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PHY101: Lecture 30 Thermodynamics

PHY101: Lecture 30 Thermodynamics. New material: Chapter 15 The dynamics of thermal processes (equilibrium, work, energy). The Ideal Gas Law (review). P V = N k B T N = number of molecules N = number of moles (n) x N A molecules/mole

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PHY101: Lecture 30 Thermodynamics

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  1. PHY101: Lecture 30Thermodynamics • New material: Chapter 15 • The dynamics of thermal processes (equilibrium, work, energy)

  2. The Ideal Gas Law(review) • P V = N kB T • N = number of molecules • N = number of moles (n) x NA molecules/mole • kB = Boltzmann’s constant = 1.38 x 10-23 J/K • P V = n R T • R= ideal gas constant = NAkB= 8.31 J/mol/K

  3. Careful with units:R = 8.31 J/mol/Km= molar mass in kg/molvrms = speed in m/s Summary of Kinetic Theory:The relationship between energy and temperature(for monatomic ideal gas) Root-mean-square speed : • Internal Energy U • = number of molecules x ave KE/molecule • = N (3/2) kBT • = (3/2) n RT = (3/2) P V (ideal gas)

  4. Thermodynamics • Dynamics of thermal processes, i.e. when a system interacts with its surroundings via thermal processes : system  surroundings P,V,T The state of the system (e.g. gas) is determined by P,V and T. Example: gas filled container with piston on open end of container: system = gas at pressure P and temperature T and volume V surroundings = piston and walls of container

  5. The Laws of Thermodynamics • Like Newton’s laws govern the dynamics of an object in motion, the three laws of thermodynamics govern the dynamics of thermal processes. 0. Law: Thermal equilibrium (Mechanics: SF = 0, a=0) Indicator of thermal equilibrium: DT=0 (like a=0) Two systems are in thermal equilibrium, if there is no net flow of heat between them if brought in thermal contact (i.e. the two systems are already in their most probable state and have thus no motivation to spontaneously change their state). If two systems are individually in equilibrium with a third system, they are in thermal equilibrium with each other.

  6. The Laws of Thermodynamics • 1. Law of Thermodynamics (energy conservation) change in heat added to/subtracted work done internal energy = from the system - by/on the system of the system D U = Q - W Q > 0 : heat is added to system Q < 0 : heat is subtracted from system W > 0 : work done by system on surroundings W < 0 : work done on system by surroundings

  7. Thermodynamic Systems and P-V Diagrams • Ideal gas law:P V = n R T • Fornfixed,Pand Vdetermine the “state” of the system • T = P V/ (n R) • U = (3/2) n RT = (3/2) P V • Examples: • which point has highest T ? • 2 • which point has lowest U ? • 3 • to change the system from 3 to 2, energy must be added to system. Work done in a thermal process is given by the area underneath the P,V graph. P 1 2 P1 P3 3 V V1 V2

  8. y W W Work Done by a System (P=constant) System: Gas Surroundings: Piston, walls Isobaric process Gas P,V2,T2 Gas P,V1,T1 • W = F s = P A D y = P V • W > 0 if V > 0 • expanding system doeswork on surroundings (W positive) • W < 0 if V < 0 • contracting system : work is done on the system by surroundings (W negative) • W = 0 if V = 0 • system with constant volume doesnowork

  9. Classification of Thermal Processes • Isobaric : P = constant, W = P DV • Isochoric : V = constant, W = 0 => DU = Q • Isothermal : T = constant, W = n R T ln(Vf/Vi) (ideal gas) • Adiabatic : Q = 0 => W = - DU= -3/2 n R DT (ideal gas)

  10. 1 P 2 4 3 P P W = PDV (>0) W = PDV = 0 V 1 1 2 2 1 P 2 4 4 3 3 Wtot > 0 4 3 DV > 0 V DV = 0 V V P P W = PDV (<0) W = PDV = 0 1 1 2 2 1 P 2 If we gothe other way thenWtot < 0 4 4 3 3 4 3 V DV = 0 V DV < 0 V Wtot = ?? Isobaric Process: P=constant Isochoric Process: V=constant

  11. Now try this:What is the total work done by system when going from state 1 to state 2 to state 3 and back to state 1 ? P 1 P1 P3 2 Area = (V2-V1)x(P1-P3)/2 3 V V1 V2

  12. P 1 P 2 V V1 V2 First Law of ThermodynamicsExample 2 moles of monatomic ideal gas is taken from state 1 to state 2 at constant pressure P=1000 Pa, where V1 =2m3 and V2 =3m3. Find T1, T2, DU, W, Q. 1. P V1 = n R T1 T1 = P V1/(nR) = 120K 2. P V2 = n RT2 T2 = P V2/(nR) = 180K 3. DU = (3/2) n R DT = 1500 J or DU = (3/2) P DV = 1500 J 4. W = P DV = 1000 J > 0 Work done by the system (gas) 5. Q = DU + W = 1500 J + 1000 J = 2500 J > 0 heat gained by system (gas)

  13. First Law of ThermodynamicsExample 2 moles of monatomic ideal gas is taken from state 1 to state 2 at constant volume V=2m3, where T1=120K and T2 =180K. Find Q. P P2 P1 2 1. P V1 = n R T1 P1 = n R T1/V = 1000 Pa 2. P V2 = n R T1 P2 = n R T2/V = 1500 Pa 3. DU = (3/2) n R DT = 1500 J 4. W = P DV = 0 J 5. Q = DU + W = 0 + 1500 = 1500 J => It requires less heat to raise T at const. volume than at const. pressure. 1 V V

  14. P(atm) P(atm) A B 4 4 A B 2 2 Case 1 Case 2 correct 3 9 3 9 V(m3) V(m3) Concept Question Shown in the picture below are the pressure versus volume graphs for two thermal processes, in each case moving a system from state A to state B along the straight line shown. In which case is the work done by the system the biggest? 1. Case 1 2. Case 2 3. Same Net Work = area under P-V curve Area the same in both cases!

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