05 Thermodynamics.pptx

THERMODYNAMICS

Thermodynamics is the study of energy relationships that involve heat,

mechanical work, and other aspects of

energy and heat transfer.

Objectives: After finishing this

unit, you should be able to:

second laws of thermodynamics. • Demonstrate your understanding

of adiabatic, isochoric, isothermal, and isobaric processes.

• Write and apply a relationship for determining the ideal efficiency of a heat engine.

• Write and apply a relationship for determining

A 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.)

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.

Usually the internal energy consists of the sum of the potential and

TWO WAYS TO

INCREASE

THE INTERNAL

ENERGY,



U.

HEAT PUT INTO

A SYSTEM

(Positive)

+



U

WORK DONE

ON A GAS

WORK DONE BY EXPANDING GAS:

W is negative

WORK DONE BY EXPANDING GAS:

W is negative

-U Decrease

-U Decrease

TWO WAYS TO

DECREASE

THE

INTERNAL ENERGY,



U.

HEAT LEAVES A SYSTEM

Q is negative

Q

hot

W

W

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

THERMODYNAMIC PROCESS

Increase in Internal Energy,



U.

Initial State:

P

V

T

n

Final State:

P

V

T

n

Heat input

Q

W

The Reverse Process

Decrease in Internal Energy,



U.

Initial State:

P

V

T

n

Final State:

P

V

T

n

Work on gas

THE FIRST LAW OF

THERMODYAMICS:

Energy can neither be created nor destroyed, it can only be transformed from one form to another.

How do you transform energy from one form to another?

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.



U = Q + W



final - initial)

SIGN CONVENTIONS

FOR FIRST LAW

• Heat Q input is positive



U =Q + W



final - initial)

• Heat OUT is negative

•

Work BY a gas is negative

•

Work ON a gas is positive

+Q

-W



U

+W

-Q

APPLICATION OF FIRST

LAW OF THERMODYNAMICS

Example 1: In the figure, the gas absorbs 400 J of heat and at the same time does 120 J of work on the piston. What is the change in internal

energy of the system?



U =Q + W

Apply First Law:

Q

400 J

Example 1 (Cont.):

Apply First Law







U = +280 J

Q

400 J

W

=120 J

U = Q + W

= (+400 J) + (-120 J) = +280 J

W is Negative: -120 J (Work OUT)

U = Q + W

Example 1 (Cont.):

Apply First Law







U = +280 J

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

Q

400 J

W

=120 J

The

increase

in

internal energy is:

Work done by expansion.

V_{0 Vf}

d

Work = force ∙ distance force = Pressure ∙ Area Work = Pressure ∙ Area ∙ distance

Work = Pressure ∙ ΔVolume

ΔVolume = V_f - V_o

Work =

-

ΔU = Q + W

V_f



U = Q + W But W = -P



V

ISOBARIC PROCESS:

CONSTANT PRESSURE,



P = 0

+U -U

QIN= 200 J Q

OUT= 125 J

Work

=

150 J

Work

ISOBARIC EXAMPLE (

Constant Pressure

):

Heat input

increases

V

with const.

P

400 J heat does 120 J of

work, increasing the

internal energy by 280 J.

B

A

P

V

V

ISOBARIC WORK

400 J

Work = Area under PV curve

B

A

P

V

V

V

V

T

=

T

P_A = P_B

Pressure

Volume

FOUR THERMODYNAMIC

PROCESSES:

• Isochoric Process:



V = 0, W

= 0

• Isobaric Process:



P = 0

• Isothermal Process:



T = 0,



U

= 0

• Adiabatic Process:



Q = 0

• Isochoric Process:



V = 0, W

= 0

• Isobaric Process:



P = 0

• Isothermal Process:



T = 0,



U

= 0

• Adiabatic Process:





Q = 0

U = Q + W

Pr

es

su

re

Volume

Is

oc

ho

ric

In a thermodynamic system, the two main

processes involved are adiabatic or isothermal. It is regarded as the former when the

transformation (fluctuations or variations in

temperature) are fast enough that no heat was significantly transferred between the outside environment and the system. When the

transformation is very slow in that same system then the process is isothermal because the



U = Q + W so that



U = Q

ISOCHORIC PROCESS:

CONSTANT VOLUME,



V = 0,



W = 0

0

+U -U

QIN QOUT

HEAT IN = INCREASE IN INTERNAL ENERGY

HEAT OUT = DECREASE IN INTERNAL ENERGY

ISOCHORIC EXAMPLE:

Heat input

increases P

with const. V

400 J heat input increases

internal energy by 400 J

and zero work is done.

