The Laws of Thermodynamics Chapter 12. Principles of Thermodynamics Energy is conserved FIRST LAW OF THERMODYNAMICS Examples: Engines (Internal -> Mechanical)

The Laws of ThermodynamicsThe Laws of Thermodynamics

Chapter 12Chapter 12

Principles of ThermodynamicsPrinciples of Thermodynamics

• Energy is conserved• FIRST LAW OF THERMODYNAMICS• Examples: Engines (Internal ->

Mechanical) Friction (Mechanical -> Internal)

• All processes must increase entropy• SECOND LAW OF THERMODYNAMICS• Entropy is measure of disorder• Engines can not be 100% efficient

ΔU =Q − PΔV

Converting Internal Energy to Converting Internal Energy to MechanicalMechanical

Work done by expansion

W =FΔx, F =PA, Δx=ΔV / A

W =PΔV

Example 12.1Example 12.1

A cylinder of radius 5 cm is kept at pressure with a piston of mass 75 kg. a) What is the pressure inside the cylinder?

b) If the gas expands such that the cylinder rises 12.0 cm, what work was done by the gas?

c) What amount of the work went into changing the gravitational PE of the piston?

d) Where did the rest of the work go?

1.950x105 Pa

183.8 J

88.3 J

Compressing the outside air

Example 12.2aExample 12.2aA massive copper piston traps an ideal gas as shown to the right. The piston is allowed to freely slide up and down and equilibrate with the outside air.

The pressure inside thecylinder is _________ thepressure outside.

a) Greater thanb) Less thanc) Equal to

Example 12.2bExample 12.2bA massive copper piston traps an ideal gas as shown to the right. The piston is allowed to freely slide up and down and equilibrate with the outside air.

The temperature inside the cylinder is __________ the temperatureoutside.


Example 12.2cExample 12.2cA massive copper piston traps an ideal gas as shown to the right. The piston is allowed to freely slide up and down and equilibrate with the outside air.

If the gas is heated by a steadyflame, and the piston rises to a new equilibrium position, the new pressure will be _________ thanthe previous pressure.a) Greater than

b) Less thanc) Equal to

Some VocabularySome Vocabulary

• Isobaric• P = constant

• Isovolumetric• V = constant

• Isothermal• T = constant

• Adiabatic• Q = 0

V

V

V

V

P

P

P

P

A massive piston traps an amount of Helium gas as shown. The piston freely slides up and down. The system initially equilibrates atroom temperature (a)Weight is slowly added to the piston, isothermally compressing the gas to half its original volume (b)

Pb is _______ Pa

Outside Air: Room T, Atm. PExample 12.3aExample 12.3a


ΔU =Q − PΔV

W = PΔV


Tb is ________ Ta

Outside Air: Room T, Atm. PExample 12.3bExample 12.3b


ΔU =Q − PΔV

W = PΔV


Wab is ________ 0

Outside Air: Room T, Atm. P

Vocabulary: Wab is work done by gas between a and b

Example 12.3cExample 12.3c


ΔU =Q − PΔV

W = PΔV


Ub is ________ Ua

Outside Air: Room T, Atm. PExample 12.3dExample 12.3d


ΔU =Q − PΔV

W = PΔV


Qab is ________ 0

Outside Air: Room T, Atm. P

Vocabulary: Qab is heat added to gas between a and b

Example 12.3eExample 12.3e


ΔU =Q − PΔV

W = PΔV

A massive piston traps an amount of Helium gas as shown. The piston freely slides up and down. The system initially equilibrates atroom temperature (a)Weight is slowly added to the piston, adiabatically compressing the gas to half its original volume (b)

Pb is _______ Pa



ΔU =Q − PΔV

W = PΔV


Wab is ______ 0



ΔU =Q − PΔV

W = PΔV


Qab is _______ 0

Outside Air: Room T, Atm. PExample 12.4cExample 12.4c


ΔU =Q − PΔV

W = PΔV


Ub is _______ Ua



ΔU =Q − PΔV

W = PΔV


Tb is _______ Ta

Outside Air: Room T, Atm. PExample 12.4eExample 12.4e


ΔU =Q − PΔV

W = PΔV

A massive piston traps an amount of Helium gas as shown. The piston freely slides up and down. The system initially equilibrates atroom temperature (a)The gas is cooled, isobarically compressing the gas to half its original volume (b)

Pb is _______ Pa



ΔU =Q − PΔV

W = PΔV

PV =nRT


Wab is _______ 0



ΔU =Q − PΔV

W = PΔV

PV =nRT


Tb is _______ Ta

Outside Air: Room T, Atm. PExample 12.5cExample 12.5c


ΔU =Q − PΔV

W = PΔV

PV =nRT


Ub is _______ Ua



ΔU =Q − PΔV

W = PΔV

PV =nRT


Qab is _______ 0

Outside Air: Room T, Atm. PExample 12.5eExample 12.5e


ΔU =Q − PΔV

W = PΔV

PV =nRT

Work from closed cyclesWork from closed cycles

Consider cycle A -> B -> A

WA->B

-WB->A


Consider cycle A -> B -> A

WA->B->A= Area


Reverse the cycle, make it counter clockwise

-WB->A

WA->B

Example 12.6Example 12.6 a) What amount of work is performed by the gas in the cycle IAFI?

b) How much heat was inserted into the gas in the cycle IAFI?

c) What amount of work is performed by the gas in the cycle IBFI?

W=3.04x105 J

W = -3.04x105 J

Q = 3.04x105 J

V (m3)

Example 12.7Example 12.7Consider a monotonic ideal gas.

a) What work was done by the gas from A to B?

b) What heat was added to the gas between A and B?

c) What work was done by the gas from B to C?

d) What heat was added to the gas beween B and C?

e) What work was done by the gas from C to A?

f) What heat was added to the gas from C to A?

