General Physics L02_paths.ppt energy transfers Thermodynamic Paths energy transfers § 15.1–15.4

Definitions System: bodies and surroundings exchanging energy State: unique set of p, V, T, (n or N) (state variables) Process: change in state of a system

Internal Energy U U  SKi + SSVij j<i Ki = kinetic energy of molecule i wrt com Vij = intermolecular potential energy of i and j Does not include potential or kinetic energy of bulk object Each thermodynamic state has a unique U (U is a state function)

Monatomic Ideal Gas U = S ½ mivi2 = S 3/2 kT = 3/2 NkT = 3/2 nRT

Question All other things being equal, adding heat to a system increases its internal energy U. True. False.

Question All other things being equal, lifting a system to a greater height increases its internal energy U. True. False.

Question All other things being equal, accelerating a system to a greater speed increases its internal energy U. True. False.

Question All other things being equal, doing work to compress a system increases its internal energy U. True. False.

Clarification Quizzes will generally be on Wednesdays First quiz Feb 8/9 (Thursday/Friday: Monday late start)

Energy Transfers Q: heat added to the system surroundings  system From a temperature difference W: work done by the system system  surroundings Achieved by a volume change

Work W The surroundings exert pressure on the system. If the system expands, it does work on the surroundings. So, W > 0, and the surroundings do negative work on the system.

First law of Thermodynamics DU = Q – W DU is path-independent

Conservation of Energy DU of a system = work done on the system + heat added to the system

Work and Heat Depend on the path taken between initial and final states.

pV Diagrams W = area under pV curve Source: Y&F, Figure 19.6a

Question What is this system doing? Expanding Contracting Absorbing heat at constant volume Absorbing heat at constant pressure Source: Y&F, Figure 19.6b

Question What is the sign of the work W for this process? + – Cannot be determined Source: Y&F, Figure 19.6b

Question What is this system doing? Expanding at constant volume Expanding at constant temperature Expanding at constant pressure Source: Y&F, Figure 19.6c

Question How is the temperature of this ideal gas changing? Increasing Decreasing Remaining constant Cannot be determined Source: Y&F, Figure 19.6c

Simple Case Expansion at constant pressure W = pDV Source: Y&F, Figure 19.6c

Question The work done by a thermodynamic system in a cyclic process (final state is also the initial state) is zero. True. False. Source: Y&F, Figure 19.12

Cyclic Process W  0 Is the system a limitless source of work? (Of course not.) W Source: Y&F, Figure 19.12

Cyclic Processes DU = U1 – U1 = 0 so Q – W = 0 Q = W Total work output = total heat input

Work out = Heat in Does this mean cyclic processes convert heat to work with 100% efficiency? (Of course not.) Waste heat is expelled, not recovered.

Types of Processes cool names, easy rules

Reversible An infinitesimal change in conditions reverses the direction Requires no non-conservative processes no friction no contact between different temperatures An ideal concept not actually possible some processes can get close

Constant pressure “Isobaric” W = PDV

Constant Volume “Isochoric” W = 0

Constant Temperature “Isothermal” Ideal gas: W = nRT ln(Vf/Vi)

General Physics L02_paths.ppt No Heat Flow “Adiabatic” Q = 0 W: more complicated PfVfg = PiVig g = heat capacity ratio CP/CV CP for constant pressure CV for constant volume

Specific Heats of Ideal Gases § 15.6

Constant Volume or Pressure Constant volume: heating simply makes the molecules go faster Constant pressure: As the molecules speed up, the system expands against the surroundings, doing work It takes more heat to get the same DT at constant pressure than at constant volume

Constant Volume U = 3/2 nRT for a monatomic gas DU = 3/2 nRDT DU = Q – W = Q Q = 3/2 nRDT Molar heat capacity CV = Q/(nDT) CV = 3/2 R

Constant Pressure U = 3/2 nRT DU = 3/2 nRDT DU = Q – W W = PDV = nRDT Q = DU + W = 3/2 nRDT + nRDT = 5/2 nRDT Molar heat capacity CP = Q/(nDT) CP = 5/2 R

Heat Capacity Ratio g = CP/CV = (5/2)/(3/2) = 5/3 For a monatomic ideal gas

Homework 1 Due Feb 1/Feb 2 pp. 442–443 Conceptual Questions 6 and 7 p. 444 Problems 1, 8, and 10 p. 444 Problem 15 Hint: Find DT in terms of DV using PV = nRT Find DU using U = 3/2 nRT p. 445 Problem 25 Hint: What is DU for A→C? p. 448 Problem 80