1. Paperwork HMWK deadline off by one hour –Everyone get some bonus? Guest Instructors –Monday – Chapter 20.1-20.3 –Week After Mon & Fri That’s when the exam is ( 2 weeks) Move exam back one week, skip one lab?

2. Schedule Short Term Today – Chapter 19 Next Week –Monday –Chapter 20.1-20.3 (Guest) –Tuesday – Lab #2 Quiz#2 [Chapter 18, Labs] –Wed – Ch. 20.4-20.5 –Friday – Practice Problems Week After –Monday – Ch. 20.6-20.7 (Guest) –Tuesday – Lab 3 & Quiz 3 –Wed (Was exam) Review 17-20 –Friday 21.1-21.3 (Guest) Then -

3. Chapter 19 1 st Law of Thermo –Q=  U+W (eq. 19.5) –Q = Heat –W = Work –U = Internal Energy [Any Guesses] Internal Energy –Sum of all KE [Thermo] –Plus Sum of all interactions [bonds] –Not U as in grav. potential energy

4. Signs & Such Q=  U+W (eq. 19.5) Talks about how heat affects a system What does Q being + mean? –Heat added to system What happens if  U is positive? –Raise temperature, change state… –Chemical bonds have negative energy –Solid to liquid means  U increases, less neg. What happens if W is positive? –System does work on its surroundings –Maybe heats up a container, etc…

5. Signs & Such Q=  U+W (eq. 19.5) Talks about how heat affects a system Q+  heat enters system  U+  Internal energy raised –Bonds broken, temperature increased W+  Work done on outside world W-  Work done on system from outside

6. Isolated System Q=  U+W System completely isolated from outside What is Q? (say as a function of time) What is W? (time dependence as well) Implications? Isolation can be attained by expansion…

7. Gas # molecules = n 0 Temperature = T 0 pressure = p 0 Volume = V 0 Discussion Q18.10 Start Vacuum

8. Gas Initial State # molecules = n 0 Temperature = T 0 pressure = p 0 Volume = V 0 Discussion Q18.10 “Sudden” Hole in wall Gas Final State # molecules = ? Temperature = ? pressure = ? Volume = ? What Happens here?

9. Changes in System Follow equation: Q=  U+W –Look at small changes dQ = dU + dW dU = dQ – dW dW = pdV [gaseous systems] dU = dQ – pdV [1 st law thermo for gas]

10. Types of Changes Adiabatic (Constant Heat) No heat transfer to/from system Q = 0 dU = dQ – dW dU = -dW As a whole:  U = -W For a gas: dU = -pdV

11. Types of Changes Isochoric (Constant Volume) No change in volume [Stiff container] dV = 0 pdV = 0 = W dU = dQ As a whole:  U = Q For a gas:  U = Q Usually implies no work done that changes volume Example: Stirring liquid usually still “isochoric”

12. Types of Changes Isobaric (Constant Pressure) No change in volume [Stiff container] p = constant dQ = dU + dW dQ = dU + pdV p is constant of integration, no V dependence Integrate both sides  Q =  U + p(  V) Example: Boiling water in an open pot

13. Types of Changes Isothermal (Constant Temperature) No change in Temp [Could add heat though…] T = constant dQ = dU + dW dQ = dU + pdV Complicated  pV = nRT p = nRT/V (Ideal Gas) So integration not trivial even for ideal gas Example: Icewater mixture, while both exist

14. Internal Energy of Ideal Gas Q=  U+W dQ = dU + dW Gas: dW = pdV What does U depend on? –Reminder about U –Measure of internal KE & PE between particles

15. Gas # molecules = n 0 Temperature = T 0 pressure = p 0 Volume = V 0 Discussion Q18.10 Start Vacuum

16. Gas Initial State # molecules = n 0 Temperature = T 0 pressure = p 0 Volume = V 0 Discussion Q18.10 Final State (Again) Q=  U+W Gas Final State # molecules = ? Temperature = ? pressure = ? Volume = ? So what is DU? Is there work done on gas? Is there heat input?

17. Internal Energy of Ideal Gas Q=  U+W What does U depend on? –Temperature –Gas doesn’t change phase (or its not a gas) –Ideal Gas: No interactions between particles –No potential energy (bonding) between particles –Diatomic molecules?

18. Ideal Gas Heat Capacities Different for different conditions dQ = nCdT [molar heat capacity] Constant Pressure: C P Constant Volume: C V Constant Temperature: C T –Wouldn’t that mean dQ = 0? Constant n: C n –n is not in C, other part of Equation

19. Relationship C P = C V + R Derivation in text Heat capacity larger for isobaric process Ratio of heat capacities  = C P /C V = 1.67 (monotomic)  = C P /C V = 1.4 (diatomic)  = C P /C V = 1.3 (“triatomic”) Look at this closer later next week…

20. Reading & Assignments Chapter 18 assignment –Up today, Due next Friday Chapter 19 Assignment –Up today, Due in 1.5 weeks Put up practice problems –Hopefully answers, etc… Read chapter 19 & sect. 20.1 to 20.3 for Monday

21. Schedule Short Term Today – Chapter 19 Next Week –Monday –Chapter 20.1-20.3 (Guest) –Tuesday – Lab #2 Quiz#2 [Chapter 18, Labs] –Wed – Ch. 20.4-20.5 –Friday – Practice Problems Week After –Monday – Ch. 20.6-20.7 (Guest) –Tuesday – Lab 3 & Quiz 3 –Wed (Was exam) Review 17-20 –Friday 21.1-21.3 (Guest) Then -