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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?. Schedule Short Term. Today – Chapter 19 Next Week
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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?
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 -
Chapter 19 • 1st Law of Thermo • Q=DU+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
Signs & Such • Q=DU+W (eq. 19.5) • Talks about how heat affects a system • What does Q being + mean? • Heat added to system • What happens if DU is positive? • Raise temperature, change state… • Chemical bonds have negative energy • Solid to liquid means DU increases, less neg. • What happens if W is positive? • System does work on its surroundings • Maybe heats up a container, etc…
Signs & Such • Q=DU+W (eq. 19.5) • Talks about how heat affects a system • Q+ heat enters system • DU+ Internal energy raised • Bonds broken, temperature increased • W+ Work done on outside world • W- Work done on system from outside
Isolated System • Q=DU+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…
Discussion Q18.10Start Gas # molecules = n0 Temperature = T0 pressure = p0 Volume = V0 Vacuum
Discussion Q18.10“Sudden” Hole in wall Gas Initial State # molecules = n0 Temperature = T0 pressure = p0 Volume = V0 Gas Final State # molecules = ? Temperature = ? pressure = ? Volume = ? What Happens here?
Changes in System • Follow equation: Q=DU+W • Look at small changes • dQ = dU + dW • dU = dQ – dW • dW = pdV [gaseous systems] • dU = dQ – pdV [1st law thermo for gas]
Types of Changes • Adiabatic (Constant Heat) • No heat transfer to/from system • Q = 0 • dU = dQ – dW • dU = -dW • As a whole: DU = -W • For a gas: dU = -pdV
Types of Changes • Isochoric (Constant Volume) • No change in volume [Stiff container] • dV = 0 • pdV = 0 = W • dU = dQ • As a whole: DU = Q • For a gas: DU = Q • Usually implies no work done that changes volume • Example: Stirring liquid usually still “isochoric”
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 = DU + p(DV) • Example: Boiling water in an open pot
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
Internal Energy of Ideal Gas • Q=DU+W • dQ = dU + dW • Gas: dW = pdV • What does U depend on? • Reminder about U • Measure of internal KE & PE between particles
Discussion Q18.10Start Gas # molecules = n0 Temperature = T0 pressure = p0 Volume = V0 Vacuum
Discussion Q18.10Final State (Again) Q=DU+W Gas Initial State # molecules = n0 Temperature = T0 pressure = p0 Volume = V0 Is there work done on gas? Is there heat input? Gas Final State # molecules = ? Temperature = ? pressure = ? Volume = ? So what is DU?
Internal Energy of Ideal Gas • Q=DU+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?
Ideal Gas Heat Capacities • Different for different conditions • dQ = nCdT [molar heat capacity] • Constant Pressure: CP • Constant Volume: CV • Constant Temperature: CT • Wouldn’t that mean dQ = 0? • Constant n: Cn • n is not in C, other part of Equation
Relationship • CP = CV + R • Derivation in text • Heat capacity larger for isobaric process • Ratio of heat capacities • g = CP/CV = 1.67 (monotomic) • g = CP/CV = 1.4 (diatomic) • g = CP/CV = 1.3 (“triatomic”) • Look at this closer later next week…
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
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 -