1. THERMODYNAMICS THE NEXT STEP

2. THERMAL PROPERTIES OF MATTER STATE VARIABLES – DESCRIBE THE SUBSTANCE –PRESSURE –TEMPERATURE –VOLUME –QUANITY OF SUBSTANCE

3. EQUATION OF STATE THE EQUATION THAT EXPRESSES THE RELATIONSHIP BETWEEN THE STATE VARIABLES V = Vo[1 + β(T – To) – k(P – Po)] β IS THE COEFFICIANT OF VOLUME EXPANSION k IS THE ISOTHERMAL COMPRESSIBILITY OF THE MATERIAL

4. IDEAL GAS LAW VOLUME IS PROPORTIONAL TO THE # OF MOLES OF THE GAS VOLUME IS INVERSLY PROPORTIONAL TO THE PRESSURE. PRESSURE IS PROPORTIONAL TO THE ABSOLUTE TEMPERATURE PV = nRT N - # MOLES R – IDEAL GAS CONSTANT = 8.314 J/mol K T - TEMP. IN KELVIN

5. REAL GASES DON’T FOLLOW THE IDEAL GAS LAW EXACTLY. AT LOW PRESSURE AND AT TEMP. THAT ARE CLOSE TO LIQUIFICATION IT APPROXIMATES THE IDEAL GAS MODEL.

6. WHAT IS AN IDEAL GAS? A LARGE NUMBER OF GAS MOLECUES IN A NEGLIGABLE VOLUME RANDOM MOTION ELASTIC COLLISIONS WITH THE WALLS OF THE CONTAINER AND WITH OTHER GAS MOLECULES TEMPERATURE OF AN IDEAL GAS IS PROPORTIONAL TO THE KINETIC ENERGY OF THE GAS MOLECULES.

7. PV DIAGRAMS EACH CURVE IS A TEMP. CONSTANT CALLED AN ISOTHERM THE AREA UNDER THE ISOTHERM EQUALS THE WORK DONE BY THE VOLUME CHANGE. ISOTHERM V2 P1P1 P2 V1 WORK

8. WORK DONE BY SYSTEM THE AREA ABOVE THE ISOTHERM EQUALS THE WORK DONE BY THE SYSTEM DUE TO PRESSURE CHANGE. ISOTHERM V2 P1P1 P2 V1 WORK

9. PATH THE SERIES OF STATES THAT A SYSTEM UNDERGOES DURING A CHANGE OF STATE POINTS 1 & 2 ARE THE INITIAL STATE POINTS 3 & 4 THE FINAL WORK DEPENDS UPON INITIAL STATE, FINAL STATE, AND PATH TAKEN. V2 P1P1 P2 V1 1 2 3 4

10. THERMODYNAMICS THE STUDY OF PROPERITES OF THERMAL ENERGY

11. LAWS OF THERMODYNAMICS ZEROETH LAW – DEALS W/TEMP, T. FIRST LAW – DEALS W/INTERNAL ENERGY, U. SECOND LAW – DEALS W/ENTROPY, S.

12. SYSTEMS ANY OBJECT OR GROUP OF OBJECTS WE ARE CONSIDERING CLOSED SYSTEM – MASS IS CONSTANT & NO ENERGY ENTERS OR LEAVES THE SYSTEM OPEN SYSTEM – MASS CHANGES & ENERGY EXCHANGE CAN OCCUR.

13. ENVIRONMENT ANYTHING THAT IS NOT PART OF THE SYSTEM THAT IS BEING CONSIDERED

14. THERMAL EQUILIBRIUM WHEN TWO OR MORE OBJECTS OF VARYING TEMPERATURES ARE ORIENTED IN SUCH A MANNER AS TO FACILITATE HEAT ENERGY TRANSFER FROM ONE TO THE OTHER UNTIL ALL OBJECTS ARE AT THE SAME TEMPERATURE.

