Physics Beyond 2000 Chapter 8 Heat and Gases Intermolecular force Intermolecular force 0 r roro...

Physics Beyond 2000

Chapter 8

Heat and Gases

Intermolecular force

Intermolecularforce

0rro

Potential energy

-ε

The intermolecular force between two gas molecules is so small that it is insignificant.

Ideal gas

• The intermolecular force = 0.

• At absolute zero, gas volume = 0.

http://www.colorado.edu/physics/2000/bec/temperature.html

http://www.colorado.edu/physics/2000/bec/temperature.html

Macroscopic point of view

• The gas laws.

• Pressure, volume, temperature and mass (mole).

http://edie.cprost.sfu.ca/~rhlogan/gas_laws.html

http://library.thinkquest.org/12354/

http://edie.cprost.sfu.ca/~rhlogan/gas_laws.html

http://library.thinkquest.org/12354/

Boyle’s Law

dry air

mercury

h

rubber tubing

Find the pressure P and the volume V of the dry air.

L

Boyle’s Law

dry air

mercury

h

rubber tubing

V = L.A

L

A

Boyle’s Law

dry air

mercury

h

rubber tubing

V = L.A

L

A

P

P = gh + PoPo

Boyle’s Law• By raising or lowering the open end, you

may change P and V.

dry air

mercury

h

rubber tubing

Boyle’s Law• For a fixed mass of gas at constant

temperature, P.V = constant.

P

0 V

P

0 V

1

Boyle’s Law• What would happen if the temperature

changes?

P

0 V

P

0 V

1

Boyle’s Law• What would happen if the temperature

changes? P.V is equal to a new constant.

P

0 V

T1 T2 T3 P

0 V

1

T3T2

T1

T1 < T2 < T3

Boyle’s Law• The following curve is called an isothermal.

P

0 V

T1 T2 T3 P

0 V

1

T3T2

T1

T1 < T2 < T3

Ideal Gas

• Not all gases obey Boyle’s law.

• Many gases follow the law well at high temperature and low pressure.

• Gases which obey Boyle’s law exactly are called ideal gases.

PV graph for oxygen

P

0V

ideal gas

real gas

high temperature

low temperature

Example 1

• Assume that water temperature does not change.

Boyle’s law calculator:http://www.1728.com/boyle.htm

http://www.1728.com/boyle.htm

Temperature and thermometer

• Name different kinds of thermometer and the substances used in the thermometers.

http://www.howstuffworks.com/therm.htm

http://www.howstuffworks.com/therm.htm

Temperature Scale

• Choose a property which changes with temperature as the thermometric substance.

• Choose two particular temperatures.– Lower fixed point– Upper fixed point

Temperature Scales

• Fixed points (i.e. some particular temperatures)

The Celsius Temperature

• Upper fixed point: the temperature of boiling pure water at one atmospheric pressure. It is called 100oC.

• Lower fixed point: the temperature of melting pure ice. It is called 0oC.

http://www.santesson.com/engtemp.html

http://www.santesson.com/engtemp.html


• Let X100 be the value of the thermometric substance at 100oC.

• Let Xo be the value of the thermometric substance at 0oC.

X100= L100

Xo= L0


X100= L100

Xo= L0

What is this temperature ?


X100= L100

Xo= L0

What is this temperature ?

X=L

CXX

XX 0

0100

0 100

Why?


X100= L100

Xo= L0

X=L

CXX

XX 0

0100

0 100

X100

Xo

X

1000C0oC


X100= L100

Xo= L0

X=L

X100

Xo

X

1000C0oC

In designing celsius temperature, it is assumed that the thermometric property changes linearly with the temperature.

Is this assumption correct?

Charles’ Law• For a fixed mass of gas at constant pressure,

the gas volume V changes linearly with the temperature t.

V

t/oC0oC

V = m.t + V0

V0

-273.15oC

Charles’ Law• For a fixed mass of gas at constant pressure,

the gas volume V is directly proportional to the temperature T in kelvin scale.