V (m3)

P (kPa)

25

50

75

0.2 0.4 0.6

A

B

C

20,000 J

20,000

-10,000 J

-25,000 J

0

15,000 J

Example ContinuedExample Continued

Take solutions from last problem and find:

a) Net work done by gas in the cycleb) Amount of heat added to gas

WAB + WBC + WCA = 10,000 JQAB + QBC + QCA = 10,000 J

This does NOT mean that the engine is 100% efficient!

Example 12.8aExample 12.8a

Consider an ideal gas undergoing the trajectory through the PV diagram.In going from A to B to C, the work done BY the gas is _______ 0.

P

V

AB

a) >

b) <

c) =

C

Example 12.8bExample 12.8b

In going from A to B to C, the change of the internal energy of the gas is _______ 0.

P

V

AB

a) >

b) <

c) =

C


In going from A to B to C, the amount of heat added to the gas is _______ 0.

P

V

AB

a) >

b) <

c) =

CD

Example 12.8dExample 12.8d

In going from A to B to C to D to A, the work done BY the gas is _______ 0.

P

V

AB

a) >

b) <

c) =

CD

Example 12.8eExample 12.8e

In going from A to B to C to D to A, the change of the internal energy of the gas is _______ 0.

P

V

AB

a) >

b) <

c) =

CD

Example 12.8fExample 12.8f

In going from A to B to C to D to A, the heat added to the gas is _______ 0.

P

V

AB

a) >

b) <

c) =

CD

EntropyEntropy• Measure of Disorder of the system

(randomness, ignorance)

• S = kBlog(N) N = # of possible arrangements for fixed E and Q

px

py

EntropyEntropy

• Total Entropy always rises! (2nd Law of Thermodynamics)

• Adding heat raises entropy

Defines temperature in Kelvin!

ΔS =Q /T

Why does Q flow from hot to cold?Why does Q flow from hot to cold?

• Consider two systems, one with TA and one with TB

• Allow Q > 0 to flow from TA to TB

• Entropy changed by:

ΔS = Q/TB - Q/TA

• If TA > TB, then ΔS > 0

• System will achieve more randomness by exchanging heat until TB = TA

ΔS =Q /T > 0

Efficiencies of EnginesEfficiencies of Engines• Consider a cycle described by:

W= work done by engineQhot= heat that flows into engine from source at Thot

Qcold= heat exhausted from engine at lower temperature, Tcold

• Efficiency is defined:

Qhot

engine

Qcold

W

Since,

=Qhot −QcoldQhot

=1−QcoldQhot

engine:

e=W

Qhot

QcoldTcold

>Qhot

Thot⇒

Qcold

Qhot>

Tcold

Thot⇒

engines :

e <1−Tcold

Thot

Carnot EnginesCarnot Engines

• Idealized engine• Most efficient possible

e=W

Qhot=1−

Tcold

Thot

Carnot CycleCarnot Cycle

Example 12.9Example 12.9An ideal engine (Carnot) is rated at 50% efficiency when it is able to exhaust heat at a temperature of 20 ºC. If the exhaust temperature is lowered to -30 ºC, what is the new efficiency.

e = 0.585

ΔS =Q /T > 0

RefrigeratorsRefrigerators

Qhot

fridge

Qcold

W

Given: Refrigerated region is at Tcold

Heat exhausted to region with Thot

Find: Efficiency

Since ,

Note: Highest efficiency for small T differences

refrigerator:

e=Qcold

W =Qcold

Qhot −Qcold=

1

Qhot /Qcold −1

QhotThot

>Qcold

Tcold⇒

Qhot

Qcold>

Thot

Tcold⇒

refrigerator:

e <1

Thot /Tcold −1

Heat PumpsHeat Pumps

Qhot

heatpump

Qcold

W

Given: Inside is at Thot

Outside is at Tcold

Find: Efficiency

Since ,

Like Refrigerator: Highest efficiency for small ΔT

heat pump:

e=Qhot

W=

QhotQhot −Qcold

=1

1−Qcold /Qhot

QhotThot

>Qcold

Tcold⇒

Qhot

Qcold>

Thot

Tcold⇒

heat pump:

e <1

1−Tcold /Thot

ΔS =Q /T > 0

Example 12.10Example 12.10

A modern gas furnace can work at practically 100% efficiency, i.e., 100% of the heat from burning the gas is converted into heat for the home. Assume that a heat pump works at 50% of the efficiency of an ideal heat pump.

If electricity costs 3 times as much per kw-hr as gas, for what range of outside temperatures is it advantageous to use a heat pump?Assume Tinside = 295 ºK.

T =29556

=245.8°K=-27°C

Example 12.11aExample 12.11a

An engine does an amount of work W, and exhausts heat at a temperature of 50 degrees C. The chemical energy contained in the fuel must be greater than, and not equal to, W.

a) Trueb) False

Example 12.11bExample 12.11b

A locomotive is powered by a large engine that exhausts heat into a large heat exchanger that stays close to the temperature of the atmosphere. The engine should be more efficient on a very cold day than on a warm day.

a) Trueb) False


An air conditioner uses an amount of electrical energy U to cool a home. The amount of heat removed from the home must be less than or equal to U.

a) Trueb) False

Example 12.11dExample 12.11d

A heat pump uses an amount of electrical energy U to heat a home. The amount of heat added to a home must be less than or equal to U.

a) Trueb) False

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The Laws of Thermodynamics Chapter 12. Principles of Thermodynamics Energy is conserved FIRST LAW OF THERMODYNAMICS Examples: Engines (Internal -> Mechanical)