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Physics 101: Lecture 30, Pg 1 PHY101: Lecture 30 Thermodynamics l New material: Chapter 15 l The dynamics of thermal processes (equilibrium, work, energy)

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Physics 101: Lecture 30, Pg 2 The Ideal Gas Law (review) l P V = N k B T è N = number of molecules »N = number of moles (n) x N A molecules/mole è k B = Boltzmanns constant = 1.38 x 10 -23 J/K l P V = n R T è R = ideal gas constant = N A k B = 8.31 J/mol/K

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Physics 101: Lecture 30, Pg 3 Summary of Kinetic Theory: The relationship between energy and temperature (for monatomic ideal gas) l Internal Energy U = number of molecules x ave KE/molecule = N (3/2) k B T = (3/2) n RT = (3/2) P V (ideal gas) Careful with units: R = 8.31 J/mol/K = molar mass in kg/mol v rms = speed in m/s Root-mean-square speed :

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Physics 101: Lecture 30, Pg 4 Thermodynamics l 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

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Physics 101: Lecture 30, Pg 5 The Laws of Thermodynamics l Like Newtons 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: F = 0, a=0) Indicator of thermal equilibrium: T=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.

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Physics 101: Lecture 30, Pg 6 The Laws of Thermodynamics l 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 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

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Physics 101: Lecture 30, Pg 7 Thermodynamic Systems and P-V Diagrams l Ideal gas law: P V = n R T l For n fixed, P and V determine the state of the system è T = P V/ (n R) è U = (3/2) n RT = (3/2) P V l Examples: è which point has highest T ? » 2» 2 è which point has lowest U ? » 3» 3 è 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. V P 1 2 3 V 1 V 2 P1P3P1P3

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Physics 101: Lecture 30, Pg 8 Work Done by a System (P=constant) W W y W = F s = P A y = P V W > 0 if V > 0 expanding system does work 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 does no work Gas P,V 1,T 1 Gas P,V 2,T 2 System: Gas Surroundings: Piston, walls Isobaric process

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Physics 101: Lecture 30, Pg 9 Classification of Thermal Processes Isobaric : P = constant, W = P V Isochoric : V = constant, W = 0 => U = Q l Isothermal : T = constant, W = n R T ln(V f /V i ) (ideal gas) Adiabatic : Q = 0 => W = - U= -3/2 n R T (ideal gas)

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Physics 101: Lecture 30, Pg 10 V P 1 2 3 4 V P W = P V (>0) 1 2 3 4 V > 0V P W = P V = 0 1 2 3 4 V = 0 V P W = P V (<0) 1 2 3 4 V < 0 V W = P V = 0 1 2 3 4 P V = 0 V P 1 2 3 4 If we go the other way then W tot < 0 V P 1 2 3 4 W tot > 0 W tot = ?? Isobaric Process: P=constant Isochoric Process: V=constant

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Physics 101: Lecture 30, Pg 11 V P 1 2 3 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 ? V 1 V 2 P1P3P1P3 Area = (V 2 -V 1 )x(P 1 -P 3 )/2

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Physics 101: Lecture 30, Pg 12 First Law of Thermodynamics Example V P 1 2 V 1 V 2 P 2 moles of monatomic ideal gas is taken from state 1 to state 2 at constant pressure P=1000 Pa, where V 1 =2m 3 and V 2 =3m 3. Find T 1, T 2, U, W, Q. 1. P V 1 = n R T 1 T 1 = P V 1 /(nR) = 120K 2. P V 2 = n RT 2 T 2 = P V 2 /(nR) = 180K 3. U = (3/2) n R T = 1500 J or U = (3/2) P V = 1500 J 4. W = P V = 1000 J > 0 Work done by the system (gas) 5. Q = U + W = 1500 J + 1000 J = 2500 J > 0 heat gained by system (gas)

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Physics 101: Lecture 30, Pg 13 First Law of Thermodynamics Example 2 moles of monatomic ideal gas is taken from state 1 to state 2 at constant volume V=2m 3, where T 1 =120K and T 2 =180K. Find Q. 1. P V 1 = n R T 1 P 1 = n R T 1 /V = 1000 Pa 2. P V 2 = n R T 1 P 2 = n R T 2 /V = 1500 Pa 3. U = (3/2) n R T = 1500 J 4. W = P V = 0 J 5. Q = U + W = 0 + 1500 = 1500 J => It requires less heat to raise T at const. volume than at const. pressure. V P 2 1 V P2P1P2P1

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Physics 101: Lecture 30, Pg 14 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 A B 4 2 39 V(m 3 ) Case 1 A B 4 2 39 V(m 3 ) P(atm) Case 2 P(atm) correct Net Work = area under P-V curve Area the same in both cases!

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