15. ZEROTH LAW OF THERMODYNAMICS IF TWO SYSTEMS ARE IN THEMAL EQUILIBRIUM WITH A THIRD SYSTEM, THEY ARE IN THERMAL EQUILIBRIUM WITH EACH OTHER.

16. INTERNAL OR THERMAL ENERGY SYMBOL – U UNIT – J THE SUM OF ALL THE ENERGY CAN NOT BE MEASURED ONLY THE CHANGE IN THE ENERGY CAN BE MEASURED

17. HOW ARE TEMPERATURE AND THERMAL ENERGY DIFFERENT? REMEMBER – TEMPERATURE AND KINETIC THEORY ARE CONNECTED. TEMPERATURE IS THE MEASURE OF THE AVERAGE KINETIC ENERGY OF INDIVIDUAL MOLECULES THERMAL ENERGY IS THE MEASURE OF ALL OF THE ENERGY OF ALL OF THE MOLECULES IN AN OBJECT.

18. INTERNAL ENERGY OF A GAS DEPENDS ON TEMPERTURE AND NUMBER OF MOLES U = 3/2nRT R=8.315J/mol K

19. FIRST LAW OF THERMODYNAMICS BASICALLY A RESTATEMENT OF THE LAW OF CONSERVATION OF ENERGY THE TOTAL INCREASE IN THE INTERNAL ENERGY OF A SYSTEM IS EQUAL TO THE SUM OF THE WORK DONE ON THE SYSTEM OR BY THE SYSTEM AND THE HEAT ADDED TO OR REMOVED FROM THE SYSTEM. CHANGES IN THE INTERNAL ENERGY OF A SYSTEM ARE CAUSED BY HEAT AND WORK.

20. 1 ST LAW EQUATION ∆U = Q + W Q = HEAT ADDED W = NET WORK DONE ON THE SYSTEM HEAT ADDED + HEAT LOST – W DONE ON THE SYSTEM + W DONE BY THE SYSTEM -

21. ISOTHERMAL PROCESS TEMP IS CONSTANT NO TEMP CHANGE = NO INTERNAL ENERGY CHANGE INTERNAL ENERGY CHANGES OCCUR ONLY WHEN THERE IS A TEMP. CHANGE. AT CONSTANT TEMPERATURE P & V DECREASE ALONG PATH STATE 1 TO STATE 2 W = nRT Vf/Vi

22. HOW DO YOU KNOW IF HEAT IS ADDED ? HEAT REMOVED –VOLUME DECREASES AND PRESSURE INCREASES HEAT ADDED –VOLUME INCREASES AND PRESSURE DECREASES

23. ISOBARIC PROCESS PRESSURE IS CONSTANT W = -P∆V ∆U = Q + W BECAUSE P IS CONSTANT THE WORK IS EQUAL TO THE AREA UNDER THE STATE 1 TO STATE 2 PATH. 1 2

24. HOW DO YOU KNOW IF HEAT IS ADDED? HEAT ADDED – –GAS EXPANDS –INCREASE IN INTERNAL ENERGY CAUSES INCREASE IN TEMPERATURE HEAT REMOVED – –GAS IS COMPRESSED

25. ISOCHORIC (ISOVOLUMETRIC) PROCESS VOLUME IS CONSTANT SINCE V IS CONSTANT NO WORK IS DONE HEAT MOVES FROM STATE 1 TO STATE 2 1 2 P V

26. AIDABATIC PROCESS(INSULATED SYSTEM) NO HEAT IS ALLOWED TO FLOW INTO OR OUT OF THE SYSTEM. OCCURS WHEN SYSTEM IS WELL- INSULATED OR THE PROCESS HAPPENS QUICKLY THE INTERNAL ENERGY AND THE TEMP DECREASE IF THE GAS EXPANDS.

