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The Laws of Thermodynamics

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Presentation on theme: "The Laws of Thermodynamics"— Presentation transcript:

1 The Laws of Thermodynamics
Physics 2053 Lecture Notes The Laws of Thermodynamics (01 of 38)

2 1) The First Law of Thermodynamics
Topics 1) The First Law of Thermodynamics 2) Work Done on a Gas 3) Pressure - Volume Graph 4) Thermodynamic Processes 5) The Second Law of Thermodynamics 6) Heat Engines 7) Carnot cycle 8) Entropy The Laws of Thermodynamics (02 of 38)

3 First Law of Thermodynamics
System U Q W Environment DU = Q + W The Laws of Thermodynamics (03 of 38)

4 Work Done on a Gas F W = F Dx DV = A Dx W= PA Dx Area Dx W = P DV DV
The Laws of Thermodynamics (04 of 38)

5 Pressure - Volume Graph
T4 T3 Isotherms (lines of constant temperature) T2 P T1 Area under curve represents work Pressure Internal energy is proportional to temperature Volume V The Laws of Thermodynamics (05 of 38)

6 Thermodynamic Processes
A. Isobaric B. Isovolumetric C. Isothermal D. Adiabatic The Laws of Thermodynamics (06 of 38)

7 Thermodynamic Processes
Isobaric Expansion (Constant Pressure) P Po a) W = PDV T4 b) DU increases T3 T2 T1 c) Q = DU + W DV V DU = Q + W The Laws of Thermodynamics (07 of 38)

8 Thermodynamic Processes
Isobaric Compression (Constant Pressure) a) W = PDV P Po b) DU decreases T4 T3 c) Q = -(DU + W) T2 T1 DV V DU = Q + W The Laws of Thermodynamics (08 of 38)

9 Thermodynamic Processes
Isovolumetric (Constant Volume) Decrease in Pressure P a) W = 0 Po Pf b) DU decreases T3 c) Q = DU T2 T1 V DU = Q + W The Laws of Thermodynamics (09 of 38)

10 Thermodynamic Processes
Isovolumetric (Constant Volume) Increase in Pressure a) W = 0 P b) DU increases Po Pf c) Q = DU T3 T2 T1 V DU = Q + W The Laws of Thermodynamics (10 of 38)

11 Thermodynamic Processes
Isothermal (Constant Temperature) Expansion Po Pf Vo Vf P b) DU = 0 T3 c) Q = -W T2 T1 V DU = Q + W The Laws of Thermodynamics (11 of 38)

12 Thermodynamic Processes
Isothermal (Constant Temperature) Compression Po Pf Vo Vf P b) DU = 0 T3 c) Q = W T2 T1 V DU = Q + W The Laws of Thermodynamics (12 of 38)

13 Thermodynamic Processes
Adiabatic (No Heat Exchange) Expansion P Po Pf Vo Vf a) W = DU b) DU decreases T2 c) Q = 0 T1 V DU = Q + W The Laws of Thermodynamics (13 of 38)

14 Thermodynamic Processes
Adiabatic (No Heat Exchange) Compression P Po Pf Vo Vf a) W = DU b) DU increases T2 c) Q = 0 T1 V DU = Q + W The Laws of Thermodynamics (14 of 38)

15 Thermodynamic Processes
An ideal gas expands to 10 times its original volume, maintaining a constant 440 K temperature. If the gas does 3.3 kJ of work on its surroundings, how much heat does it absorb? T P V Po Pf Vo Vf In an isothermal process DU = 0 The Laws of Thermodynamics (15 of 38)

16 Thermodynamic Processes (n/a)
Po 2Po Vo 2Vo 3Vo II An ideal gas initially has pressure Po, at volume Vo and absolute temperature To. It then undergoes the following series of processes: III I IV I. Heated, at constant volume to pressure 2Po II. Heated, at constant pressure to pressure 3Vo III. Cooled, at constant volume to pressure Po IV. Cooled, at constant pressure to volume Vo The Laws of Thermodynamics (16 of 38)

17 Thermodynamic Processes (n/a)
Po 2Po Vo 2Vo 3Vo 2To II 6To III I To IV 3To Find the temperature at each end point in terms of To The Laws of Thermodynamics (17 of 38)

18 Thermodynamic Processes (n/a)
II 2Po III I Find the net work done by the gas in terms of Po and Vo Po IV Vo 2Vo 3Vo Net work equals net area under curve The Laws of Thermodynamics (18 of 38)

19 Thermodynamic Processes (n/a)
Po 2Po Vo 2Vo 3Vo II III I Find the net change in internal energy in terms of Po and Vo IV The Laws of Thermodynamics (19 of 38)

20 The Second Law of Thermodynamics
The second law of thermodynamics is a statement about which processes occur and which do not. There are many ways to state the second law; here is one: Heat will flow spontaneously from a hot object to a cold object. It will not flow spontaneously from a cold object to a hot object. The Laws of Thermodynamics (20 of 38)