V

t/oC0oC

V = m.T

V0

V

0K T/K

Disagreement in Celsius Temperature

• Celsius temperature shows disagreement between two thermometers using different thermometric substances.

• Example: Length of mercury column

and Resistance.

• Note that both temperature readings are correct according to their own scales.

Constant volume gas thermometer

• Standard thermometer in physics.

• The thermometric substance is the gas pressure.

• %100

0100

0

PP

PP

Ideal Gas Scale Temperature

• The product P.V is chosen as the thermometric substance.

• Lower fixed point: absolute zero. 0 K.

• Upper fixed point: triple point of water.

Triple point of water Ttr

• This is the temperature at which ice, water and water vapour exist in equilibrium in the absence of air.

• The triple point of water is 0.01oC higher than the ice point.

• The triple point of water is 273.16 K.

http://www.physics.reading.ac.uk/units/flap/glossary/ii/idlgstsl.htm

http://www.physics.reading.ac.uk/units/flap/glossary/ii/idlgstsl.htm


Ptr

273.16K0 K

Ideal gas temperature/K

Pressure of an ideal gas

P

T

trP

PT 16.273

Triple pointAbsolute zero


• Kelvin scale

• T/K = /oC + 273.15/oC

Example 2

• Calculate the unknown temperature by using a constant volume gas thermometer.

General Gas Law

• For a fixed mass of gas,

T

VP.= constant

http://www.jersey.uoregon.edu/vlab/Piston/

http://www.jersey.uoregon.edu/vlab/Piston/

Example 3

• Apply general gas law.

Ideal gas calculator:http://library.thinkquest.org/2923/ideal.html

http://library.thinkquest.org/2923/ideal.html

The universal gas constant R

0

Amount n/mol

PV/J

P1.V1

n1

P2.V2

n2

For an ideal gas at constant temperature, P.V is directly proportional the amount of gas.

P.V n

The universal gas constant R• For a fixed mass of gas,

P.V T

• For a gas at constant temperature,

P.V n

• Combining the above two relations,

P.V n.T

or P.V = R.n.T where R is a constant.

The universal gas constant R

• P.V = nRT

• R = 8.31 J K-1 mol-1.

• R is called the universal gas constant.

http://www.treasure-troves.com/physics/UniversalGasConstant.html

http://www.treasure-troves.com/physics/UniversalGasConstant.html

Example 4

• standard temperature and pressure (0oC and 1.03 105 Pa).

• Find the gas volume at s.t.p.

Notations

• Avogadro constant NA = 6.02 1023 mol-1

• So one mole of substance contains 6.02 1023 elementary units.

Notations

• In this text,n = quantity in moles

N = number of moleculesN = n.NA

Notations• In this text,

Mm = molar mass

M = mass of the gas m = mass of a gas moleculesM = Nm = n.NA.m

and

Mm = NA.m

Example 5

• Find the mass of gas at s.t.p.

Avogadro’s Law

• With the same temperature, pressure and volume, two ideal gases contain the same number of molecules.

• How to prove it?

http://www.avogadro.co.uk/definitions/avogconst.htm

http://www.avogadro.co.uk/definitions/avogconst.htm

Example 6

• Use the Avogadro’s law.

Dalton’s law of partial pressure

• In a mixture of ideal gas, the total pressure equals to the sum of individual gas pressures.

+

P1 + P2= P

http://edie.cprost.sfu.ca/~rhlogan/dalton.html

http://edie.cprost.sfu.ca/~rhlogan/dalton.html

Dalton’s law of partial pressure

+

P1 + P2= P

V

RTnP 1

1 V

RTnP 2

2 V

nRT

V

RTnnP

)( 21

Typical examples

• Mixture of gases from two chambers

P1

V1

T1

P2

V2

T2

Two gases with different pressure, volume and temperature.

Mixture of gases from two chambers

PV1

T1

PV2

T2

When they are mixed, they will have the samepressure (common pressure) P.What is this pressure P?

