States of Matter Gasses

8/11/2019 States of Matter Gasses

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Notes: States of Matter & Gas Laws

States:

There are three common states of matter: solid, liquid and gas. Each state has some general traits whichall tyes of matter in this state follow.

Phases Characteristics:!ensity Shae Thermal E"ansion #omressi$le

Solid %igh fi"ed ery Little ery Little

Liquid Medium ta'es shae of the container ery Little ery Little

Gas E() Low ta'es shae of the container Lots ery Easily

*hat determines a su$stance+s state of matter There are three factors. The first to come to mind is

ro$a$ly temerature. -f you heat something eentually it melts or $oils. The ne"t is not so o$ious, the

ressure. /or e"amle, with a change in atmosheric ressure water $oils at a lower temerature in

!ener than it does in San /rancisco. 0ro$a$ly the most imortant factor is what su$stance we aretal'ing a$out. *ater $oils at a lot lower temerature than titanium.

Phase Diagram:

The hase diagram is a grahical tool used to determine what state of matter a su$stance will $e in a gien

temerature and ressure.

The hase diagram is a series of lines generated from actual e"erimental data. -n other words, someonetoo' a su$stance and chec'ed its state at many different com$inations of temerature and ressure. They

chec'ed the state when the su$stance was at 1# and at 234 mm%g, then at 1# and 235 mm%g, then at

1# and 236 mm%g and so on. 7s you can see it would ta'e a while to chec' all the temerature8ressure

com$inations there are for any gien su$stance.

The lines on the hase diagram diide the three hases. -f you cross a line, a hysical rocess hasoccurred.


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Generally, heating a solid will cause it to melt to a liquid 9at the melting or freeing oint;. #ontinued

heating 9to the $oiling oint; roduces a gas. /or most su$stances, at higher ressures, the freeingand $oiling oints are higher.

There are two secific areas of interest, the trile oint and the critical temerature. The trile oint

occurs at a secific temerature and ressure at which all three hases e"ist in equili$rium. *hen

o$sering a su$stance at its trile oint it aears to $e $oiling freeing and melting all at the same

time.


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Units of Gasses:

a$$reiation name normal atmosheric ressure

torr Torricelli 234

mm%g millimeters of mercury 234

si ounds er square inch 5?.2

7tm atmosheres 5

@ar @ar 5.45

'0a 'iloascal 545.A61

in%g inches of mercury 6B.B

ft %6< feet of water AA

Nature of Gas Pressure:

The ressure of a gas is a measure of force alied oer a gien area. /or e"amle, ounds er square

inch. The force $eing alied is due to the imact of gas articles on some o$>ect.

(ight this second, as you read these notes, $illions of gas articles C N 6,


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Evangelista Torricelli's Barometer:

The $arometer icture here was inented $y Eangelista Torricelli.This was one of the first methods used to measure atmosheric

ressure, and it is still used today. Today, most instruments named

$arometers are not this large and contain no mercury, they wor' onother rincials of hysics. @ut the name has $een alied to all

instruments used to measure atmosheric ressure.

The aaratus is set u $y ta'ing a tu$e filled with %g, lacing your

thum$ oer the to, inerting the tu$e and su$merging the oen endin another container, a dish, also filled with %g. The %g in the tu$e

falls, leaing a acuum at the to of the tu$e. @ut not all of the %g

drains out of the tu$e. *hat force is holding this dense liquid u

against graity+s ull 7ir ressure. Normal atmosheric ressurewill hold a column of %g 234 mm tall. This is one of our units listed

a$oe for atmosheric ressureF it is equialent to the Torr,

o$iously named after Torricelli.

Gas Las:

There are seeral laws used to descri$e the $ehaior of gasses. *hat follows are descritions of the most

imortant. There a$out laws we will discuss. These laws we will $e discussing ressure, temerature,olume and moles or num$ers of gas articles.

