Lecture 26Purdue University, Physics 2201 Lecture 26 Thermodynamics I PHYSICS 220

Lecture 26Purdue University, Physics 2202

Lecture 26Purdue University, Physics 2203 First Law of Thermodynamics  U = Q - W Heat flow into system Increase in internal energy of system Work done by system Equivalent ways of writing 1st Law: Q =  U + W The change in internal energy of a system (  U) is equal to the heat flow into the system (Q) minus the work done by the system (W) Energy Conservation

Lecture 26Purdue University, Physics 2204

Lecture 26Purdue University, Physics 2205 Signs Example You are heating some soup in a pan on the stove. To keep it from burning, you also stir the soup. Apply the 1 st law of thermodynamics to the soup. What is the sign of A) Q B) W C)  U Positive, heat flows into soup Negative, stirring does work on soup Positive, soup gets warmer

Lecture 26Purdue University, Physics 2206 W = P  V W > 0 if  V > 0 expanding system does positive work W < 0 if  V < 0 contracting system does negative work W = 0 if  V = 0 system with constant volume does no work Work Done by a System M M yy W = F d cos  = P A d = P A  y = P  V

Lecture 26Purdue University, Physics 2207 PV Diagram Ideal gas law: PV = nRT For n fixed, P and V determine “state” of system T = PV/nR U = (3/2)nRT = (3/2)PV Examples: – which point has highest T? B – which point has lowest U? C – to change the system from C to B, energy must be added to system V P A B C V 1 V 2 P1P3P1P3

Lecture 26Purdue University, Physics 2208 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. pV 1 = nRT 1  T 1 = pV 1 /nR = 120K 2. pV 2 = nRT 2  T 2 = pV 2 /nR = 180K 3.  U = (3/2) nR  T = 1500 J  U = (3/2)  pV = (3/2) ( )J =1500 J (has to be the same) 4. W = p  V = 1000 J 5. Q =  U + W = = 2500 J Example (Isobaric)

Lecture 26Purdue University, Physics 2209 Example (Isochoric) 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. pV 1 = nRT 1  p 1 = nRT 1 /V = 1000 Pa 2. pV 2 = nRT 2  p 2 = nRT 2 /V = 1500 Pa 3.  U = (3/2) nR  T = 1500 J 4. W = p  V = 0 J 5. Q =  U + W = = 1500 J Requires less heat to raise T at const. volume than at const. pressure C V =C P -R V P 2 1 V P2P1P2P1

Lecture 26Purdue University, Physics 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? A) Case 1 B) Case 2 C) Same A B V(m 3 ) Case 1 A B V(m 3 ) P(atm) Case 2 P(atm) Net Work = area under P-V curve Area the same in both cases !

Lecture 26Purdue University, Physics 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 change in internal energy of the system the biggest? A) Case 1 B) Case 2 C) Same A B V(m 3 ) Case 1 A B V(m 3 ) P(atm) Case 2 P(atm)  U = 3/2  (PV) Case 1:  (PV) = 4x9-2x3=30 atm*m 3 Case 2:  (PV) = 2x9-4x3= 6 atm*m 3

Lecture 26Purdue University, Physics iClicker 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 heat added to the system the biggest? A) Case 1 B) Case 2 C) Same A B V(m 3 ) Case 1 A B V(m 3 ) P(atm) Case 2 P(atm) Q =  U + W W is same for both  U is larger for Case 1 Therefore, Q is larger for Case 1

Lecture 26Purdue University, Physics Q =  U + W Heat flow into system Increase in internal energy of system Which part of cycle has largest change in internal energy,  U ? 2  3 (since U = 3/2 PV) l Which part of cycle involves the least work W ? 3  1 (since W = P  V) l What is change in internal energy for full cycle?  U = 0 for closed cycle (since both P & V are back where they started) l What is net heat into system for full cycle (positive or negative)?  U = 0  Q = W = area of triangle (>0) Some questions: First Law Questions Work done by system V P V 1 V 2 P1P3P1P3

Lecture 26Purdue University, Physics Special PV Cases Constant Pressure –isobaric Constant Volume –isochoric Constant Temp  U = 0 –isothermic Adiabatic Q=0 (no heat is transferred) V P W = P  V (>0)  V > 0V P W = P  V =  V = 0

Lecture 26Purdue University, Physics st Law Summary 1st Law of Thermodynamics: Energy Conservation Q =  U + W Heat flow into system Increase in internal energy of system Work done by system V P l Point on P-V plot completely specifies state of system (PV = nRT) l Work done is area under curve l U depends only on T (U = 3nRT/2 = 3PV/2) For a complete cycle  U=0  Q=W

Lecture 26Purdue University, Physics Reversible? Most “physics” processes are reversible, you could play movie backwards and still looks fine. (drop ball vs throw ball up) Exceptions: –Non-conservative forces (friction) –Heat Flow Heat never flows spontaneously from a colder body to a hotter body.

Lecture 26Purdue University, Physics Summary of Concepts First Law of thermodynamics: Energy Conservation –Q =  U + W Heat Engines –Efficiency = 1-Q C /Q H Refrigerators –Coefficient of Performance = Q C /(Q H – Q C )