1. Physics 101: Lecture 25, Pg 1 Work Done by a System W W yy 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

2. Physics 101: Lecture 25, Pg 2 Homework Problem: Thermo I 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 = ??

3. Physics 101: Lecture 25, Pg 3 V P 1 2 3 Now try this: V 1 V 2 P1P3P1P3 Area = (V 2 -V 1 )x(P 1 -P 3 )/2

4. Physics 101: Lecture 25, Pg 4 First Law of Thermodynamics Energy Conservation  U = Q - W Heat flow into system Increase in internal energy of system Work done by system V P 1 2 3 V 1 V 2 P1P3P1P3 Equivalent ways of writing 1st Law: Q =  U + W Q =  U + p  V

5. Physics 101: Lecture 25, Pg 5 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. 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) p  V = 1500 J (has to be the same) 4. W = p  V = 1000 J 5. Q =  U + W = 1500 + 1000 = 2500 J

6. Physics 101: Lecture 25, Pg 6 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. pV 1 = nRT 1  p 1 = nRT 1 /V = 1000 Pa 2. pV 2 = nRT 1  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 = 0 + 1500 = 1500 J requires less heat to raise T at const. volume than at const. pressure V P 2 1 V P2P1P2P1

7. Physics 101: Lecture 25, Pg 7 Chapter 15, Problem 1 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!

8. Physics 101: Lecture 25, Pg 8 Chapter 15, Problem 2 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? 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  U = 3/2  (pV) Case 1:  (pV) = 4x9-2x3=30 atm-m 3 Case 2:  (pV) = 2x9-4x3= 6 atm-m 3

9. Physics 101: Lecture 25, Pg 9 Chapter 15, Problem 2 (followup) 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? 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 Q =  U + W W is same for both  U is larger for Case 1 Therefore, Q is larger for Case 1

10. Physics 101: Lecture 25, Pg 10 Q =  U + W Heat flow into system Increase in internal energy of system Work done by system V P 1 2 3 V 1 V 2 P1P3P1P3 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?  U = 0  Q = W = area of triangle (>0) Some questions:

11. Physics 101: Lecture 25, Pg 11Recap: è 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