B

A

P

V

= V

P

P

P

T

=

T

400 J

No Change in volume:

P_A V_A

n_AT_A

P_B V_B

ISOTHERMAL PROCESS:

CONST. TEMPERATURE,



T = 0,



U = 0

NET HEAT INPUT = WORK OUTPUT



U = Q + W AND



Q = - W

U = 0 U = 0

Q

Work

In

Work Out

Q

ISOTHERMAL EXAMPLE

(Constant T):

P

V

=

P

V

Slow compression at

constant temperature:

---

No change in U

.



U =



T

= 0

B

A

P

V

V

ISOTHERMAL EXPANSION (

Constant T)

:

400 J of energy is absorbed

by gas as 400 J of work is

done on gas.

T = U = 0



U

=



T

= 0

B

A

P

V

V

P

P

V

= P

V

T

= T

U = Q + W ; W = U or U = W

ADIABATIC PROCESS:

NO HEAT EXCHANGE, Q = 0

Work done at EXPENSE of internal energy

INPUT of Work INCREASES internal energy

Work Out

Work

In

U +U

ADIABATIC EXAMPLE:

Insulated

Walls:

Q = 0

B

A

P

V

V

P

Expanding gas does

work with zero heat

ADIABATIC EXPANSION:

400 J of WORK is done,

DECREASING the internal

energy by 400 J: Net heat

exchange is ZERO.

Q = 0



Q = 0

B

A

P

V

V

P

P

V

P

V

T

=

T

A A B B

Pr

es

su

re

Volume

Is

oc

ho

ric

HEAT ENGINES

• Absorbs heat

Q

• Performs work

W

• Rejects heat

Q

A heat engine is any

device which through

a cyclic process:

Cold Res. T

Engine

Hot Res. T

Q

W

out

THE SECOND LAW OF

THERMODYNAMICS

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.

Not only can you not win (1st law);

(can’t get out more energy than you put in)

You can’t even break even (2nd law)!

(There will always be wasted energy)

W

Cold Res. T

Engine

Hot Res. T

Q

THE SECOND LAW OF

THERMODYNAMICS

Cold Res. T_C Engine

Hot Res. T_H

400 J

300 J

100 J

• A possible engine.

• An IMPOSSIBLE

engine.

Cold Res. T_C Engine

Hot Res. T_H

EFFICIENCY OF AN ENGINE

Cold Res. T_C Engine

Hot Res. T_H

Q

W

Q

The efficiency of a heat engine

is the ratio of the net work

done W to the heat input Q

.

e = 1 -

Q

Q

e = =

W

Q

Q

- Q

EFFICIENCY EXAMPLE

Cold Res. T_C Engine Hot Res. T_H

800 J

W

600 J

An engine absorbs 800 J and

wastes 600 J every cycle. What

is the efficiency?

e = 1 -

600 J

800 J

e = 1 -

Q

Q

e = 25%

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.

e = 1 -

T

T

e =

T

- T

T

Cold Res. T_C Engine Hot Res. T_H

Q

W

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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?

e = 1 -

T

T

e = 1 -

300 K

500 K

e = 40%

Actual e = 0.5e

= 20%

e =

W

Q

Cold Res. T_C Engine Hot Res. T_H

Q

W

Q

Cold Res. T_C Engine Hot Res. T_H

Q_{H W}

Q_W

Cold Res. T_C Engine

WARM Res. TW Q_{W W}

Q_C

e = 1 -

T

T

Heat flows from

HOT to COLD, not

REFRIGERATORS

A refrigerator is an engine

operating in reverse:

Work is done on gas

extracting heat from cold

reservoir and depositing

heat into hot reservoir.

W

+ Q

= Q

W

= Q

- Q

Cold Res. T

Engine

Hot Res. T

Q

Q

THE SECOND LAW FOR

REFRIGERATORS

It 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!

Cold Res. T_C Engine

Hot Res. T_H

Q

Summary



Q =



U +



W



final - initial)

The

First 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.

• Isochoric Process:



V = 0,



W = 0

• Isobaric Process:



P = 0