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HEATCHAPTER 4

1

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4.1: ThermalEquilibrium

1. Temperature is a measure of the degree of

hotness of an object.

2. A hot body has high temperature whereas a coldbody has a low temperature.

. !eat is a form of energy being transferred from ahot body to a cold body.

4. !eat and temperature are two di"erent physical

quantities2

4.1.1: Relationship between Temperature and Heat

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3

Temperature !eat

The degree of hotnessof an object A form of energy

#ase quantity $eri%ed quantity

&nit: 'el%in (') ofdegree *elcius (o*)

&nit: +oule (+)

*an be measuredusing thermometer

,o speci-c measuringequipment

le 4.1: The di"erence between heat and tempera

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4

4.1.2: Thermal Equilibrium

1. n -gure body A has higher temperature thanbody #.

2. The net /ow of heat is from body A to body #.

ster rate of heat transfer

lower rate of heat transfer

#ody A

#ody #

hot cold

#ody A

,et heat/ow from

A to #

#ody #

equi%alent

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. Temperature of body A will decrease while thetemperature of body # will increase.

4. The net heat will /ow from A to # until thetemperature of A is the same as thetemperature of #.

. n this situation3 the two bodies are said toha%e reached thermal equilibrium.

. 5hen thermal equilibrium is reached3 the netrate heat of /ow between two bodies is 6ero.

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!

7. Two bodies are said to be in thermalequilibrium when:

a) they are at same temperatureb) the net heat of /ow between the two

bodies is 6ero

8. Application of thermal equilibrium:

a) 9icrowa%e o%en

b) efrigerator

c) Thermometer

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libration of liquid;in;glass thermometer

1. n the construction of a thermometer3 two -<edtemperatures are always chosen:

a) A lower -<ed point3 ice point:

The temperature at which pure ice melts at=o* under standard atmospheric pressure

b) The upper -<ed point3 steam point:

The temperature of steam from water that isboils at 1==o* under standard atmosphericpressure

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Thermometer without scale

ice

l=

x = > = o* > l= c

x =

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x 1== > 1== o* > l1== cm

#oiling water

l1==

x 1==

steam

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lθ

warm water

θ = ? oC = lθ cm

θ

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l=

l&ice3 =o*

team point31==o*

warm water3 θ

l1==

l=

l&

l1==

l ?cm

θ?o*

θ 1===

C100o

0100

0 ×

−−

=l l

l l θ

θ

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E'ample 4.1.2

2.7 cm 1. cm

=o* θ 1==o*

1. 0igure shows a thermometer. 5hat is thetemperature3 @ recorded by thermometer?

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2. 0igure shows the graph of length of mercurycolumn against temperature for a mercury;

in;glass thermometer.

5hat is the temperature s?

2

2=

27

cm

o*s 1==

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1. 0igure shows a thermometer. 5hat is thetemperature3 @ recorded by thermometer?

2 cm

8 cm

=o* θ 1==o*

24 cm

E'er(ise 4.1.2

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2. 0igure shows the graph of length of mercurycolumn against temperature for a mercury;

in;glass thermometer.

5hat is the %alue of l?

2

l

2

cm

o*4 1==

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2. Biquid;in;glass thermometer maCes used thechange in %olume of -<ed mass of liquid (mercuryor alcohol) to measure temperature.

. The glass bulb is thin so that the heat transfer byconduction is faster and thermal equilibrium can

reached faster.

4. The narrow bore of the capillary tube maCes thethermometer more sensiti%e.

. Bower temperature can be measured by liquid;in;glass thermometer -lled with alcohol.

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Mercury Alcohol

!as free6ing point of;o* and boiling pointof 7o*

!as free6ing point of;112o* and boilingpoint of 78o*

ensiti%e to changesin temperature

eact slowly tochanges in

temperature$o not sticC to theglass wall of thethermometer

ticC to the glass wallof the thermometer

E<pand uniformly $oes not e<panduniformly

t does not wet the

tube

t wets the tube Table 4.1: comparison between mercury

and alcohol liquid;in;glass

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4.2: &nderstanding peci-c !eat *apacity

1.The heat capacity of an object is the amountof heat required to increase the temperature

of the object by 1o*.

2.The heat capacity of an object depends on

the

4.2.1!eat *apacity

turein temperachange

object by thereleasedorabsorbedHeatcapacityHeat =

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4.2.2: peci-c !eat *apacity

1. The speci-c heat capacity3 c3 of a substanceis the quantity of heat needed to increase thetemperature of a mass of 1 Cg by 1o* or 1 '.

