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=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

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…

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

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…

Gas # molecules = n0

Temperature = T0

pressure = p0

Volume = V0

Discussion Q18.10Start

Vacuum

Gas Initial State# molecules = n0

Temperature = T0

pressure = p0

Volume = V0

Discussion Q18.10“Sudden” Hole in wall

Gas Final State# molecules = ?Temperature = ?pressure = ?Volume = ?

What Happens here?

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 [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: U = -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: U = Q• For a gas: U = 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 = U + p(V)• 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=U+W• dQ = dU + dW• Gas: dW = pdV

• What does U depend on?– Reminder about U– Measure of internal KE & PE between

particles

Gas # molecules = n0

Temperature = T0

pressure = p0

Volume = V0

Discussion Q18.10Start

Vacuum

Gas Initial State# molecules = n0

Temperature = T0

pressure = p0

Volume = V0

Discussion Q18.10Final 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?

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?

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• = CP/CV = 1.67 (monotomic)• = CP/CV = 1.4 (diatomic)• = 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 -