Lecture 26 Purdue University, Physics 220 1

Lecture 26

Thermodynamics I

PHYSICS 220 Final Exam

Dec. 16 10:20 AM - 12: 20 PM

Half is 3rd Test plus half is Comprehensive

2/3 questions Chapt. 11 - 16 18 Q’s

1/3 questions Chapt. 1 - 10 7 Q’s

You can bring 3 crib sheets

#2 pencil, calculator, & I.D.

In three rooms by lab

"PRACTICE EXAM FOR THE FINAL"

on the homework page.

Locations by Lab InstructorWTHR 200 - Mandal, Nirajan

Lin, Tzu-Ging

Jung, Yookyung

CL50 224 - Cervantes, Mayra

Mukundan, Vineetha

Li, Xin-A

Lilly 1105 -- Setty, Chandan

Shao, Siyi

Zhang, Shunyuan

See

Home

Page

For lab

hours

Lecture 26 Purdue University, Physics 220 4

First Law of Thermodynamics

!U = Q - W

Heat flow into system

Increase in internalenergy of system

Work done by system

Equivalent ways of writing 1st Law:

Q = !U + W

The change in internal energy of a system (!U) is

equal to the heat flow into the system (Q) minus the

work done by the system (W)

Energy Conservation

Lecture 26 Purdue University, Physics 220 5 Lecture 26 Purdue University, Physics 220 6

Signs Example

You are heating some soup in a pan on the stove.

To keep it from burning, you also stir the soup.

Apply the 1st law of thermodynamics to the soup.

What is the sign of

A) Q

B) W

C) !U

Positive, heat flows into soup

Negative, stirring does work on soup

Positive, soup gets warmer

Lecture 26 Purdue University, Physics 220 7

PV Diagram

• Ideal gas law: PV = nRT

• For n fixed, P and V determine “state” of system

T = PV/nR

U = (3/2)nRT = (3/2)PV

• Examples:

– which point has highest T?

• B

– which point has lowest U?

• C

– to change the system from C to B,

energy must be added to system

V

P

A B

C

V1 V2

P1

P3

Lecture 26 Purdue University, Physics 220 8

V

P

1 2

V1 V2

P

2 moles of monatomic ideal gas is takenfrom state 1 to state 2 at constantpressure p=1000 Pa, where V1 = 2m3

and V2 = 3m3. Find T1, T2, !U, W, Q.

1. pV1 = nRT1 " T1 = pV1/nR = 120K

2. pV2 = nRT2 " T2 = pV2/nR = 180K

3. !U = (3/2) nR !T = 1500 J

!U = (3/2) ! pV = (3/2) (3000-2000)J =1500 J

(has to be the same)

4. W = p !V = 1000 J

5. Q = !U + W = 1500 + 1000 = 2500 J

/8.31

A B

J KR N k

m= =

/8.31

A B

J KR N k

m= =

Example (Isobaric)

Lecture 26 Purdue University, Physics 220 9

Example (Isochoric)

2 moles of monatomic ideal gas is takenfrom state 1 to state 2 at constant volumeV=2m3, where T1 = 120K and T2 = 180K.Find Q.1. pV1 = nRT1 " p1 = nRT1/V = 1000 Pa

2. pV2 = nRT2 " p2 = nRT2/V = 1500 Pa

3. !U = (3/2) nR !T = 1500 J

4. W = p !V = 0 J

5. Q = !U + W = 1500 + 0 = 1500 J

V

P

2

1

V

P2

P1

Requires less heat to raise T at const. volume

than at const. pressure CV=CP-R

Why QP = nCP!T = !U + !PV = nCV!T+ nR !T

Lecture 26 Purdue University, Physics 220 10

QuestionShown 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?

A) Case 1

B) Case 2

C) Same

A

B4

2

3 9 V(m3)

Case 1

A

B

4

2

3 9 V(m3)

P(atm)

Case 2

P(atm)

Net Work = area under P-V curve

Area the same in both cases!

Lecture 26 Purdue University, Physics 220 11

QuestionShown 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?

A) Case 1

B) Case 2

C) Same

A

B4

2

3 9 V(m3)

Case 1

A

B

4

2

3 9 V(m3)

P(atm)

Case 2

P(atm)

!U = 3/2 !(PV)

Case 1: !(PV) = 4x9-2x3=30 atm*m3

Case 2: !(PV) = 2x9-4x3= 6 atm*m3

Lecture 26 Purdue University, Physics 220 12

Q = !U + W

Heat flowinto system Increase in internal

energy of system

! Which part of cycle has largest change in internal energy, !U ?

2 # 3 (since U = 3/2 PV)

! Which part of cycle involves the least work W ?

3 # 1 (since W = P!V)

! What is change in internal energy for full cycle?

!U = 0 for closed cycle (since both P & V are back where they started)

! What is net heat into system for full cycle (positive or negative)?

!U = 0 " Q = W = area of triangle (>0)

Some questions:

First Law Questions

Work done by system

V

P

1 2

3

V1 V2

P1

P3

Lecture 26 Purdue University, Physics 220 13

Special PV Cases

• Constant Pressure

– isobaric

• Constant Volume

– isochoric

• Constant Temp !U = 0

– isothermic

• Adiabatic Q=0 (no heat is transferred)

V

P

W = P!V (>0)1 2

34

!V > 0

V

P

W = P!V = 01 2

34

!V = 0

Lecture 26 Purdue University, Physics 220 14

1st Law Summary1st Law of Thermodynamics: Energy Conservation

Q = !U + W

Heat flow into system

Increase in internalenergy of system

Work done by system

V

P

! Point on P-V plot completely specifies state of system (PV = nRT)

! Work done is area under curve

! U depends only on T

(U = 3nRT/2 = 3PV/2)

! For a complete cycle !U=0 " Q=W

Lecture 26 Purdue University, Physics 220 15

Reversible?

• Most “physics” processes are reversible, you

could play movie backwards and still looks

fine. (drop ball vs throw ball up)

• Exceptions:

– Non-conservative forces (friction)

– Heat Flow

Heat never flows spontaneously from a colder body

to a hotter body.

Demo 3E - 09

Runs on heat

flow

Can be reversed

The world’s best engine?

Lecture 27 Purdue University, Physics 220 17

Gasoline engine

•Work done is area

inside closed region

W = 0 if no exhaust

Must dump some

energy

Lecture 26 Purdue University, Physics 220 19

TH

TC

QH

QC

W

HEAT ENGINEThe objective: turn heat from hotreservoir into work

The cost: “waste heat”

1st Law: QH -QC = W

Efficiency e $ W/QH

=W/QH

= (QH-QC)/QH

= 1-QC/QH

Heat Engine: Efficiency