5here m > mass

D > heat supplied

θ > change in temperature

2. The unit of speci-c heat capacity is + Cg;1 o*;1

or + Cg;1 ' ;1.

θ ∆=m

Q

c

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. The quantity of heat absorbed or lost from abody is gi%en by:

4. A material which has a high speci-c heatcapacity has the following properties:

a) t taCes a longer time to be heated

b) t does not lose heat easily

c) t is usually used as a heater insulator

d) t is a poor heat conductor

θ ∆= mcQ

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22

4.2.: Application of peci-c !eat *apacity

1. A metal has low speci-c heat capacity and itstemperature increases easily when heated.

2. 9etals are normally used as pans and heating-lament of Cettles so that food and water can

be heated faster.

. The handles of pots and pans are usually madeof materials of high speci-c heat capacity or

poor heat conductor.

*ooCing &tensils

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24

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2

ea #ree6e

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2!

1. $uring the day3 the land and the sea recei%e thesame amount of heat from the sun.

2. The land is heated to a higher temperature thanthe sea because water has higher speci-c heatcapacity than earth.

. This cause the air abo%e the land to be hotterthan the air abo%e the sea.

4. The hot air abo%e the land /ows up and the coolair from the sea /ows towards the land.

. The mo%ement of air causes wind to blow from

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2"

Band #ree6e

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2#

1. At night3 the air abo%e the land and the searelease the heat to the atmosphere.

2. Temperature of the sea decreases more slowlythan temperature of the land because water hashigher speci-c heat capacity than the earth.

. The land become colder than the sea.

4. The hot air abo%e the sea /ows upwards the

cool air from the land /ows towards the sea.

. The land bree6e is produced due to the

mo%ement of air from the towards the sea.

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2$

1.A hot water tanC contains 8= Cg of water. Thewater is initially at =o*.

Fspeci-c heat capacity for water > 42== + Cg;1

o*;1Ga)*alculate the amount of energy that must

be transferred to the water to raise the

temperature to 7=o*.b)*alculate the time this will taCe when a

C5 electric water heater is used.

E'ample 4.2.2

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3%

1.!ow much heat energy is required to raisethe temperature of 4 Cg of iron bar from 2o*to 2o*?

Fspeci-c heat capacity for iron > 4 + Cg;1 o*;

1G

2.An electric Cettle has a power rating of 2.4C5. !ow long does it taCe the Cettle to heatup 4. Cg of tap water from 28o* to 1==o*?

Fspeci-c heat capacity for water > 42== + Cg;

1 o*;1G

E'er(ise 4.2.2

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31

2.A hot water tap deli%ers water at 8=H*. Tenlitres of hot water is added to a pailcontaining = litres of water at 2=H*. 5hat

will be the -nal temperature of the water inthe pail?

.= grams of iron is heated o%er a /ame forse%eral minutes. t is then plunged into aclosed container containing 1.= litre of coldwater3 originally at 1H*. After the

temperatures equali6e3 the water is now

9*pe(i+i( heat (apa(it, o+ water 42%% /01 ;C1<

*pe(i+i( heat (apa(it, o+ iron 44% /01 ;C1=

E'ample 4.2.2

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32

.A steel ball bearing of mass =.=2 Cg and

temperature H* is placed into =.2 Cg ofwater at 24H* in a polystyrene cup. 5hat isthe temperature when the ball bearing and

water ha%e come to thermal equilibrium? Assume that the e<change of heat is betweenthe ball bearing and water only.Fpeci-c heat capacity of water > 42== + Cg;1

H*;1I peci-c heat capacity of steel > 4= +Cg;1 H*;1G

E'er(ise 4.2.2

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33

1.1== g of water is heated to change itstemperature from 1H* to 2=H*. !ow muchenergy has been transferred to the waterJ

2.1= g of water is heated from 1=H* to =H*.5hat amount of energy is requiredJ

.A 2.= Cg blocC of copper at 1==H* is put intoa large pot containing .= litres of water at2=H*. Assuming no energy is lost to the

surrounding en%ironment3 what is the -naltem erature of the Kmi<tureLJ

>or +ollowin0 questions re+er to table 4.2 +or spe(i+i( heat (apa(ities

Homewor/

are heated through the same temperature

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34

are heated through the same temperaturerange. 5hich one requires more energyJ(b) 5hat is the ratio of the speci-c heat

capacity of aluminium to that of waterJ(c) Equal amounts of energy are absorbed

by equal masses of aluminium and water.5hat is the ratio of the temperature rise of

the aluminium to that of waterJ

.A 2=== watt Cettle holds .= Cg of water at

2=H*. t is turned on for minutes. f all theenergy supplied is used to heat the water3will it boilJ Answer

