Interaction Between Macroscopic Systems

1. Interaction BetweenMacroscopic Systems

2. We’ve been focusing on isolated Macroscopic Systems. So far, we’ve been interested in the statistical treatment of the dependence of the number of accessible states Ω(E) on the system energy E. We’ve found that Ω(E)  Ef δE (f = # of degrees of freedom of system ~ 1024). • Now, we want to focus on how to characterize the Macroscopic properties of the system & to do this especially when it isn’t isolated, but is allowed to interact with another macroscopic system  “The Outside World”.

3. The System & It’s Surroundings orThe System & The Universe! Abstract SketchAn Example

4. 3 General Kinds of Systems 1. Open Systems • Systems that can exchange both matter & energy with their surroundings. 2. Closed Systems • Systems that can exchange energy with their surroundings, but not matter. 3. Isolated Systems: • Systems that do not exchange matter or energy with their surroundings.

5. We now want to characterize the macroscopic properties of a system, when it isn’t isolated, but is interacting with another macroscopic system “The Outside World”. • To describe the system’s properties, we specify it’s Average EnergyĒ • Plus some (usually a small number n) of measurable External Parameters: x1,x2,x3,…xn • Of course, the quantum mechanical energy levelsof the systemdepend on these external parameters, through the equations of motion.

6. Examples of External Parametersx1,x2,x3,…xn • Example 1: System = A mass m(or more than one mass) interacting with its environment: The position coordinates x1,x2,x3 are external parameters.

7. Examples of External Parametersx1,x2,x3,…xn • Example 1: System = A mass m(or more than one mass) interacting with its environment: The position coordinates x1,x2,x3 are external parameters. • Example 2: System = A gas confined to a container: The container volume V is an external parameter.

8. More Examples of External Parametersx1,x2,x3,…xn • Example 3: System = An electric charge q (or more than one charge) interacting with its environment: An Applied Electric Field E is an external parameter.

9. More Examples of External Parametersx1,x2,x3,…xn • Example 3: System = An electric charge q (or more than one charge) interacting with its environment: An Applied Electric Field E is an external parameter. • Example 4: System = A magnetic dipole(or more than one dipole): An Applied Magnetic Field B is an external parameter.

10. The energy of a system’s many body, quantum mechanical Microstate, labeled r, is specified by it’s quantized energies: Er(x1,x2,…xn) • The Macrostate of the same system can be Defined by specifying the system’s Average EnergyĒ. • For an ensemble of similar systems, all in the same Macrostate, we can find any one of these systems in a HUGE NUMBER of different Microstates. • From our previous discussion, these are characterized by Ω(E) = AEf

11. Consider 2 macroscopic systems A & A', interacting with each other & in thermal equilibrium (note that we still haven’t yet rigorously defined thermal equilibrium!). It is reasonable that Interaction Exchange of Energy. • We assume that the total system Ao= A + A', is isolated & at equilibrium. A & A' are interacting.

12. A' (N',V',E') A (N,V,E) • The situation is shown schematically in the figure, assuming the external parameter is the volume V. • Because it is isolated, the combined system A0 has a constant energy E0: • E0 = E + E' = constant

13. Interaction Exchange of Energy. • Assume that the total system Ao= A + A', is isolated & at equilibrium. A & A' are interacting. An ensemble of similar systems is shown schematically in the figure. We focus attention on system A. A A' A A' A A'

14. There are 2 General Kinds of Interactions between systems. These are: 1. Thermal Interactions • External parameters x1,x2,x3,…xn remain fixed. The quantum energy levels Er(x1,x2,…xn) are unchanged. But, the POPULATIONS of these levels change, so the occupation probability of these levels also changes. 2. Mechanical Interactions • External parameters x1,x2,x3,…xnDO change The quantum energy levels Er(x1,x2,…xn) are shifted.

