The First Law of Thermodynamics

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1 Thermodynamics The First Law of Thermodynamics Thermodynamic Processes (isobaric, isochoric, isothermal, adiabatic) Reversible and Irreversible Processes Heat Engines Refrigerators and Heat Pumps The Carnot Cycle Entropy (The Second Law of Thermodynamics) The Third Law of Thermodynamics 1

2 The Zeroth Law of Thermodynamics If A is in thermal equilibrium with C and B in thermal equilibrium i with C then A and B have to be in thermal equilibrium. No heat flows! 2

3 Internal Energy From In thermodynamics, the internal energy of a thermodynamic system, or a body with well-defined boundaries, denoted by U, or sometimes E, is the total of the kinetic energy due to the motion of molecules (translational, rotational, vibrational) and the potential energy associated with the vibrational and electric energy of atoms within molecules or crystals. It includes the energy in all the chemical bonds, and the energy of the free, conduction electrons in metals. 3

4 The First Law of Thermodynamics The first law of thermodynamics says the change in internal energy of a system is equal to the heat flow into the system plus the work done on the system. ΔU Q W 4

5 First Law of Thermodynamics The change in a systems internal energy is related to the heat and the work. ΔU U f U i Q W Where: U f internal energy of end U i internal energy of start Q net thermal energy flowing into system during process Positive when system gains heat Negative when system loses heat W net work done by the system Positive when work done by the system Negative when work done on the system 5

6 Thermodynamic Processes A state variable describes the state of a system at time t, but it does not reveal how the system was put into that state. Examples of state variables: pressure, temperature, volume, number of moles, and internal energy. Thermal processes can change the state of a system. We assume that thermal processes have no friction or other dissipative forces. In other words: All processes are reversible (Reversible means that it is possible to return system and surroundings to the initial states) REALITY: irreversible 6

7 Humpty Dumpty sat on a wall. Humpty Dumpty had a great fall All the king s horses and all the king s men Couldn t put Humpty Dumpty together again * Martin Schullinger-Krause (PH202 Winter 2008) 7

8 A PV diagram can be used to represent the state changes of a system, provided the system is always near equilibrium. The area under a PV curve gives the magnitude of the work done on a system. W<0 for compression and W>0 for expansion. 8

9 To go from the state (V i, P i ) by the path (a) to the state (V f, i i f P f ) requires a different amount of work then by path (b). To return to the initial point (1) requires the work to be nonzero. The work done on a system depends on the path taken in the PV diagram. The work done on a system during a closed cycle can be nonzero. 9

10 An isothermal process implies that both P and V of the gas change (PV T). 10

11 Specific Heats under constant pressure and constant volume Specific heat Q m c ΔT For a gas we use Molar specific heat Q n C ΔT Constant Volume: C V Constant Pressure : C P 11

12 Thermodynamic Processes for an Ideal Gas No work is done on a system when its volume remains constant (isochoric process). For an ideal gas (provided the number of moles remains constant), Q W Q the change in internal ΔU n V ΔT energy is ΔU 0 Q C 12

13 For a constant pressure e (isobaric) process, the change in internal energy is Δ U Q W where W PΔV nrδt and Q nc P Δ T. C P is the molar specific heat at constant pressure. For an ideal gas C P C V +R. 13

14 For a constant temperature (isothermal) process, ΔU 0 and the work done on an ideal gas is V V nrt V V NkT W i i. ln ln f f W Q U i i Δ 0 14

15 We have found for a monoatomic gas ΔU 3/2nRΔT Constant volume: ΔU Q 3/2 n R ΔT n C V ΔT C V 3/2 R Constant pressure: Q ΔU + W n C P ΔT 3/2 n R ΔT + n R ΔT C P 5/2 R C V C P R (always valid for any ideal gas) 15

16 Adiabatic ( not passable ) processes (no heat is gained or lost by the system Q0 0, i.e. system perfectly isolated ) Q0 and so ΔU -W P V constant (isothermal) P V γ constant (adiabatic) γ C P /C V For a monoatomic gas therefore γ 5/3 16