1. 21%% 2. 22%% 3. 22.#"oC

4. -a water -b 3 : 14 -( 14 : 3. 8ater will not boil be(ause the temperature rise is onl, !".!oC

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3

4.: &nderstanding Batent !eat

1.All matter e<ists in three phaseI solid3 liquidand gas.

2.trong intermolecular forces hold the%ibrating molecules together in solid.

.Adding heat energy can cause a breaCdownof this forces.

4. The solid melts and becomes a liquid and

similarly3 heat may also cause a liquid to

4..1: Mhase *hange

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3!

1. The heat supplied to a substance duringchange of phase does not cause a change inthe temperature of the substance.

2.5hen a solid melts and liquid is boiling3 heatis absorbed but the temperature remainconstant.

. The heat absorbed or heat released atconstant temperature during change of phase is Cnown as latent heat.

4. The four main changes of phaseI melting3

4..2: Batent !eat

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3"

6?575@

7atent heat absorbed

C?@E@*AT5?@

7atent heat released

)E7T5@

7atent heat absorbed

*?755>5CAT5?@

7atent heat released

as

solid liquid

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3#

solid

olidNliquid

liquid

Biquid Ngas

gas

emperature3 o*

Time0igure 4.1: The heating cur%e

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3$

solid

olid

Nliquid

liquid

Biquid Ngas

gas

emperature3 o*

Time0igure 4.2: The cooling cur%e

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4%

. There are three common characteristics inthe four processes of change of phase

a)A substance undergoes a change of phaseat a particular temperature

b)!eat energy is transferred during changeof phase

c)$uring change of phase the temperatureremains e%en though there is transfer ofheat

. The transfer of heat does not cause a changein the Cinetic energy of the molecules

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41

7atent heat o+ +usion

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42

1.$uring melting3 the heat absorbed is used tobreaC up the bonds between the particles.

2.The heat absorbed by melting solid is Cnownas the latent heat of fusion.

.0or a liquid to solidify at its free6ing point3latent heat of fusion has to remo%e from it.

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43

7atent heat o+ Baporisation

1.5hen a liquid boils3 the heat absorbed isused toa)*ompletely breaC the force between

particlesb)$o worC against atmospheric pressure

when the gasses e<pands into theatmosphere

2. The heat absorbed during boiling is Cnown asthe latent heat of %aporisation.

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44

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4

1. The amount of heat required to change thephase of substance depends on the mass andthe type of material of the substance.

2. The speci-c latent heat3 l3 of substance is theamount of heat required to change the phaseof 1 Cg of substance at a constant temperature.

8here Q latent heat absorbed or released b, the substan(e

m mass o+ substan(e

with unit in /01

4..: peci-c Batent !eat

rearrangemQl = ml Q =

The speci-c latent heat of fusion l of

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4!

. The speci-c latent heat of fusion3 lf 3 of

substance is de-ned as the amount of heatrequired to change 1 Cg of substance from a

solid to liquid phase without change oftemperature.

4. The speci-c latent heat of %aporisation3 lv 3 of

substance is de-ned as the amount of heatrequired to change 1 Cg of substance from aliquid to gas phase without change oftemperature.

rearrange

rearrangem

Ql f =

m

Ql v = vml Q =

f ml Q =

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4"

.f heat is supplied electrically to change thephase of substance3 the equation Q = ml canbe written as:

here3 P > power of heater3 unit in watt3 W

t > time the heater is on3 unit in seconds

ml Pt =

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4#

ubstan

ce

9elting

point ?o* lf ? + Cg;1

#oiling

point ?o* lv ? + Cg;1

5ater = . <1=

1== 2.2 <1=

9ercury ; 1.14 <1=4

7 2. <1=

Ethanol ;114 1.=8 <

1=

78 8. <

1=

Oold 1= .28 <1=4

28=8 1.72 <1=

*opper 1=8 2.=7 <

1=

2 4.7 <

1=

Table 4.3: The speciic latent heats o some common substance

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4$

1.!ow much energy has to be remo%ed from 2 Cgof water at 1=o* to produce a blocC of ice at=o*?