27. WORK? VOLUME INCREASES SO WORK IS DONE ON THE ENVIRONMENT. AREA UNDER STATE 1 TO STATE 2 WORK IS NEGATIVE DECREASE IN INTERNAL ENERGY 1 2

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29. HOW DO YOU KNOW IF HEAT IS ADDED? HEAT ADDED – PRESSURE INCREASES TEMPERTURE INCREASES WHEN THE INTERNAL ENERGY INCREASES

30. SECOND LAW OF THERMODYNAMICS IN NATURAL PROCESSES, HEAT CANNOT FLOW FROM A COLD TO A HOT SUBSTANCE NATURAL PROCESSES INCREASE THE ENTROPY(DISORDER OR CHAOS) OF THE UNIVERSE; WITH TIME, DISORDER CANNOT BECOME ORDER A HEAT ENGINE CANNOT CONVERT ALL ITS HEAT TO MECHANICAL ENERGY. NO MACHINE IS EVER 100% EFFICIENT

31. WHY CAN WE NOT REACH ABSOLUTE ZERO? HEAT MUST BE REMOVED TO LOWER TEMPERATURE HEAT TRAVELS FROM WARM TO COLD THERE IS NOTHING COLDER THAN ABSOLUTE ZERO AND SO THERE IS NO OBJECT(HEAT SINK) FOR THE EXCESS HEAT TO ENTER IN ORDER FOR ANOTHER OBJECT TO BECOME ABSOLUTE ZERO

32. ENTROPY DISORDER, CHAOS SYMBOL – S EQUATION - ∆S=Q/T T IS TEMP. IN KELVIN THE CHANGE IN ENTROPY IS WHAT IS IMPORTANT

33. ENTROPY REVERSIBLE PROCESSES ENTROPY IS CONSTANT…IF ENTROPY INCREASES IN THE SYSTEM IT DECREASES IN THE ENVIRONMENT IRREVERSIBLE PROCESSES (REAL LIFE) ENTROPY EITHER IS UNCHANGED OR INCREASES.

34. HEAT ENGINES AUTOMOBILE ENGINES – THERMAL ENRGY FROM A HIGH HEAT SOURCE IS CONVERTED INTO MECHANICAL ENERGY AND EXHAUST IS EXPELLED REFRIG. – THERMAL ENERGY IS REMOVED FROM A COLD BODY AND TRANSFERRED TO A HOT BODY. THIS REQUIRES WORK BECAUSE ENERGY IS FORCED AGAINST THE GRADIENT.

35. PURPOSE OF HEAT ENGINE TO TRANSFORM AS MUCH THERMAL HEAT INTO WORK AS POSSIBLE WORK NET = ENERGY ADDED AS HEAT – ENERGY REMOVED AS HEAT

36. PURPOSE OF A REFRIGERATOR TO TRANSFER HEAT FROM THE LOW TEMP TO THE HIGH TEMP RESERVOIR DOING AS LITTLE WORK ON THE SYSTEM AS POSSIBLE. NO PERFECT REFRIGERATOR CONVERTS MECHANICAL ENERGY INTO THERMAL ENERGY

37. EFFICIENCY OF A HEAT ENGINE THE RATIO OF THE WORK THE ENGINE DOES TO THE HEAT INPUT A THE HIGH TEMP. E = W/Q h = Q h – Q c /Q h = 1 – Q c /Q h E = NET WORK DONE BY ENGINE______ ENERGY ADDED TO ENGINE AS HEAT OR E = ENERGY ADDED AS HEAT – ENERGY REMOVED AS HEAT/ENERGY ADDED AS HEAT OR E = 1 – ENERGY REMOVED AS HEAT ENERGY ADDED AS HEAT

38. CARNOT (IDEAL) EFFICIENCY THEORETICAL LIMIT TO EFFICIENCY DEFINED IN TERMS OF OPERATION TEMPERATURES e ideal = (T H -T L )/T H

39. HOMEWORK AP HEAT SAMPLE PROBLEMS 1-4 AP THERMODYNAMICS PROBLEMS 1-4 DUE NEXT CLASS EXAM THURSDAY