Chap 4-W6 Thermodynamics Cheong.pptx

8/10/2019 Chap 4-W6 Thermodynamics Cheong.pptx

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WEEK 6

THERMODYNAMICS

PHYSICS 3

FIS 0334


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Chapter 1

Thermodynamic system and its surrounding

Thermodynamic process

The first law of thermodynamics

Work done during volume changes P-V diagram

Kinds of thermodynamic process

Adiabatic, Isochoric, Isobaric and Isothermal


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Learning Outcomes

By the end of this chapter, students will be able to do the following:

Describe thermodynamics process by involving the change of state

of a thermodynamic system.

Use the first law of thermodynamics to relate heat transfer, work

done and internal energy change of a thermodynamics system withits surroundings.

Calculate the work done by a thermodynamics system when its

volume changes.

Distinguish among adiabatic, isochoric, isobaric and isothermal

processes. Sketch and analyze the P-V diagram for a thermodynamics process.


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IntroductionExample 1

Thermodynamics of Popping Popcorn

A quantity of popcorn kernels in a

pot with a lid is initially placed on a

stove.

Heat energy is added to the

popcorn by conduction.

As the popcorn pops and expands,

it does work on its surroundings

(by exerting an upward force on thelid and moves through a

displacement).

FsW


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Thermodynamics of Popping Popcorn

The stateof popcorn changes in

this process, since the volume,temperature and pressure of

popcorn all change as it pops.

Such process involves the change

in state of a thermodynamic

system (Popcorn)is calledthermodynamic process.


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Thermodynamics of Gasoline Car Engine

If you put a tiny amount of high energy

fuel (i.e. gasoline) in a small and

enclosed space and ignite it.

Turn to create an incredible amount of

energy released in the form of

expanding gas (heated gas).

The heated gas pushes on the piston

within cylinder, doing mechanical work

that used to propel the car in distance.

Here, the thermodynamic system is the

quantity of gas enclosed in a cylinder

with a movable piston.


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Principle behind Gasoline Car Engine

Four-Stroke Combustion Cycle (Intake, Compression, Combustion and Exhaust stroke)

* Get the online animation to enhance your understanding


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Step 1: Intake Stroke

The piston starts at top, when the intake valve open, the piston moves down to let the

engine take in a cylinder-full of air and gasoline.

Step 2: Compression Stroke

The piston moves back up to compress the mixture of air/gasoline mixture. As

compression makes the explosion more powerful.

Step 3: Combustion Stroke

When the piston reaches top, the spark plug emit a spark to ignite the gasoline. The

gasoline in the cylinder explodes, driving the piston down.

Step 4: Exhaust Stroke

Once the piston hits the bottom surface, exhaust valve opens and the exhaust leaves

the cylinder.


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Introduction

THERMODYNAMICS has its roots in many practical problems

essentially in the gasoline engine in an automobile which use the

heat of combustion of their fuel to perform mechanical work in

propelling the vehicles.

Thus, we able to describe the energy relationships in any

thermodynamic processin term of the quantity of heat Qadded to

the systemand the work Wdone by the system which

consequently resulting the change of internal energy Uof system.

Internal Energy, U is the total energy of all of molecules in the

system ( due to their individual random motions) excluding for the

kinetic and potential energy due to external force.


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The change in internal energy of a closed system will be equal

to the heat energy added to the system minus the work done by

the system on its surroundings.

** Be careful with the sign of heat energy, Q and work done, W

The First Law of Thermodynamics

WQU


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The First Law of Thermodynamics

WQU

Quantity Sign Situation

U+ Increasein internal energy

- Decreasein internal energy

Q+ Heat flows in tothe system

- Heat flows out of the system

W+ Work done bythe system (Expanding gas)

- Work done onthe system (Compression gas)


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The First Law of ThermodynamicsExample 1

Q = 150J W = 100J

U = Q W = + 50J

More heat is added to

system than systemdoes work.

Internal energy of

system increases.

system

surroundings

system

surroundings

Q = -150J W = -100J

U = Q W = - 50J

More heat flows out of

system than work isdone on system.

Internal energy of

system decreases.

U = Q W = 0J

Heat added to system

equals work done bysystem.

Internal energy of

system unchanged.

system

surroundings

Q = 150J W = 150J


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The First Law of ThermodynamicsExample 2

Your body is a thermodynamic system

When you exercise, your body does work. W > 0

(Since the work is done by your body)

Your body also warms up during exercise; and by perspiration and

other means the body releases the heat. Q < 0 Since Q is negative and W is positive,

The bodys internal energy decreases.

Thats why exercise helps you to lose weight since some of the

internal energy stored in your body in form of fat is used up.

0 WQU


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Question 1

Rank the following thermodynamic processes according to the

change in internal energy in each process, from most positive to

most negative.