Each chamber is kept at its own temperature.


• Find the total number of moles n when they are still separated.

P1.V1=n1RT1P2.V2=n2RT2

21

12221121 TRT

TVPTVPnnn

P1

V1

T1

P2

V2

T2


• Find the total number of moles n after they are mixed.

P.V1= RT1P.V2= RT2

21

1221'2

'1 TRT

TPVTPVnnn

'1n

'2n

PV1

T1

PV2

T2


21

1221'2

'1 TRT

TPVTPVnnn

21

12221121 TRT

TVPTVPnnn

Compare these two equations.

1221

122211

TVTV

TVPTVPP

Escape of gas from a container

Po = atmospheric pressureTo = external temperature

n

PVT

If we open the tap for a while, how much gas would escape?



PV=nRT

There are n moles of gas.n

PVT



The chamber is kept at constant temperature T.

Po

VT



There remains n’ moles of gas in the container.

PoV=n’RTPo

VT

Po

VT

n’



Po

VT

Po

VT

n’

The amount of gas escaped is

)(,oPP

RT

Vnnn


Po

VT

Po

VT

n’

The volume occupied by the escaped gas V’=

o

oo

o

o

P

RT

RT

PPV

P

TRn.

)(..

Kinetic theory model

• A microscopic point of view.

• How does the behaviour of gas molecules relate to its pressure, temperature, volume and mass?

http://www.phy.ntnu.edu.tw/java/idealGas/idealGas.html

http://www.phy.ntnu.edu.tw/java/idealGas/idealGas.html

Kinetic theory model• Ideal gas

– 1. There is not any intermolecular force between two gas molecules.

– 2. The volume of each molecule is zero.– 3. The collisions between molecules and the

container are perfectly elastic.– 4. The gas molecules are in random motion.

Pressure from kinetic theory

• Suppose that there are N gas molecules each of mass m in a cubical box of side L.

L

L

L

Resolve a vector• Revision: Resolve a vector c into 2

components (2-dimensional case).y

x

c

What is the relation between c and its 2 components?


x

y

z

Take one molecule, the ith molecule which has speedci.

ci


x

y

z

Resolve its velocity ci intothe three components ui, vi and wi.

ciui

vi

wi

2222iiii wvuc


• Consider its motion along the x-axis.

ui

x

L

LL



ui

x

L

LL

The molecule hitsthe right wall withspeed ui after travellinga distance L.



uix

L

LL

The moleculerebounds at thesame speed ui

because the collisionis perfectly elastic.



ui

x

L

LL

The molecule hitsthe right wall againafter travellinga distance 2L.


ui

x

L

LL

-ui

What is the change of momentum of the ith molecule along x-axis?

(mui) = (-mui)-(mui)= -2mui


What is the time elapsed between two successive impacts on the right wall?

uix

L

LL

Pressure from kinetic theoryWhat is the time elapsed between two successive impacts on the right wall?

ui

x

L

LL

iu

Lt

2


ui

x

L

LL

-ui

What is the average force fi’ on the ith molecule by the right wall?

(mui) = -2mui

iu

Lt

2

L

mu

t

muf iii

2, )(

fi’


ui

x

L

LL

-ui

What is the average force fi on the right wall by the ith molecule?

By Newton’s 3rd lawfi = -fi’

fi

L

mui2


• As there are N molecules, the total force on the right wall is Fx = f1 + f2 + … + fN

L

L

LF

N

iiuL

m

1

2.


