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Thermodynamics is the study of energy relationships that involve heat, mechanical work, and other aspects of energy and heat transfer.

Central Heating

THERMODYNAMICSA THERMODYNAMIC SYSTEM

- A system is a closed environment in which heat transfer can take place. (For example, the gas, walls, and cylinder of an automobile engine.)

Work done on gas or work done by gas

Usually the internal energy consists of the sum of the potential and kinetic energies of the working gas molecules.INTERNAL ENERGY OF SYSTEM

- The internal energy U of a system is the total of all kinds of energy possessed by the particles that make up the system.

+U

WORK DONE ON A GAS (Positive)

TWO WAYS TO INCREASE THE INTERNAL ENERGY, U.HEAT PUT INTO A SYSTEM (Positive)

Wout

Qout

hot

hot

HEAT LEAVES A SYSTEM

Q is negative

TWO WAYS TO DECREASE THE INTERNAL ENERGY, U.-U

Decrease

WORK DONE BY EXPANDING GAS: W is positive

THERMODYNAMIC STATE

The STATE of a thermodynamic system is determined by four factors:

- Absolute Pressure P in Pascals
- Temperature T in Kelvins
- Volume V in cubic meters

- Number of moles, n, of working gas

THE FIRST LAW OF THERMODYAMICS:

- The net heat put into a system is equal to the change in internal energy of the system plus the work done BY the system.

Q = U + W final - initial)

- Conversely, the work done ON a system is equal to the change in internal energy plus the heat lost in the process.

Wout =120 J

Qin

400 J

Apply First Law:

Q = U + W

APPLICATION OF FIRST LAW OF THERMODYNAMICSExample 1:In the figure, the gas absorbs400 Jof heat and at the same time does120 Jof work on the piston. What is the change in internal energy of the system?

Wout =120 J

Qin

400 J

U = +280 J

Example 1 (Cont.): Apply First LawDQ is positive: +400 J (Heat IN)

DW is positive: +120 J (Work OUT)

Q = U + W

U = Q - W

U = Q - W

= (+400 J) - (+120 J)

= +280 J

Wout =120 J

Qin

400 J

U = +280 J

Example 1 (Cont.): Apply First LawEnergy is conserved:

The 400 J of input thermal energy is used to perform 120 J of external work, increasing the internal energy of the system by 280 J

The increase in internal energy is:

Q = U + WFOUR THERMODYNAMIC PROCESSES:

- Isochoric Process: V = 0, W = 0
- Isobaric Process: P = 0
- Isothermal Process: T = 0, U = 0
- Adiabatic Process: Q = 0

0

QIN

QOUT

-U

+U

HEAT IN = INCREASE IN INTERNAL ENERGY

HEAT OUT = DECREASE IN INTERNAL ENERGY

ISOCHORIC PROCESS: CONSTANT VOLUME, V = 0, W = 0Q = U + W so that Q = U

No Work Done

No Change in volume:

P2

B

PA P B

=

TA T B

P1

A

V1= V2

400 J

ISOCHORIC EXAMPLE:400 J heat input increases internal energy by 400 J and zero work is done.

Heat input increases P with const. V

QIN

QOUT

Work Out

Work In

HEAT IN = Wout + INCREASE IN INTERNAL ENERGY

ISOBARIC PROCESS: CONSTANT PRESSURE, P = 0Q = U + W But W = P V

-U

+U

HEAT OUT = Win + DECREASE IN INTERNAL ENERGY

B

A

VA VB

P

=

TA T B

V1 V2

400 J

ISOBARIC EXAMPLE (Constant Pressure):400 Jheat does 120 J of work, increasing the internal energy by 280 J.

Heat input increases Vwith const. P

Q = U + W AND Q = W

QIN

QOUT

Work Out

Work In

NET HEAT INPUT = WORK OUTPUT

ISOTHERMAL PROCESS: CONST. TEMPERATURE, T = 0, U = 0U = 0

U = 0

WORK INPUT = NET HEAT OUT

A

PA

B

PB

V2 V1

PAVA =PBVB

ISOTHERMAL EXAMPLE (Constant T):U = T = 0

Slow compression at constant temperature: ----- No change in U.

U = -W

W = -U

Work Out

Work In

U

+U

Q = 0

Work done at EXPENSE of internal energyINPUT Work INCREASES internal energy

ADIABATIC PROCESS: NO HEAT EXCHANGE, Q = 0Q = U + W ; W = -U or U = -W

A

PA

B

PB

V1 V2

ADIABATIC EXAMPLE:Expanding gas does work with zero heat loss. Work = -DU

Insulated Walls: Q = 0

AB: Heated at constant V to 400 K.Example Problem:

A 2-L sample of Oxygen gas has an initial temp-erature and pressure of 200 K and 1 atm. The gas undergoes four processes:

- BC: Heated at constant P to 800 K.
- CD: Cooled at constant V back to 1 atm.
- DA: Cooled at constant P back to 200 K.

B

400 K

800 K

PB

A

200 K

1 atm

2 L

PV-DIAGRAM FOR PROBLEMHow many moles of O2 are present?