(i) As you do 250J of work on a system, it transfers 250J of heat to

its surroundings;

(ii) as you do 250J of work on a system, it absorbs 250J of heat from

its surroundings;

(iii) as a system does 250J of work on you, it transfers 250J of heat

to its surroundings;

(iv) as a system does 250J of work on you, it absorbs 250J of heatfrom its surroundings.


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Work Done in Volume Changes

Motion of piston Figure shows a system whose volume canchange in a cylinder with a movable piston.

The cylinder has cross-sectional areaA

and the pressure exerted by system at

piston is P. The total force exerted bysystem on piston is F = PA.

When the piston moves out an infinitesimal

distance , the work done by this force isd

PdVdW

PAdFddW


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P-V diagram (Pressure vs. volume)

If the gas expands / compressed from volume VAto VB, the work

done by the gas is equal the area under the curve;

Work doneby gas

Work done

on gas

VB

PB

PA

VA A

B

V

V

V

V

V

VnRTW

V

dVnRT

PdVW

B

A

B

A

ln

B

A

V

V

PdVAreaWork 0 B

A

V

V

PdVAreaWork 0

A

B


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P-V diagram (Pressure vs. volume)

If the volume is constant (does not change), VA= VB, hence no work

is done, W = 0.

For a gas that expands / compressed with constant pressure,

Work done

by gasWork done

on gas

)(AB

VVPVPW


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Last Updated:5 November 2014 LMS SEGi education group 18

A process in which the temperatureof the system remains constantis

called an isothermal process. On a p-V diagram, a path representing a

constant temperature process is called an isotherm.

Internal energy U depends only on temperature, so if the temperature is

constant, the internal energy is also constant;

According to Boyles law , PV = constant.

If the gas is expanded, heat flows intothe gas from the water.

If the gas is compressed, heat flows outof the gas into the water.

The First Law of Thermodynamics Applied:

Isothermal process (T=0)

QW

U

0


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Isothermal process (T=0)

PV diagram for an ideal gasundergoing isothermal

processes at two different

temperatures.

The ideal gas in the cylinder

is expanding isothermally at

temperature T.

Isotherms


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Adiabatic process (Q=0)

Defined as one with no heat transferred into or out of a system.

System expands adiabat ical ly, W > 0, U < 0.

System is com pressed adiabat ical ly, W < 0 , U > 0.

To do work , the gas uses up some of its internal energy as heat

is not allow to enter the gas. When its internal energy decrease,

its temperature decreases too.

WU

Q

0


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Adiabatic process (Q=0)


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Isochoric process (V=0)

QU

WQU

W

0


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Isobaric process (P=0)

An isobaric process occurs at constant pressure.

The pressure P experience by the substance is always the same

and is determined by the external atmosphere and the weight of

the piston and the block resting on it.

Heating the substance makes it expands and do work W in lifting

the piston and block through the displacement s.

The expression of the work at constant pressure;

)( if VVPVPW


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Isobaric process (P=0)

Positive W value for the work done by

the gas when it expands.Negative W value for the work done

on the gas to compress it.


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Describe the changes when an ideal gas changes from state WX Y Z

and back to W.

The First Law of Thermodynamics Applied

Net change in heat

Qnet= Q WX+ Q XY- Q YZ- Q ZW

Net change in work

Wnet= W WX- W YZ

Net change in internal energy

Unet

=0

(For a cyclic process, that eventually return a system

to its initial state, Uf= Ui; Unet=0)

Qnet= Wnet= shaded area


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Change of

state

Equation Explanation

WX U= Q - W Expansion at constant pressure, temperature increases.

Work done, heat absorbed.

XY U= Q Pressure increase at constant volume. Temperature increase.

Hence U>0 , Q>0. Heat absorbed.

YZ U= Q - W Compression at constant pressure, temperature decreases.

Work done on the gas, heat lost from the gas.

ZW U= Q Pressure decrease at constant volume, temperature decreases.

Hence U


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Describe the changes when an ideal gas changes from state AD B and

back to A.


Net change in heat

Qnet= Q DB- QAD- Q BA

Net change in internal energy

Unet= 0

(For a cyclic process, that eventually return a system

to its initial state, Uf= Ui; Unet=0)

Net work = shaded area

Isobaric


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Change of

state

Equation Explanation

AD W=0

U= Q - W

U= Q

As temperature decreases, the internal energy decreases too

(U0 and W>0.

Temperature increases. Work is done by the gas, heat is absorbed

by the gas.

BA U= Q - W

U=0, T=0

Isothermal compression at constant temperature.

Since W


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The following is a simple summary of the various thermodynamic

processes.



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References

University Physics, H.D. Young and R.A. Freedman

College Physics,A. Giambattista, B.M. Richardson and R.C.

Richardson

SerwaysEssentias of College Physics, R.A. Serway and C. Vuille

Chap 4-W6 Thermodynamics Cheong.pptx

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