L

L

LF

N

iix u

L

mF

1

2.Define the mean square speed as

N

uu

N

ii

1

2

2


L

L

LFx

Define the mean square speed as

N

uu

N

ii

1

2

2

L

uNmFx

2


L

L

LFx

L

uNmFx

2

Pressure on the right wallis

V

uNm

L

uNm

L

F

A

FP x

2

3

2

2


L

L

LFx

Pressure on the right wall isV

uNmP

2


x

y

z

ciui

vi

wi

Pressure on the right wall isV

uNmP

2

How does relate to c?2u


x

y

z

ciui

vi

wi

2222iiii wvuc

2222

2222

wvuc

N

w

N

v

N

u

N

c iiii


x

y

z

ciui

vi

wi

2222 wvuc

As the gas molecules aremoving randomly,

222 wvu

22 3uc


L

L

L

Pressure isV

cNm

V

uNmP

3

22

2

3

1cNmPV


L

L

L

Pressure is

22

3

1

3

1cc

V

NmP

where is the density of gas

Root mean square speed cr

• Definition of r.m.s. speed of gas

2ccr

22rcc

Root mean square speed cr

2

3

1rNmcPV

or

2

3

1rcP

r.m.s.speed at s.t.p.

• Given: Molar mass of O2 gas = 0.032 kg

• Find the r.m.s. speed of oxygen at s.t.p.

• Use 2

3

1rAmcnNnRTPV

Average r.m.s. speed of air at room temperature

• Given: The average density of air at room temperature is 1.2 kg m-3.

• The atmospheric pressure = 1.01 105 Pa.

• Use2

3

1rcP

P

cr3

Distribution of molecular speed

• The r.m.s. speed of ideal gas molecules is given by

• The r.m.s speed increases with temperature and decreases with molecular mass.

2

3

1rmcnMnRTPV

Some r.m.s. speeds:http://www.physics.mun.ca/~gquirion/P2053/html21/slide9.html

http://www.physics.mun.ca/~gquirion/P2053/html21/slide9.html


• The actual speed of each gas molecules varies.• If temperature is kept constant, the speeds of gas molecules follow the Max

wellian distribution.

No. of molecules with speeds of 0.5 ms-1 of speed v

speed v

dv

dN

low temperature

high temperature



speed v

dv

dN

low temperature

high temperature

• The area under the graph represents the total number of gas molecules.



speed v

dv

dN

low temperature

high temperature

• As the temperature increases, the r.m.s. speed increases.

• The peak shifts to the right (higher speed).



speed v

dv

dN

low temperature

high temperature


• The range of speeds spreads out.



speed v

dv

dN

low temperature

high temperature


• The amplitude is lower because the area, representing the total number of molecules, under the curve must be the same.

http://cyniska.ubishops.ca/0002933/3dspeed.html

http://cyniska.ubishops.ca/0002933/3dspeed.html

Example 7

• Double the gas temperature.

• What would happen to its r.m.s. speed?

• Use

mr M

RTc

3

Escape speed and atmosphere


speed v

dv

dN

escapespeed

These moleculesare able to escape.


• On earth– The escape speed is 11.2 103 ms-1

– The r.m.s. speed of air molecules is 502 ms-1

• Do the air molecules on earth escape easily?


• On the moon– The escape speed is 2.38 103 ms-1

– The r.m.s. speed of air molecules is 558 ms-1

• Do the air molecules on the moon escape easily?

Translational kinetic energy

• The average translational kinetic energy is

221

2

2

1

2

121

r

N

ii

k mccmN

mcE


• From the ideal gas equation,

TN

Rmc

RTN

NNmc

nRTPV

Ar

Ar

2

3

2

1

3

1

2

2


• So the average translational kinetic energy is

kTTN

RmcE

Ark 2

3

2

3

2

1 2

whereAN

Rk is the Boltzmann constant

k=1.38 10-23 JK-1

Translational kinetic energy• So the average translational kinetic energy is di

rectly proportional to the ideal gas temperature.

kTEk 2

3

kE

T/K0

Translational kinetic energy• At the same temperature, all gas molecules hav

e the same average translational kinetic energy.

kTEk 2

3

kE

T/K0

Translational kinetic energy• At absolute zero, all molecules are at rest.

kTEk 2

3

kE

T/K0

Example 8

• Find the total translational kinetic energy of 1 mole of O2 gas at room temperature 27oC.