17 Example: An ideal gas is in contact with a heat reservoir so that it remains at constant temperature of K. The gas is compressed from a volume of 24.0 L to a volume of 14.0 L. During the process, the mechanical device pushing the piston to compress the gas is found to expend 5.00 kj of energy. How many moles of the ideal gas are in the system? How much heat flows between the heat reservoir and the gas, and in what direction does the heat flow occur? W nrt V f W 5000J ln( ) n V V i f 8.31 ln(14/ 24) RT ln( ) V i 3.7 mol This is an isothermal process, so ΔU Q - W 0 (for an ideal gas) and W Q kj. Heat flows from the gas to the reservoir. 20

18 An ice cube placed on a countertop in a warm room will melt. The reverse process cannot occur: an ice cube will not form out of the puddle of water on the countertop in a warm room. 21

19 Any process that involves dissipation of energy is not reversible. Any process that involves heat transfer from a hotter object to a colder object is not reversible. The second law of thermodynamics (Clausius Statement): Heat never flows spontaneously from a colder body to a hotter body. 22

20 Heat Engines A heat engine is a device designed to convert disordered energy into ordered d energy. The net work done by an engine during one cycle is equal to the net heat flow into the engine during the cycle (ΔU 0). W Q net Q net 23

21 e The efficiency of an engine is defined as net work done by the engine heat input W net Q H (e.g. a efficiency of e0.8 means 80% of the heat is converted to mechanical work) net work output W e heat input Q QH QC QC 1. Q H Q H Note: Q net Q in - Q out net H 24

22 Refrigerators and Heat Pumps Here, heat flows from cold to hot but with work as the input. Pump Refrigerator K Coefficient of performance 25

23 26

24 Reversible Engines and Heat Pumps A reversible engine can be used as an engine (heat input from a hot reservoir and exhausted to a cold reservoir) or as a heat pump (heat is taken from cold reservoir and exhausted to a hot reservoir). 28

25 From the second law of thermodynamics, no engine can have an efficiency ygreater than that of an ideal reversible engine Carnot engine that uses the same two reservoirs. The efficiency of this ideal reversible engine is e r T 1 C. T H 29

26 Details of the Carnot Cycle The ideal engine of the previous section is known as a Carnot engine. The Carnot cycle has four steps: 1. Isothermal expansion: takes in heat from hot reservoir; keeping the gas temperature at T H. 2. Adiabatic expansion: the gas does work without heat flow into the gas; gas temperature decreases to T C. 3. Isothermal compression: Heat Q C is exhausted; gas temperature remains at T C. 4. Adiabatic compression: raises the temperature back to T H. 30

27 The Carnot engine model was graphically expanded upon by Benoit Paul Émile Clapeyron in 1834 and mathematically elaborated upon by Rudolf Clausius in the 1850s and 60s from which the concept of entropy emerged The Carnot cycle illustrated 31

28 The Otto cycle Its power cycle consists of adiabatic compression, heat addition at constant volume, adiabatic expansion and rejection of heat at constant volume and characterized by four strokes, or reciprocating movements of a piston in a cylinder: intake/induction stroke compression stroke power stroke exhaust stroke 32

29 Entropy Heat flows from objects of high temperature to objects at low temperature because this process increases the disorder of the system. Entropy is a measure of a system s disorder. Entropy is a state variable. 35

30 If an amount of heat Q flows into a system at constant temperature, then the change in entropy is Q ΔSS. T Every irreversible ibl process increases the total t entropy of the universe. Reversible processes do not increase the total entropy of the universe. 36

31 The second law of thermodynamics (Entropy Statement): The entropy of the universe never decreases. 37

32 Example: An ice cube at 0.0 C is slowly melting. What is the change in the ice cube s entropy for each 1.00 g of ice that melts? To melt ice requires Q ml f joules of heat. To melt one gram of ice requires J of energy. The entropy change is ΔS Q T J 273 K 1.22 J/K. 38

33 300K Δ Δ S ht hot S cold Q T Q T 300 J 1J/K. 300 K J 60 J/K. 5 K Q 5 K t / t h? tf23tq 39

34 Statistical Interpretation of Entropy A microstate specifies the state of each constituent particle in a thermodynamic system. A macrostate is determined by the values of the thermodynamic state variables. 40

35 probabilit y of a macrostate number of microstates corresponding to the macrostate total number of microstates for all possible macrostates 41

36 The number of microstates for a given macrostate is related to the entropy. S k ln Ω where Ω is the number of microstates. 42

37 The Third Law of Thermodynamics It is impossible to cool a system to absolute zero. 46

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