15. Section 2.6: Thermal Interactions • Consider 2 macroscopic systems A & A', interacting with each other & in thermal equilibrium. Consider the case where there are Thermal Interactions Only,no mechanical interactions. See the Figure A A' A' • We’ll now focus on system A with mean internal energy Ē(x1,x2,x3,…xn)

16. A A' A' Thermal Interactions Only: No Mechanical Interactions • A’smean internal energy = Ē(x1,x2,x3,…xn) • No mechanical interaction so all external parameters x1,x2,x3,…xn remain fixed (no mechanical work is done!) •  xi = constant, i = 1,…n • The total system Ao = A + A', is isolated & at equilibrium. So •  The energy of system Ao is conserved: • Ēo = Ē + Ē' = constant

17. A Very Important Definition! • Heat≡The mean energy transferred from one system to another as a result of a purely thermal interaction. • More precisely, due to it’s interaction with A', the mean energy of A is changed by Ē Ē ≡ Q ≡heat absorbed (emitted) by A (Q can be positive or negative)

18. Similarly, for , A', due to its interaction with A, its mean energy change is Ē'≡ Q' ≡heat absorbed (or emitted) by A' • Now, consider the total isolated systemAo. • Its subsystems A' & A' are interacting & exchanging energy as described. • Because Ao is isolated, its total energy is conserved. That is: Ēo = Ē + Ē' = const

19. Because Ao is isolated, its total energy is conserved. That is: Ēo = Ē + Ē' = const  Ēo = 0 = Ē+ Ē' Or, from the definition of heat: Q + Q' = 0; Q = - Q'  The heat absorbed (given off) byA = - heat given off (absorbed) by A'. Conservation of Total System Energy!!

20. Section 2.7: Mechanical Interactions • Consider again 2 macroscopic systems A & A', interacting with each other & in thermal equilibrium. • Consider the case where there are Mechanical Interactions only, & no Thermal Interactions. This requires that they are thermally isolated (insulated) from each other. This is achieved by surrounding the systems with an “Adiabatic Envelope”

21. Macroscopic systems A & A', interacting & in equilibrium. Mechanical Interactions only, no Thermal Interactions. This requires them to be completely thermally isolated (insulated) from each other. This is achieved by surrounding the systems with an “Adiabatic Envelope” . ≡An ideal partition which separates the 2 systemsA & A', in which external parameters are fixed & each of which is in internal equilibrium, such that each subsystem remains in its Macrostate indefinitely. Obviously, this is an idealization!

22. Physically, this Adiabatic Envelope is such that no energy (heat) transfer can occur between the two systems A & A'. This is clearly, an idealization!!! But many materials behave approximately as is shown in the Figure: A' A

23. A' A • Adiabatic Envelope No energy (heat) transfer can occur between the two systems A & A'. Idealization!!! Many materials behave approximately as in the Figure: • When 2 systems, A, A', are thermally insulated from each other, they are STILL capable of interacting. How? Through • Changes in their External Parameters • ≡ Mechanical Interaction • In this case, the 2 systems do MECHANICAL WORK on each other & This Work Can Be Measured.

24. w gravity s A Freshman Physics Example! • A gas is enclosed in a vertical cylinder closed by a piston of weight W. The piston is thermally insulated from the gas. System A gas + cylinder System A'piston + weight • Consider system A which has a mechanical interaction with A' gas • A’s external parameters change:  So does it’s mean • energy. Call this change xĒ. The macroscopic • work done ON the system is then defined as W  xĒ.

25. w gravity s Freshman Example continued • Initially, the piston is clamped in position at height si. It is released & the height changes to some final height sf(higher or lower than si). System A gas + cylinder System A'piston + weight • Their interaction involves changes in system’s external parameters (gas volume, piston height s). gas • A’s external parameters change • due to its interaction with A'. • So, it’s mean energy change is • xĒ. The macroscopic work done • ON the system is:W  xĒ • The macroscopic work done BYthe system is defined to be the negative of this: • W = -W  - xĒ

26. Freshman Example, Continued • Macroscopic work done BY the system: W = -W  - xĒ • Energy Conservation for combined system: W = -W ', • or W + W ' = 0. Mechanical interaction between the systems due to changes in their external parameters causes changes in their energy levels & also changes in their occupancy. See figure!