• NA = 6.02 1023

• k = 1.38 10-23 JK-1

Internal energy of an ideal gas

• Internal energy of an object

= total kinetic energy of the molecules

+ total potential energy of the molecules

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/inteng.html

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/inteng.html


• There is not any intermolecular force among ideal gas molecules

Potential energy of ideal gas = 0

Internal energy of ideal gas

= Total kinetic energy of ideal gas molecules

Internal energy of an ideal gas• Consider an ideal gas of monatomic molecu

les only. There is not any rotational kinetic energy.

The ideal gas has translational kinetic energy only.

Internal energy U of a monatomic ideal gas

= total translational kinetic energy of molecules.


TN

RnNkTNENU

AAk .

2

3).(

2

3..

nRTU2

3

where n is the number of moles


nRTU2

3

Apply PV=nRT

PVU2

3 for a monatomic ideal gas

Example 9

• He gas with volume 1 litre (0.001 m3) and at 2 atmospheric pressure.

• Assume that it is an ideal gas.

• Find its internal energy.

Real Gases

• Ideal gas• Molecules do not have

size.

• Real gases• There is a finite size

for each molecule.

Real Gases

• Ideal gas• There is not any

intermolecular force among ideal gas molecules

• Real gases• When they are far

apart, there is attraction among the molecules.

• When they collide, there is repulsion.

Real Gases

• Ideal gas• May apply the equatio

n

P.V = nRT

• Real gases• Need to correct the

equation

Gas equation for real gases

• Effect of attractive intermolecular force

Gas pressure P is due to the collision of moleculeson the wall of the container.

P piston



For real gas, the collision of a molecule on the wallis retarded by the attraction of all other gas molecules.

piston



As a result, the measured real gas pressure is less thanthe pressure of an ideal gas under the same condition.

piston



Add a positive correction k to the measured pressureP. (P + k) is the pressure due to an ideal gas underthe same condition.

pistonP+k


• Effect of finite size of gas molecules

P+k piston

Some portion of volume is occupied by the molecules. The measured volume is greater thanthe volume of an ideal gas.


• Effect of finite size of gas molecules

P+k piston

Add a negative correction –nb to the measured volume V. (V - nb) is volume of the ideal gasunder the same condition. n = number of moles.

V - nb

van der Waals equation

• (P + k).(V – nb) = n.R.T

• k =

• a and b are constants. Their values depend on the kind of real gas.

2

2

V

an

van der Waals equation

• (P + k).(V – nb) = n.R.T

P

V

P.V = nRT

Critical temperature

• An ideal gas has no forces among molecules.

• So it cannot become a liquid.

• At infinite high pressure or at the absolute zero, the volume of an ideal gas become zero.


• Real gas has forces among molecules.

• It can become liquid.

• If the temperature is below its freezing point, it becomes liquid.


• Experiments show that when a gas is below a certain temperature (which is above the boiling point), it can become liquid by applying high pressure without further decreasing the temperature.

• The temperature is called critical temperature.


• Critical temperature

It is the temperature above which a gas cannot become liquid by applying high pressure only.

http://www.chem.purdue.edu/gchelp/liquids/critical.html

http://www.chem.purdue.edu/gchelp/liquids/critical.html

Isothermals for CO2

Critical temperature of CO2 is 31.2oC

Isothermals for CO2

At high temperature above the critical temperatureCO2 is gas.

Isothermals for CO2

In this region, CO2 could be compressed into liquid. It is called vapour.

Isothermals for CO2In this region, it is difficult to change its volume by applying pressure. It is in either liquid or solid state.

Isothermals for CO2In this region, both liquid and vapour exist in equilibrium.The vapour is called the saturated vapour.

Average separation of gas molecules

• When liquid changes into gas under one atmospheric pressure, its volume increases by 750 times.

• What is the average separation D of gas molecules under one atmospheric pressure in terms of the diameter ro of each molecule?

