ub, phy101: chapter 15, pg 1 physics 101: chapter 15 thermodynamics, part i l textbook sections 15.1...

UB, Phy101: Chapter 15, Pg 1

Physics 101: Physics 101: Chapter 15 Chapter 15Thermodynamics, Part IThermodynamics, Part I

Textbook Sections 15.1 – 15.5


The Ideal Gas LawThe Ideal Gas Law(review)(review)

pV = NkBTN = number of molecules

» N = number of moles (n) x NA molecules/molekB = Boltzmann’s constant = 1.38 x 10-23 J/K

pV = nRTR = ideal gas constant = NAkB = 8.31 J/mol/K

= .0823 atm-l/mol/K


Summary of Kinetic Theory:Summary of Kinetic Theory:The relationship between energy and The relationship between energy and

temperaturetemperature(for monatomic ideal gas)(for monatomic ideal gas)

Tk2

3 vm

2

1 eKE/molecul ave B

2

3RT

M

T3k v v B2

rms

Internal Energy U

= number of molecules x ave KE/molecule

= N (3/2)kBT

= (3/2)nRT = (3/2)pV

Careful with units:R = 8.31 J/mol/Km = molar mass in kgvrms = speed in m/s


Thermodynamic Systems and p-V Thermodynamic Systems and p-V DiagramsDiagrams

ideal gas law: pV = nRT for n fixed, p and V determine “state” of system

T = pV/nRU = (3/2)nRT = (3/2)pV

examples: which point has highest T?

» 2 which point has lowest U?

» 3 to change the system from 3 to 2,

energy must be added to systemV

P1 2

3

V1 V2

P1

P3


Work Done by a SystemWork Done by a System

W

W y

W = p V W > 0 if V > 0

expanding system does positive work

W < 0 if V < 0

contracting system does negative work

W = 0 if V = 0

system with constant volume does no work


Problem: Thermo IProblem: Thermo I

V

P 1 2

34

V

PW = PV (>0)1 2

34

V > 0 V

PW = PV = 01 2

34

V = 0

V

PW = PV (<0)1 2

34

V < 0 V

W = PV = 01 2

34

P

V = 0 V

P 1 2

34

If we gothe other way thenWtot < 0

V

P 1 2

34

Wtot > 0

Wtot = ??


V

P

1 2

3

Now try this:

V1 V2

P1

P3

Area = (V2-V1)x(P1-P3)/2


First Law of ThermodynamicsFirst Law of ThermodynamicsEnergy ConservationEnergy Conservation

U = Q - W

Heat flow into system

Increase in internalenergy of system

Work done by system

V

P

1 2

3

V1 V2

P1

P3

Equivalent ways of writing 1st Law:

Q = U + W

Q = U + p V


First Law of ThermodynamicsFirst Law of ThermodynamicsExampleExample

V

P

1 2

V1 V2

P2 moles of monatomic ideal gas is takenfrom state 1 to state 2 at constant pressure

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) p V = 1500 J (has to be the same)

4. W = p V = 1000 J 5. Q = U + W = 1500 + 1000 = 2500 J


First Law of ThermodynamicsFirst Law of ThermodynamicsExampleExample

2 moles of monatomic ideal gas is takenfrom state 1 to state 2 at constant volume

V=2m3, where T1=120K and T2 =180K. Find

Q.

1. pV1 = nRT1 p1 = nRT1/V = 1000 Pa

2. pV2 = nRT1 p2 = nRT2/V = 1500 Pa

3. U = (3/2) nR T = 1500 J 4. W = p V = 0 J 5. Q = U + W = 0 + 1500 = 1500 J

requires less heat to raise T at const. volume than at const. pressure

V

P

2

1

V

P2

P1


Preflight Preflight

Shown 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?1. Case 1 2. Case 2 3. Same

A

B4

2

3 9 V(m3)

Case 1

A

B

4

2

3 9 V(m3)

P(atm)

Case 2

P(atm)

correct

Net Work = area under P-V curve

Area the same in both cases!


Preflight Preflight

Shown 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?1. Case 1 2. Case 2 3. Same

A

B4

2

3 9 V(m3)

Case 1

A

B

4

2

3 9 V(m3)

P(atm)

Case 2

P(atm)correct

U = 3/2 (pV)

Case 1: (pV) = 4x9-2x3=30 atm-m3

Case 2: (pV) = 2x9-4x3= 6 atm-m3


Preflight (followup)Preflight (followup)

Shown 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 heat added to the system the biggest?1. Case 1 2. Case 2 3. Same

A

B4

2

3 9 V(m3)

Case 1

A

B

4

2

3 9 V(m3)

P(atm)