Consider point A: PV = nRT

B

400 K

800 K

PB

A

200 K

1 atm

PA P B

=

2 L

TA T B

1 atmP B

P B = 2 atm

=

200 K400 K

or 203 kPa

PROCESS AB: ISOCHORICWhat is the pressure at point B?

B

400 K

800 K

PB

C

200 K

VB V C

1 atm

D

=

TB T C

2 L

4 L

2 LV C

V C = V D = 4 L

=

400 K800 K

PROCESS BC: ISOBARICWhat is the volume at point C (& D)?

B

400 K

800 K

2 atm

C

200 K

1 atm

2 L

4 L

W = +400 J

FINDING W FOR PROCESS BC.Work depends on change in V.

P = 0

Work = PV

W = (2 atm)(4 L - 2 L) = 4 atm L = 400 J

B

400 K

800 K

PB

C

A

200 K

D

1 atm

PC P D

=

2 L

TC T D

2 atm1 atm

T D = 400 K

=

800 KTD

PROCESS CD: ISOCHORICWhat is temperature at point D?

400 K

800 K

2 atm

A

200 K

400 K

1 atm

D

2 L

4 L

W = -200 J

FINDING W FOR PROCESS DA.Work depends on change inV.

P = 0

Work = PV

W= (1 atm)(2 L - 4 L) = -2 atm L = -200 J

+404 J

B

C

B

C

-202 J

2 atm

2 atm

Neg

1 atm

1 atm

2 L

4 L

2 L

4 L

B

C

2 atm

Area = (1 atm)(2 L)

Net Work = 2 atm L = 200 J

1 atm

2 L

4 L

NET WORK FOR COMPLETE CYCLE IS ENCLOSED AREAHot Res. TH

Qhot

Wout

Engine

Qcold

Cold Res. TC

HEAT ENGINESA heat engine is any device which through a cyclic process:

- Absorbs heat Qhot
- Performs work Wout
- Rejects heat Qcold

Hot Res. TH

It is impossible to construct an engine that, operating in a cycle, produces no effect other than the extraction of heat from a reservoir and the performance of an equivalent amount of work.

Qhot

Wout

Engine

Qcold

Cold Res. TC

THE SECOND LAW OF THERMODYNAMICSHot Res. TH

Hot Res. TH

400 J

400 J

100 J

400 J

Engine

Engine

300 J

Cold Res. TC

Cold Res. TC

- An IMPOSSIBLE engine.

- A possible engine.

Hot Res. TH

QH

W

W

QH

QH- QC

QH

Engine

e = =

QC

Cold Res. TC

QC

QH

e = 1 -

EFFICIENCY OF AN ENGINEThe efficiency of a heat engine is the ratio of the net work done W to the heat input QH.

Hot Res. TH

800 J

W

QC

QH

Engine

e = 1 -

600 J

600 J

800 J

Cold Res. TC

e = 1 -

e = 25%

EFFICIENCY EXAMPLEAn engine absorbs 800 J and wastes 600 J every cycle. What is the efficiency?

Question: How many joules of work is done?

Hot Res. TH

QH

W

Engine

TH- TC

TH

QC

e =

Cold Res. TC

TC

TH

e = 1 -

EFFICIENCY OF AN IDEAL ENGINE (Carnot Engine)For a perfect engine, the quantities Q of heat gained and lost are proportional to the absolute temperatures T.

TC

TH

e = 1 -

W

QH

e =

300 K

500 K

e = 1 -

Work = 120 J

Example 3:A steam engine absorbs 600 J of heat at 500 K and the exhaust temperature is 300 K. If the actual efficiency is only half of the ideal efficiency, how much work is done during each cycle?

Actual e = 0.5ei = 20%

W = eQH = 0.20 (600 J)

e = 40%

Hot Res. TH

Qhot

Win

Engine

Qcold

Cold Res. TC

WIN = Qhot - Qcold

REFRIGERATORSA refrigerator is an engine operating in reverse: Work is done on gas extracting heat from cold reservoir and depositing heat intohot reservoir.

Win + Qcold = Qhot

Hot Res. TH

Qhot

Engine

Qcold

Cold Res. TC

THE SECOND LAW FOR REFRIGERATORSIt is impossible to construct a refrigerator that absorbs heat from a cold reservoir and deposits equal heat to a hot reservoir with W = 0.

If this were possible, we could establish perpetual motion!

TheFirst Law of Thermodynamics:The net heat taken in by a system is equal to the sum of the change in internal energy and the work done by the system.

Q = U + W final - initial)

Summary- Isochoric Process: V = 0, W = 0
- Isobaric Process: P = 0
- Isothermal Process: T = 0, U = 0
- Adiabatic Process: Q = 0

TheSecond Law of Thermo:It is impossible to construct an engine that, operating in a cycle, produces no effect other than the extraction of heat from a reservoir and the performance of an equivalent amount of work.

Hot Res. TH

Qhot

Wout

Engine

Qcold

Cold Res. TC

Summary (Cont.)
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