D = 9.1 ro

Thermodynamics

• To study the laws governing– conversion of energy– direction of heat flow– work

Heat and Internal Energy

• Heat flow– Heat is the energy transfer from a body at high temperature to

another body at low temperature.

hightemperature

lowtemperature

heat flow

Heat and Internal Energy

• Internal energy– Sum of kinetic energy and potential energy of the molecules of a

substance.

hightemperature

lowtemperature

heat flow

Lose internalenergy

Gain internalenergy

Potential energy of gas molecules

• For a real gas, its potential energy is a maximum because they are separated by an infinite distance. The potential energy is zero.

• For other states, its potential energy is negative.

Internal energy of Ideal gas

• Potential energy of an ideal gas is zero.

• The internal energy of an ideal gas is the sum of kinetic energy of its molecules.

• When the internal energy of an ideal gas increases, its temperature rises up.

• When the internal energy of an ideal gas decreases, its temperature drops down.

Internal energy of gas

• Heating the gas Increasing the internal energy (kinetic energy) of the gas Rise in gas temperature

• Cooling the gas Decreasing the internal energy (kinetic energy) of the gas A drop in gas temperature

Internal energy of gas

• Doing work on the gas (or the gas does a negative work e.g. compression of a gas) Increasing the internal energy (kinetic energy) of the gas Rise in temperature

• The gas does a positive work (e.g. expansion of a gas) Decreasing the internal energy (kinetic energy) of a gas A drop in temperatrue.

Work done by a gas

• Expansion of a gas

The gas does a positive work

The gas loses internal energy.

gasF

F = P.A

Work done by a gas


The gas does a positive work

The gas loses internal energy.

gas

x

F

F = P.A

Work done by a gas

• compression of a gas

The gas does a negative work

The gas gains internal energy.

gasF

F = P.A

Work done by a gas

• compression of a gas

The gas does a negative work

The gas gains internal energy.

gas

x

F

F = P.A

Work done by a gas


Work done by the gas W = F. x = P.V

gas

x

F

F = P.A

Work done by a gas


Work done by the gas W =

gas

x

F

F = P.A

2

1

.V

VdVPdW

Work done by a gas

gas

x

F

F = P.A

2

1

.V

VdVPdW• W =

• To evaluate the integration, it is necessary to express P in terms of V.

Example 10

• Assume that the force is a constant. Evaluate the integration.

• As the work done by the gas is positive, it loses internal energy.

The indicator diagram• The state of a fixed mass of gas is described

by its pressure, volume and temperature.

• The state can be represented by a point in the P-V graph. (P1, V1, T1)

P

VT1

P1

V1

0

The indicator diagram• In the following diagram, it shows that the

gas expands at a constant temperature.

P

VTi

Pi

Vi

0Pf

Vf

Initial state: Pi , Vi , Ti

Final state: Pf , Vf , Tf

Tf = Ti

The indicator diagram• The area indicates the work done by the

gas.

• It is also the loss of internal energy of the gas.



Tf = Ti

P

VTi

Pi

Vi

0Pf

Vf

The indicator diagram• In the following diagram, it shows that the

gas is compressed at a constant temperature.

P

VTi

Pf

Vf

0Pi

Vi



Tf = Ti

The indicator diagram• The area indicates the negative work done

by the gas.

• It is also the gain of internal energy of the gas.

P

VTi

Pf

Vf

0Pi

Vi



Tf = Ti

Work done by the gas

• W > 0 if its volume expands Loss of internal energy

Vf > Vi

• W < 0 if its volume is compressed Gain of internal energy

Vf < Vi

Work done by a gas

• Example 11

• Find the gas pressure first

• Calculate the work done by the gas.

Changes of states of a gas

gas

heat reservoir

weight A gas is trapped inside acylinder with a movable piston.The gas temperature canbe regulated by a heatreservoir.

Movable piston

Isobaric process

• The pressure P is kept constant.

