreversible process is either spontaneous (ΔS universe > 0) or does not occur (ΔS universe < 0)

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1 Summarizing, Key Point: An irreversible process is either spontaneous (ΔS universe > 0) or does not occur (ΔS universe < 0) Key Point: ΔS universe allows us to distinguish between reversible and irreversible processes

2 Heat Engine with Overall Spontaneous Process Heat withdrawn from hot reservoir. Entropy decreases by q h /T h. - q q c h S T c T h > 0 for spontaneous If T c <T h, entropy change can be positive even if q c <q h. The difference q h -q c can be withdrawn as work, w. Heat added to cold sink. Entropy increases by q c /T c.

3 Ideal Gas Equation of State

4 Entropy change in a general process with an ideal gas V 1, T 1 V 2, T 2 (n mol) Path may be reversible or irreversible Since entropy is a state function and ds=dq rev /T, we simply find a calculable reversible path that connects the two V,T VT points it and calculate lt S. Construct a two component reversible path: Isothermal (expan or compr) from V 1, T 1 to V 2, T 1 Isochoric heating or cooling from V 2, T 1 to V 2, T 2 P P 1 P 2 Path 1 Path 2 V 1 V V2

5 Entropy change in a general process with an ideal gas V 1 1, T 1 V 2 2, T 2 Make a two component reversible path: Isothermal (expan or compr) from V 1, T 1 to V 2, T 1 S1 nrln V 2 V 1 P P 1 P 2 Path 1 Path 2 V 1 V V2

6 Entropy change in a general process with an ideal gas V 1, T 1 V 2, T 2 Make M k a two component reversible path: P P 1 P 2 Isothermal (expan or compr) from V 1,T 1 to V 2,T 1 Isochoric heating or cooling from V 2,T 1 to V 2,T 2 V 1 V Path 1 V2 Path 2 du dq PdV dq du CV dt dqrev dt ds C V T T T2 C T V 2 S 2 dt CV ln T T T 1 1

7 Entropy change in a general process with an ideal gas V 1 1, T 1 V 2 2, T 2 Make a two component reversible path: Isothermal (expan or compr) from V 1, T 1 to V 2, T 1 Isochoric heating or cooling to V 2, T 2 P V T S S S nr ln C ln V V T 1 1 P 1 P 2 Path 1 Path 2 V 1 V V2

8 What if T 1 = T 2 V 1, T 1 V 2 2, T 1 Make a one component reversible path: Isothermal (expansion or compression) from V 1, T 1 to V 2, T 1 V 2 T 2 S S1S2 nrln CV ln V1 T1 P P 1 Path 1 ln V 2 1 V 1 S nr V P 2 V 1 V V2

9 Adiabatic expansion or compression ConcepTest #1 The system is an ideal gas. In an adiabatic expansion, what happens to the entropy of the system? a. Increases b. Decreases c. Is unchanged d. Either a or c, depending on reversibility of path

10 Adiabatic expansion or compression ConcepTest #2 The system is an ideal gas. In an adiabatic expansion, what happens to the entropy of the surroundings? a. Increases b. Decreases c. Is unchanged d. Either a or c, depending on reversibility of path

11 Irreversible Processes Spontaneous processes are irreversible and always accompanied by a net increase in entropy Statement of First & Second Laws: The energy of the universe is a constant; the entropy of the universe tends always toward a maximum

12 Irreversible Processes Spontaneous Not Spontaneous

13 A generalized reversible heat engine

14 A generalized reversible,cyclic heat engine We will see that such an engine can be considered to be a composite of many Carnot cycle heat engines.

15

16 Carnot Cycle (1825) Model for Steam Engine All steps are reversible! 1. Isothermal Expansion 2. Adiabatic Expansion 3. Isothermal Compression 4. Adiabatic Compression HOT COLD

17 Net work = work done by system work done on system

18 Efficiency of Carnot Engine Efficiency,, of Engine: Work Performed Heat Absorbed w q h We will see shortly that this is equivalent to T h T For now, we see that this expression passes the smell test! T h c

19 ConcepTest #3 This diagram represents a reversible cycle for an ideal gas. What is the thermodynamic efficiency of the engine? A. 80 % B. 75 % C. 60 % D. 30 % E. 20 %

20 ConcepTest #3 This diagram represents a reversible cycle for an ideal gas. What is the thermodynamic efficiency of the engine? A. 80 % B. 75 % C. 60 % D. 30 % E. 20 %

21 Steps of Carnot Cycle q=0 w done on system (+) q h absorbed (+) from hot reservoir T h w done by system ( ) q=0 w done by system ( ) q c discarded ( ) into cold q c ( ) reservoir T c w done on system (+)

22 Parameters for this Carnot Cycle State P(atm) V(L) T(K) A B C D

23 Steps in Carnot Cycle Step U q rev w rev S AB 0 RT h lnv B /V A J RT h lnv B /V A 1678 J RlnV B /V A 4.19 J/K BC C v (T c -T h ) 1247 J 0 C v (T c -T h ) 1247 J 0 CD 0 RT c lnv D /V C 1259 J RT c lnv D /V C J RlnV D /V C 4.19 J/K DA C v (T h -T c ) J 0 C v (T h -T c ) J 0 Net 0 R(T h -T c )lnv B /V A +419 J R(T h -T c )lnv B /V A 419 J 0

24 Net work = work done by system work done on system In example above, net work = = 1678 J J J 1247 J = 419 J What is the efficiency?

25 Efficiency of Carnot Engine The efficiency is the ratio of the usable work out to the total heat input to the system. Waste heat is not usable in this context! So in this case the efficiency is 419 J/1678 J = Compare with (T h T c )/T h = ( )/400 = 0.25 Multiple Carnot engines can be combined to describe any reversible heat engine!

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