Case 2

P(atm)correct

Q = U + W

W is same for both

U is larger for Case 1

Therefore, Q is larger for Case 1


Q = U + W



Work done by system

V

P

1 2

3

V1 V2

P1

P3

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 = pV)

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?U = 0 Q = W = area of triangle (>0)

Some questions:

DEMO


Recap:Recap:

1st Law of Thermodynamics: Energy Conservation

Q = U + W



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


Thermodynamics, part 2Thermodynamics, part 2

engines and refrigerators 2nd law of thermodynamics


TH

TC

QH

QC

W

HEAT ENGINE

TH

TC

QH

QC

W

REFRIGERATOR

system

Engines and Refrigerators

system taken in closed cycle Usystem = 0 therefore, net heat absorbed = work done

QH - QC = W (engine)

QC - QH = -W (refrigerator)

energy into green blob = energy leaving green blob


TH

TC

QH

QC

W

HEAT ENGINE

The objective: turn heat from hot reservoir into work

The cost: “waste heat”

1st Law: QH -QC = W

efficiency e W/QH

=W/QH

= (QH-QC)/QH

= 1-QC/QH


TH

TC

QH

QC

W

REFRIGERATOR

The objective: remove heat from cold reservoir

The cost: work

1st Law: QH = W + QC

coeff of performance CPr QC/W

= QC/W

= QC/(QH - QC)


Preflights Preflights Consider a hypothetical device that takes 1000 J of heat from a hot reservoir at 300K, ejects 200 J of heat to a cold reservoir at 100K, and produces 800 J of work.Does this device violate the first law of thermodynamics ? 1. Yes 2. No

This device doesn't violate the first law of thermodynamics because no energy is being created nor destroyed. All the energy is conserved.

correct

W (800) = Qhot (1000) - Qcold (200) Efficiency = W/Qhot = 800/1000 = 80%

80% efficient


New concept: Entropy (S)New concept: Entropy (S)

A measure of “disorder”

A property of a system (just like p, V, T, U)

related to number of different “states” of system

Examples of increasing entropy:

ice cube melts

gases expand into vacuum

Change in entropy:

S = Q/T» >0 if heat flows into system (Q>0)

» <0 if heat flows out of system (Q<0)


Second Law of ThermodynamicsSecond Law of Thermodynamics

The entropy change (Q/T) of the system+environment 0

never < 0

order to disorder

Consequences

A “disordered” state cannot spontaneously transform into an “ordered” state

No engine operating between two reservoirs can be more efficient than one that produces 0 change in

entropy. This is called a “Carnot engine”


TH

TC

QH

QC

W

HEAT ENGINE

The objective: turn heat from hot reservoir into work

The cost: “waste heat”

1st Law: QH -QC = W

efficiency e W/QH =W/QH = 1-QC/QH

S = QC/TC - QH/TH 0

S = 0 for CarnotTherefore, QC/QH TC/ TH

QC/QH = TC/ TH for Carnot

Therefore e = 1 - QC/QH 1 - TC/ TH

e = 1 - TC/ TH for Carnot

e = 1 is forbidden!e largest if TC << TH

Engines and the 2nd Law


Preflight Preflight

Consider a hypothetical device that takes 1000 J of heat from a hot reservoir at 300K, ejects 200 J of heat to a cold reservoir at 100K, and produces 800 J of work.Does this device violate the second law of thermodynamics ? 1. Yes 2. No

.

correct

W (800) = Qhot (1000) - Qcold (200) Efficiency = W/Qhot = 800/1000 = 80% Max eff = 1 - 100/300 = 67%


Second Law of ThermodynamicsSecond Law of Thermodynamics

The entropy change (Q/T) of the system+environment 0

never < 0

order to disorder

Consequences

A “disordered” state cannot spontaneously transform into an “ordered” state

No engine operating between two reservoirs is more efficient than one that produces 0 change in entropy

(“Carnot engine”)Heat cannot be transferred spontaneously from cold to

hot


PreflightPreflight

Consider a hypothetical refrigerator that takes 1000 J of heat from a cold reservoir at 100K and ejects 1200 J of heat to a hot reservoir at 300K.

1. How much work does the refrigerator do?2. What happens to the entropy of the universe?3. Does this violate the 2nd law of thermodynamics?

Answers:200 JDecreasesyes

TH

TC

QH

QC

W

QC = 1000 J

QH = 1200 JSince QC + W = QH, W = 200 J

SH = QH/TH = (1200 J) / (300 K) = 4 J/K

SC = -QC/TC = (-1000 J) / (100 K) = -10 J/K

STOTAL = SH + SC = -6 J/K decreases (violates 2nd law)

ub, phy101: chapter 15, pg 1 physics 101: chapter 15 thermodynamics, part i l textbook sections 15.1...

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