P

V

Po

Vi

Ti

Vf

Tf

Pi = Pf = Po

ExpansionW > 0


gas

heat reservoir

weight

Movable piston

Change the temperature of the heat reservoir fromTi to Tf. The gas expands with constant pressure.Pf = Pi = Po

P

V

Po

Vi

Ti

Vf

Tf


gas

heat reservoir

weight

Movable piston

There is heat flow to the gas while the pressure is kept at a constant.

P

VVi

Ti

Vf

Tf

Po


gas

heat reservoir

weight

Movable piston

The area represents the net work done by the gas.Heat is absorbed.The internal energy of the gas increases.

P

VVi

Ti

Vf

Tf

Po


gas

heat reservoir

weight

Movable piston

W1 = Po.(Vf – Vi) is the work done by the gas.Q1 is the heat flow to the gas.U is the increase in the internal energy of the gas.

Q1

P

VVi

Ti

Vf

Tf

Po


gas

heat reservoir

weight

Movable piston

Q1 = U + W1 or U = Q1 - W1 > 0

The heat flow increases the internal energy.The work done by the gas decreases the internal energy.

Q1

P

V

Po

Vi

Ti

Vf

Tf


gas

heat reservoir

weight

Movable piston

There may be another way to reach the same final state.

P

V

Po

Vi

Ti

Vf

Tf

P’T’


gas

heat reservoir

weight

Movable piston

Step 1: Decrease both the pressure and temperature while keeping the volume constant.

P

V

Po

Vi

Ti

Vf

Tf

P’

1

T’


gas

heat reservoir

weight

Movable piston

Step 2: Increasing both the volume and temperature while keeping the pressure constant.

P

V

Po

Vi

Ti

Vf

Tf

P’ 2T’


gas

heat reservoir

weight

Movable piston

P

Step 3: Increasing both the pressure and temperature while keeping the volume constant.

V

Po

Vi

Ti

Vf

Tf

P’

3

T’


gas

heat reservoir

weight

Movable piston

The area represents the work done by the gas.Heat is absorbed by the gas.The internal energy of the gas increases.

P

V

Po

Vi

Ti

Vf

Tf

P’T’


gas

heat reservoir

weight

Movable piston

W2 = P’.(Vf – Vi) is the work done by the gas.Q2 is the heat flow to the gas.U is the increase in the internal energy of the gas.

Q2

P

V

Po

Vi

Ti

Vf

Tf

P’T’


gas

heat reservoir

weight

Movable piston

Q2 = U + W2

Q2

P

V

Po

Vi

Ti

Vf

Tf

P’T’

Changes of states of a gas• Comparison of the t

wo processes:• Both U are equal because they have the s

ame initial and final states.

• W1 = Po (Vf – Vi) > W2 = P’ (Vf – Vi) Q1 > Q2

P

V

Po

Vi

Ti

Vf

Tf

P’T’

process 1: process 2:

Q = U + W

First Law of Thermodynamics

Q = U + W

The heat supplied to the gas, Q, enables it to do work W and to increase its internal energy U .

Example 12

• Assume that the gas volume decreases in the process.

Example 13

• On expansion, there is loss of internal energy.

• By heating, the internal energy is supplemented.

Adiabatic process• In this process, there is not any heat transfer.

• Q = 0 U + W = 0

U = - W

• On adiabatic expansion

W > 0 U < 0

• On adiabatic compression

W < 0 U > 0

Adiabatic Expansion• Q = 0• W > 0 U < 0

P

VLow T

High T

Pi

Vi

Pf

Vf

The shaded area representsthe positive work done W.

Adiabatic Compression• Q = 0• W < 0 U > 0

P

VLow T

High T

Pf

Vf

Pi

Vi

The shaded area representsthe negative work done W.

Adiabatic process

• Using a container with good heat insulation adiabatic process Q = 0.

• Change the gas volume quickly adiabatic process Q = 0. (I) Quick compression of gas A rise in

temperature. (II) Quick expansion of gas A drop in

temperature.