2.An electric Cettle contains .4 Cg of water at=o*.(a) *alculate the amount of heat required to

boil away all the water after the boiling pointhas been reached.(b) f the power of the heater is 2.4 C53 whatis the time taCen?

efer %alue of c3 lf and l% from table 4.2 andE'ample 4.3

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%

1.5ater of mass =. Cg at 2H* is put into the

free6er compartment of a refrigerator.*alculate the amount of heat that must beremo%ed to change the water completely intoice.

Fpeci-c heat capacity3 c > 42== + Cg;1 H*;1

peci-c latent heat of fusion3 lf > . < 1= +

Cg;1G

2.!ow much heat must be supplied to 2.= Cg ofwater at 28H* to change it to steam at 1==H*JFpeci-c heat capacity3 c > 42== + Cg;1 H*;1

peci-c latent heat of %aporisation3 lv

> 2.2 < ;1

E'er(ise 4.3

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1

1. A %.2 /0 steel ball at #%;C is pla(ed on a lar0e blo(/ o+ i(e.

Cal(ulate the mass o+ i(e melted b, the heat +rom the steelball. 9*pe(i+i( heat (apa(it, o+ steel 42% /01 oC1< *pe(i+i(

latent heat o+ +usion o+ i(e 33! %%% /01=

Homewor/

2.

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2

2. >i0ure shows how the temperature o+ a substan(e (han0es

with time when it (han0es phase +rom a liquid to solid.

a. 8hat is the meltin0 point o+ the substan(eD

b. E'plain wh, the temperature remains (onstant durin0solidi+i(ation thou0h heat is bein0 released to the

surroundin0s.

(. The mass o+ the substan(e is %.% /0 and heat is lost to

the surroundin0s at an aBera0e rate o+ 2 s1

. Cal(ulatethe spe(i+i( latent heat o+ +usion o+ the substan(e.

3. An i(e (ube is ta/en out o+ the +reeer (ompartment o+ a

re+ri0erator. E'plain wh, the sur+a(e o+ the i(e is initiall, dr, but

be(omes wet a short while later.

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3

4.*alculate the total amount of heat required tochange an ice cube of mass =.==8 Cg andtemperature ;7 H* to steam at 1==H*.Fpeci-c heat capacity of ice > 2=== + Cg;1

peci-c latent heat of fusion of ice > . <1= + Cg;1

peci-c heat capacity of water > 42== + Cg;1

H*;1 peci-c latent heat of %aporisation of water >2.2 < 1= + Cg;1G Answer

1. %.%2 /0

2. -a $oC -( 1% %%% /01

4. 2424%

4 4: Fnderstandin0 the as 7aws

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4

4.4: Fnderstandin0 the as 7aws

4.4.1: Mroperties of Oas

1. 0or a gas in enclosed container3 the(a) number of molecules is constant

(b) mass of the gas is constant(c) beha%ior of the gas depends on the%olume3 temperature and pressure of the

gas

2. The %olume3 temperature and pressure of agas can be e<plained using the Cinetic theory

of gas.

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1. According to the Cinetic theory of gases3 gasmolecules mo%e freely3 rapidly and randomly inthe space of container.

2. The molecules collide with one another and thecontainer walls. All the collision are elastic.

. The pressure of a gas is due to the collision ofthe gas molecules with the container walls.

Mressure of gas

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!

1. The %olume of gas is the space occupied by its

molecules.

2. Therefore3 a%erage distance between the gasmolecules determines its %olume.

Qolume of gas

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"

1. Oas molecules are in constant random motion

and ha%e Cinetic energy.

2. 5hen a gas at higher temperature3 the gasmolecules mo%e with a greater %elocity andha%e more Cinetic energy.

. The temperature of the gas is proportional to

the a%erage Cinetic energy of the gasmolecules.

Temperature of gas

4 4 2 O B

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#

4.4.2: Oas Baw

#oyleLs law

1. #oyleLs law state that the pressure of the -<ed massof gas in in%ersely proportional to the %olume of the

gas3 pro%ided the temperature of the gas is constant.

2. 0rom mathematical equation3

or PV > constant3 C

V P

1∝

. This means for particular sample of gas at constant

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$

temperature3 an increase in pressure from P1 to P2

causes corresponding decrease in %olume from V 1 to V 2.