27. Section 2.8: General Interaction

28. Consider 2 interacting macroscopic systems A, A', in thermal equilibrium. Thermal InteractionInteraction with no mechanical work.  The energy exchange between A, A' is Heat Exchange: • Conservation of energy for system Ao gives: Ēo = 0 = Ē+ Ē' Ē ≡ Q =heat absorbed (emitted) by A Ē'≡ Q' =heat absorbed (emitted) by A' So, Q + Q' = 0 or Q = - Q'

29. Consider 2 interacting macroscopic systems A, A', in thermal equilibrium. Mechanical Interaction One in which A, A', are thermally insulated from each other.  No heat exchange is possible. • They interact by doing MECHANICAL WORK on each other through changes in their external parameters. The work done ONA is W xĒ. The work done BYA: W = -W - xĒ. • Conservation of energyfor the combined system gives xĒo = 0 = xĒ+ xĒ' So, W + W' = 0

30. The most general case is one with Thermal & • Mechanical Interactions at the same time. • The external parameters are NOT fixed. A, A' are NOTthermally insulated from each other. • As a result of this General Interaction, the mean energy of A is changed BOTH by a change in external parameters ANDby a transfer of thermal energy. This mean energy change may be written: • Ē = W + QorĒ = - W + Q Q Q = Ē + W

31. Summary • In the most general case with Thermal & Mechanical Interactions at the same time, the mean energy of A can be changed BOTH by a change in external parameters AND by a transfer of thermal energy. The resulting change in the mean energy of A can be written: Q Q = Ē + W

32. Q Q = Ē + W • This result is a statement of The First Law of Thermodynamics The physics of this law is simply Conservation of Total Energy Total Energy Change ≡ Heat + Mechanical Work • This result could also be viewed as a definition of the heat absorbed (emitted) by a system.

33. Q Q = Ē + W ≡The 1st Law of Thermodynamics Physics: Conservation of Total Energy • So, for two interacting systems at equilibrium, The 1st Law of Thermo says that total energy is conserved, • Note that it says nothing about the DIRECTION of energy transfer!! For that, we will need the 2nd Law of Thermodynamics!

34. Q Q = Ē + W • In this course we are studying this law in order to obtain a fundamental understanding of the relation between thermal & mechanical interactions. This type of study is called Classical Thermodynamics

35. Q Q = Ē + W Comments on Units • Work, Heat, & Internal Energy obviously all have the same units. The SI Energy units are Joules (J). • But, the old units for heat are calories (C) & sometimes we’ll use them. • Using calories for heat units is widespread in Biology, the Life Sciences, Medicine, Chemistry & some Engineering disciplines.

36. Moveable Piston • As a simple Example, consider two gases, A & A' confined to a container & separated by a moveable piston, as shown in the figure. s A' A Q Q = Ē + W • Now, we’ll analyze this system using the 1st Law of thermodynamics in 4 different situations.

37. Q Q = Ē + W A A' Example • At least 4 possible cases to consider: • 1. The piston is clamped & thermally insulating. •  A, A'don’t interact. • That is, nothing happens! Moveable Piston

38. Q Q = Ē + W A A' Example • At least 4 possible cases to consider: Moveable Piston 2. The piston is clamped & NOT thermally insulating.  A, A'interact thermally. So,the pressures change. In this case, A, A'exchange heat Q, butno mechanical work W is done. Q Q = Ē

39. Q Q = Ē + W Example A A' • At least 4 possible cases to consider: • 3. The piston is thermally insulating & free to move.  A, A' interact mechanically. So, the pressures & Moveable Piston the volumes both change. In this case, no heat Qis exchanged between A, A',but one of them does mechanical work W on the other. 0 0 = Ē + W

40. Q Q = Ē + W Example A A' • At least 4 possible cases to consider: • 4. The piston is NOT thermally insulating & is free to move. •  A, A'interact both thermally & Moveable Piston mechanically. So, the pressures & volumes both change. A, A' exchange heat Q, and one of them does mechanical work W on the other. Q Q = Ē + W

41. The 1st Law of Thermodynamics • Q = Ē + W • Ē = Change in internal energy of the system. • Q = NETheat transferred to the system. • W =Work done BY the system. • The 1st Law is deceptively simple looking. It’s • obviously a form of the general • Law of Conservation of Total Energy. • But • Be careful about the sign conventions! • Positive Q is heat transferred to the system. • Positive W is work done by the system.

42. The 1st Law of Thermodynamics In words: “The change in the internal energy OFa system depends only on theNETheat transferred to the system & the net work done BYthe system, & is independent of the processes involved”.