21 The Second Law of Thermodynamics
Direction of Time The Laws of Thermodynamics (21 of 38)

22 Heat Engines In a heat engine; mechanical energy can be obtained from thermal energy when heat flows from a higher temperature to a lower temperature. Work done by engine: High Temp. Th Qh Thermal efficiency: Engine W Qc Low Temp. Tc The Laws of Thermodynamics (22 of 38)

23 Work done by engine each cycle
Heat Engines Th= 550 K Tc Engine Qh Qc W = 630 J = 920 J For the engine Work done by engine each cycle The efficiency of the engine The Laws of Thermodynamics (23 of 38)

24 P V The Carnot Cycle Isothermal Expansion Th Adiabatic Expansion
Adiabatic Compression Tc V Isothermal Compression The Laws of Thermodynamics (24 of 38)

25 P Work V The Carnot Cycle High Temp. Th Qh Th Engine W Qc Low Temp. Tc
Carnot efficiency: The Laws of Thermodynamics (25 of 38)

26 Work done by engine each cycle
The Carnot Cycle Th= 550 K Tc Engine Qh Qc W = 470 J = 890 J For the engine Work done by engine each cycle The efficiency of the engine The Laws of Thermodynamics (26 of 38)

27 Temperature of the cool reservoir based on a ratio
The Carnot Cycle Th= 550 K Qh = 890 J Temperature of the cool reservoir based on a ratio Engine W = 420 J Qc = 470 J Tc The engine undergoes 22 cycles per second, its mechanical power output (N/A) The Laws of Thermodynamics (27 of 38)

28 A carnot engine absorbs 900 J of heat each cycle and provides 350 J
The Carnot Cycle Th A carnot engine absorbs 900 J of heat each cycle and provides 350 J of work Qh = 900 J Engine W = 350 J The efficiency of the engine Qc Tc The heat ejected each cycle The Laws of Thermodynamics (28 of 38)

29 A carnot engine absorbs 900 J of heat each cycle and provides 350 J
The Carnot Cycle Th A carnot engine absorbs 900 J of heat each cycle and provides 350 J of work Qh = 900 J Engine W = 350 J The engine ejects heat at 10 oC The temperature of the hot reservoir Qc =550 J Tc = 283 K The Laws of Thermodynamics (29 of 38)

30 The Carnot Cycle Th= 650 K Tc= 300 K Engine Qh Qc W = ? = 400 J A carnot engine operates between a hot reservoir at 650 K and a cold reservoir at 300 K. If it absorbs 400 J of heat at the hot reservoir, how much work does it deliver? = The Laws of Thermodynamics (30 of 38)

31 Entropy The Laws of Thermodynamics (31 of 38)

32 Another statement of the second law of thermodynamics:
Entropy Definition of the change in entropy S when an amount of heat Q is added:DD Another statement of the second law of thermodynamics: The total entropy of an isolated system never decreases. When an irreversible process occurs in a closed system, the entropy S of the system always increases: it never decreases. The Laws of Thermodynamics (32 of 38)

33 Natural processes tend to move toward a state of greater disorder.
Entropy Entropy is a measure of the disorder of a system. This gives us yet another statement of the second law: Natural processes tend to move toward a state of greater disorder. Example: If you put milk and sugar in your coffee and stir it, you wind up with coffee that is uniformly milky and sweet. No amount of stirring will get the milk and sugar to come back out of solution. The Laws of Thermodynamics (33 of 38)

34 Thermal equilibrium is a similar process –
Entropy Another example: when a tornado hits a building, there is major damage. You never see a tornado approach a pile of rubble and leave a building behind when it passes. Thermal equilibrium is a similar process – the uniform final state has more disorder than the separate temperatures in the initial state. The Laws of Thermodynamics (34 of 38)

35 Another consequence of the second law:
Entropy Another consequence of the second law: In any natural process, some energy becomes unavailable to do useful work. If we look at the universe as a whole, it seems inevitable that, as more and more energy is converted to unavailable forms, the ability to do work anywhere will gradually vanish. This is called the heat death of the universe. The Laws of Thermodynamics (35 of 38)

36 SUMMARY First Law of Thermo Thermo Processes
Isothermal = temp is constant Adiabatic = no heat exchanged Isobaric = pressure is constant Isochoric (isovolumetric) = volume is constant

37 SUMMARY cont. Heat engines change heat into useful work (requires a temperature difference) Efficiency of a heat engine: Carnot effeciecy:

38 Second law of thermodynamics:
Summary Second law of thermodynamics: heat flows spontaneously from a hot object to a cold one, but not the reverse Thermal energy cannot be changed entirely to work natural processes tend to increase entropy. Change in entropy: Entropy is a measure of disorder. As time goes on, less and less energy is available to do useful work. The Laws of Thermodynamics (37 of 38)

39 END


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