ThereforeI

4. The following graph show the #oyleLs law

P

V

P

V

1

!!11 V P V P =

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!%

1. The air in the foot pump has an initial %olumeof 28== cm and pressure 1== CMa. The outletof the pump is closed and piston pushed

inward until the %olume of the air become 7==cm. what is the pressure of the compressed isin the pumpJ

2. A bubble of air with %olume 4 cm is releasedfrom a submarine at the depth = m. what willbe the %olume of the bubble when it has riseto a depth of 14 mJ

atmos heric ressure > 1= m water

E'ample 4.4

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!1

1. An air bubble released by a di%er has a%olume of cm at a depth m. what is the%olume of the bubble at a depth of 2 mJ

Fatmospheric pressure > 1= m waterG

2. A balloon is -lled with helium gas to %olumeof = litres at 1 atm pressure. The balloon isreleased and rises to an altitude where its%olume e<pands to 18= litres. *alculate thepressure being e<erted on the balloon at this

altitude.

E'er(ise 4.4

*h l L l

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!2

*harlesL law

1. *harlesL law states that for a -<ed mass of anideal gas3 the %olume is directly proportionalto its absolute temperature at constantpressure.

2. $eri%ation of *harlesLs law equationI> CT 3 where C constant3 therefore

T V ∝

!

!

1

1

T

V

T

V =

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!3

V

T ? '

4. The following graph shows the *harlesL law

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!4

Absolute ero and the elBin s(ale o+ temperature

1. The %alue P27o* is equi%alent to = ' . Thistemperature is Cnown as absolute 6ero because it is the lowest possible temperature

to be reached.

2. Theoretically3 when the gas reaches thistemperature3 the %olume and pressure

become 6ero.

. Actually all gas will become liquid before

reaching this temperature.

4 f the absolute 6ero is taCen as origin the

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!

4. f the absolute 6ero is taCen as origin3 thetemperature scale is Cnown as absolutescale or 'el%in scale.

. The temperature in this scale is Cnown asabsolute temperature and the unit is theCel%in (').

. elationship between the temperature in the*elsius scale and 'el%in scale is as follows:T " #θ $ !%3& '

θ " #T ( !%3& oC

here T > absolute temperature in 'el%in (')

@ > temperature in degree *elsius (o

*)

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!!

. A gas at 2=o* has a %olume 21. litres. Thetemperature is increased to o*. 5hat%olume will the gas occupy at thistemperature

. 0igure shows air trapped inn an emptycontainer /oating on the sea. The %olume ofthe air is 1= cm. what is the %olume of thetrapped air when the temperature of theseawater dro s from 7.=o* to 1.o*J

E'ample 4.4

E'er(ise 4.4

Mressure law

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!"

Mressure law

1. Mressure law states that the pressure of anideal gas is directly proportional to itsabsolute temperature at constant %olume.

2. n short3 or

thereforeI

T P ∝ ) constant*=T P

!

!

1

1

T

P

T

P =

Th f ll i h h th M l

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!#

P

T ? '

. The following graphs show the Mressure law

P

θ + oC

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!$

4. The pressure of nitrogen gas in light bulb is= CMa at 2=o*. *alculate the temperature ofthe gas when the pressure inside the bulbrises to = CMa after the bulb is lighted up.

4. A -<ed %olume gas has pressure of =.8 atmat 2=o*. The gas is heated to 8o*. 5hat isthe new pressureJ

E'ample 4.4

E'er(ise 4.4

Homewor/ Answer

3 213 " ml

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"%

Homewor/

1. 5hat happen to an air bubble released by adi%er under water slowly approaches thesurfaceJ E<plain.

2. Cetch appropriate graphs that show #oyleLsBaw.

. A sample of air at room temperature has a

%olume = ml at CMa pressure. f the airremains at room temperature3 what %olumewill it occupy if the pressure is changed to 4=CMaJ

3 213." ml

At 1 atm pressure and 2o* a gas;-lled balloon

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. At 1 atm pressure and 2 *3 a gas -lled balloonhas a %olume of 2= litres. The balloon is cooleddown to P o*. 5hat would be the %olume of

the balloon now3 assuming the pressureremains constantJ

. Cetch the appropriate graph that shows therelationship between pressure andtemperature in degree *elsius of a gas atconstant %olume.

7. At 2o*3 a sample of gas occupies a containerof constant %olume at a pressure of 82= mm!g. The gas is heated and the pressure Answer

phyf4_chap4

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