Adiabatic process: Experiment

• Change the gas volume quickly

adiabatic process Q = 0.

(I) Quick compression of gas A rise in temperature.

(II) Quick expansion of gas A drop in temperature.

Example 14

• Expansion of gas quickly.

Constant volume process• The gas volume does not change W = 0.• U = Q.• Heat is supplied to the gas and

increases its internal energy (kinetic energy) Temperature rises up.

• Heat is extracted from the gas and decreases its internal energy (kinetic energy) Temperature drops down.

Heating under constant volume

P

VLow T

High T

Pf

Vo

Pi

•Vi = Vf = Vo

•W = 0 U = Q

Heating under constant volume

P

VLow Ti

High Tf

Pf

Vo

Pi

•For monatomic ideal gas, U = nRT2

3

)(2

3if TTnRUQ

Vi = Vf = Vo

Cooling under constant volume

P

VLow Tf

High Ti

Pi

Vo

Pf

•For monatomic ideal gas, U = nRT2

3

)(2

3if TTnRUQ

Vi = Vf = Vo

Example 15

• Monatomic ideal gas.

Isothermal process

• Temperature is kept constant. U =0.

• Q = W.

• Q > 0 W > 0 Expansion of gas

• Q < 0 W < 0 Contraction of gas

Isothermal process

• Temperature is kept constant. U =0.• Q = W > 0 Expansion

P

VTo

Pi

Vi

0Pf

Vf

The shaded area represeentsthe work done.

Ti = Tf = To

Isothermal process• Temperature is kept constant. U =0.• Q = W > 0 Expansion• If , find W.

nRTPV

2

1

2

1

)ln(1

2V

V

V

V V

VnRTdV

V

nRTPdVW

P

VTo

Pi

Vi

0Pf

Vf

Ti = Tf = To

Isothermal process• Temperature is kept constant. U =0.• Q = W < 0 Contraction• If PV = nRT , find W.

P

VTo

Pf

Vf

0Pi

Vi

2

1

2

1

)ln(1

2V

V

V

V V

VnRTdV

V

nRTPdVW

Ti = Tf = To

Example 16

• Compression of a gas without temperature changes.

Isobaric process

• The pressure is kept constant.

P

V

Po

Vi

Ti

Vf

Tf

Isobaric expansion

• The pressure is kept constant on expansion.

P

V

Po

Vi

Ti

Vf

Tf

W > 0 Q > 0 and U > 0

Isobaric expansion


P

V

Po

Vi

Ti

Vf

Tf

If it is a monatomic ideal gas,find W, U and Q.

Isobaric expansion


P

V

Po

Vi

Ti

Vf

Tf

W = Po(Vf – Vi)

)(2

3ifo VVPU

)(2

5ifo VVPQ

Example 17

• Expansion of a monatomic ideal gas without change in pressure.

Cycle• A gas changes from one state to another

state and back to the initial state through another path.

U = 0.

P

V

P1

V1

0P2

V2

CycleU = 0.

• In the following process,

W > 0 Q > 0.

P

V

P1

V1

0P2

V2

CycleU = 0.

• In the following process

W < 0 Q < 0.

P

V

P1

V1

0P2

V2

Example 18

• Discussion of Figure 24.

Typical exampleA cylinder with a movable piston is immersed inice-water.

gas ice-waterOoC

Typical example1. A quick compression (adiabetic process)

gasice-water

P

VV1V2

1

0oC

T2>0oC

Typical example1. A quick compression (adiabetic process)2. Allow the gas to cool down at constant vol

ume.

gasice-water

P

VV1V2

1

0oC

T2>0oC2

Typical example1. A quick compression (adiabetic process)2. Allow the gas to cool down at constant volume.3. Expands the gas along an isothermal.

gasice-water

P

VV1V2

1

0oC

T2>0oC2

3

Example 19

• Discuss the energy change.

Physics Beyond 2000 Chapter 8 Heat and Gases Intermolecular force Intermolecular force 0 r roro...

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