Final Review: December 5th, 5:30-7:00pm, 130 Nicholson Hall by Ryan Gibson

pV = nRT pV = NkBT

Boltzmann Constant

kB = 1.38×10-23 J/K

R = kBNA

gas constant

R = 8.315 J/(mol⋅K)

= 1.99 calories/(mol⋅K)

n = number of moles N = number of particles

€

vrms =3RTM

(v2 )avg = vrmsRMS = Root-Mean-Square

€

Wby = pdV∫= area under p −V graph

3)Cyclical process (closed cycle) ΔEint,closed cycle =0 net area in p-V curve is Q

€

ΔE int = 0⇒Q =W

€

ΔE int = −W[ ]adiabatic

Adiabatic expansion/contraction - NO TRANSFER OF ENERGY AS HEAT Q = 0

€

ΔE int =Q

Constant-volume processes (isochoric)- NO WORK IS DONE W = 0

€

Wby = pdVVi

V f =Vi

∫ = 0

QΔV = 0 = nCVΔT

€

Wby = pdVVi

V f =Vi

∫ = pΔV

QΔP= 0 = nCPΔT

Constant-pressure processes

€

ΔE int =Q − pΔVCV = CP − R

example: monatomic gas γ = 5/3

€

p1V1γ = p2V2

γ

T1V1γ −1 =T2V2

γ −1

Compare with Isothermal Expansion (ΔT = 0)

€

T1 =T2 ⇔ p1V1 = p2V2[ ] isothermal

€

KE = 32kBT The KE of all ideal gas molecules

depends only on the temperature (not mass!)

Monoatomic ideal gas : He, Ar, Ne, Kr… (no potential energies)

€

Eint,monotonic−1mole = NA32kBT

=

32RT

The internal energy of an ideal gas depends only on the temperature

€

ΔE int,monotonic =32 nR ΔT( )

€

KE = 12m vrms( )

2=12m

3RTM

€

M = mNA

€

k = R NAB

0th law Thermal Equilibrium: A = B & B = C then A = C

Q = nCΔT

Q → 0 as ΔT → 0

1st law Conservation of energy: ΔEint = Q - Wby= Q + Won

Change in Internal energy = heat added minus work done by

The entropy of a closed system (no energy and no mass comes in and out) never decreases. It either stays constant (reversible process) or increases (irreversible process).

€

0 ≤ ΔStotal

2nd law

1) For reversible process:

2) For isothermal process (T=constant):

3) In general for a “small” change in temperature:

€

ΔScycle,rev = 0 =dQT

∫

€

ΔS = S f − Si ≈QTave

€

ΔSisothermal =QT

Q is total energy transferred as heat during the process (note: heat must be transferred from reservoir to keep temperature constant

1) For reversible process:

2) For isothermal process:

3) In general for gas, using 1st law:

4) For adiabatic (reversible) adiabatic compression/expansion (Q=0):

€

ΔScycle,rev = 0 =dQT

∫

€

ΔSrev,isothermal =QT

€

ΔE int = 0 ⇒ Q =W

W = nRT lnVfVi

€

⇒ ΔSisothermal = nR lnVfVi

Now integrate:

€

ΔSrev,gas = S f − Si = nR lnVfVi

+ nCV ln

TfTi

⇐ pV = nRT[ ] reversible€

dE int = dQ − dWnCV dTT

=dQT

−pdVT

& p = nRTV

€

ΔSrev,adiabatic = 0

€

⇒dQT

∫ = nRTV

dVT

∫ +nCV dTT

∫

Entropy is a State Function

Entropy and Engines

- Easy to produce thermal energy by doing work. How?

- Much harder to get work from thermal energy ⇒ engine

What is the thermal efficiency, ε, of an engine?

€

ε =energy we get

energy we pay for=WQH

€

ε =QH − QLQH

=1−QLQH

Perfect engine, ε=1

Conservation of energy

|QH| = |QL| + |W|

|QH| = heat added|QL| = heat released

€

0 = ΔScycle = ΔSH + ΔSL

0 =QHTH

−QLTL

Carnot Engine: An “Ideal” Engine

1) a → b: Isothermal expansion

2) b → c: adiabatic expansion

3) c → d: Isothermal compression

4) d → a: adiabatic compression

€

ΔE int = 0 ⇒ QH =W = nRTH lnVbVa

> 0

€

ΔSH = QHTH

= nR ln VbVa

> 0 positive

€

ΔE int = 0 ⇒ QL =W = nRTL lnVdVc

< 0

€

ΔSL = QLTL

= nR ln VdVc

< 0 negative

€

Q = 0 ⇒ ΔSb→c = 0

€

Q = 0 ⇒ ΔSd→a = 0

€

εcarnot =1−TLTH

All Carnot engines operating between two constant temperatures TH & TL have the same efficiency.

An irreversible engine is less efficient

All processes are reversible and no wasteful energy transfers occur

€

ε =QH − QLQH

=1−QLQH

Sample Problem

Imagine a Carnot engine that operates between the temperatures of TH = 850 K and TL = 300 K. The engine performs 1200 J of work each cycle, which takes 0.25 s.

a)  What is the efficiency of this engine?

b)  What is the average power of this engine?

c)  How much energy |QH| is extracted as heat from the high temperature reservoir every cycle?

d)  How much energy |QL| is delivered as heat to the low temperature reservoir every cycle?

e)  What is the entropy change ΔS of the working substance for the energy transfer to it from the high-temperature and low-temperature reservoir?

The efficiency, ε, of a Carnot engine is ONLY determined by the ratio TL/TH:

Power is found from Work done per cycle/time per cycle:

For any engine, the efficiency, ε, is defined as the work the engine does per cycle divided by the energy it absorbs as heat per cycle:

€

ε = 1− TLTH

= 1− 300 K850 K

≅ 65%

€

Pave =W Δt =1200 J

0.25 s( ) = 4800 W

Along the hot and cool isotherms, the entropy changes are:

€

ε =WQH

→ QH =Wε

=1200 J65%

≅1855 J

For a Carnot engine, from the 1st Law of thermodynamics says that the net heat transfer per cycle is equal to the net work done:

€

W = QH − QL → QL = QH −W = 1855 J −1200 J = 655 J

€

ε =WQH

=QH − QLQH

=1−QLQH

€

εcarnot =1−TLTH

€

ΔSL =QLTL

=−655 J300 K

= −2.18 J/K

€

ΔSH =QHTH

=1855 J850 K

= 2.18 J/K

€

→ ΔStot = 0

Stirling Engine

From a to b is isothermal Q=W: QH transferred to Engine

From b to c constant V Heat flows out, W=0

From c to d is isothermal Q=W: QL flows out of the engine

From d to a constant volume from low TL to TH Heat Q flows into the engine.

About Final Examination (200 pts) There are two parts: 1)  Part A covering Chapters 16 to Ch. 20 (100 pts) 2)  Part B is the cumulative exam from 1-16 (100 pts)

Chapter 1 MeasurementChapter 2 Motion along a Straight Line Chapter 3 VectorsChapter 4 Motion in 2 and 3 DimensionsChapter 5 Force and Motion IChapter 6 Force and Motion II Chapter 7 Kinetic Energy and WorkChapter 8 Potential Energy and Conservation of EnergyChapter 9 Center of Mass and Linear MomentumChapter 10 RotationChapter 11 Rolling Torque and Angular MomentumChapter 12 Equilibrium and ElasticityChapter 13 GravitationChapter 14 FluidsChapter 15 Oscillations

Exam1

Exam2

Exam3

Final