*hen doing these ro$lems, 'ee in mind that there are only ? arameters to consider: 0, T, , and n. /or

many of these ro$lems you will $e 'eeing two of the four arameters steady as you modify one of these

four and calculate that modification of the fourth arameter.

Bo!le's La:

This law defines the relationshi $etween ressure and olume if temerature and amount of gas is held

constant. -f the olume of a container is increased, the ressure decreases. -f the olume of a container is

decreased, the ressure increases. The law is descri$ed $y the following equation:

055H 066

E"amle: 7 samle of gas is in a 6.44 L contain at a ressure of 2?4.4 mm%g. *hat is the new ressure

of the samle if the container+s olume is reduced to 5.61 L

7nswer: This ro$lem is soled $y inserting alues into the gien equation:

92?4.4 mm%g; 96.44 L; H9I; 95.61 L;

Soling for I will gie you a new ressure of 55? mm%g.


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Charles's La:

This law defines the relationshi $etween olume and temerature if ressure and amount of articles areheld constant. -f the temerature of a gas is increased, the olume of the gas will increase. -f the

temerature of a gas is decreased, the olume of the gas will decrease. This is a direct relationshi.


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The answer is A55.A K, $ut the question as's for #elsius, so you su$tract 62A to get the final answer of

A.AJ#, $ut then you 'new that. (ight

#voga$ro's La:

This law gies the relationshi $etween olume and num$er of gas articles when ressure andtemerature are held constant. (emem$er the num$er is measured in moles. -f the amount of gas in a

container is increased, the olume increases. -f the amount of gas in a container is decreased, the olume

decreases. 7nother direct relationshi.

The olume of a container holding a gas will increase with increasing num$ers of gas articles $ecause

there are more articles imacting the wall of the container.

E"amle: 7 1.44 L samle of a gas is 'nown to contain 4.B31 mol. -f the amount of gas in this container

is increased to 5.4 mol, what new olume will result 9at an unchanged temerature and ressure;

7nswer:

5n6 H 6n5

91.44 L; 95.4 mol; H 9"; 94.B31 mol;

Com%ine$ Gas La:

To derie the #om$ined Gas Law, follow these stes:

Ste 5: *rite @oyles Law055H 066

Ste 6: Multily $y #harles+s Law

5T6H 6T5

05568 T5H 066

68 T6

Ste A: Multily $y Gay=Lussacs Law05T6H 06T5

0565

68 T56H 06

6668 T6

6

Ste ?: Ta'e the square root to get the com$ined gas law:

0558 T5H 0668 T6

7s a side note:-f we include 7ogadro+s Law the following equation is generated:

0558 n5T5H 0668 n6T6


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E"amle:

7 6.44 L samle of a gas is collected at 61.4J# and 2?1.4 mm%g. *hat is the olume at ST0

)ou hae to recognie that fie alues are gien in the ro$lem and the si"th is the only un'nown. 7lso,

remem$er to change the #elsius temeratures to Kelin.

*hen ro$lems li'e this are soled it is ery helful to write out all the aria$les in the equation as shown

$elow:

Ne"t fill in the data gien in the ro$lem. %ere is the right=hand side filled in with the ST0 alues:

)ou can $e retty sure that the term ST0 will aear in these tyes of ro$lems. - recommend youmemorie these standard conditions.

%ere are all the gien alues:

-nsert the alues in their roer laces in the com$ined gas law equation:

0558 T5H 0668 T6

and sole for ".

P& n(T: The )$eal Gas La:

The deriation of this law is a lot of math. So, - will >ust gie you the equation and e"amles of how touse it.

0 H n(TThe Numerical alue for (:

(s alue can $e determined many ways. This is >ust one way:


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*e will assume we hae 5.444 mol of a gas at ST0. The olume of this amount of gas under the

conditions of ST0 is 'nown to a high degree of recision. *e will use the alue of 66.?5? L.

@y the way, 66.?5? L at ST0 has a name. -t is called molar olume. -t is the olume of 7N) ideal gas at

standard temerature and ressure. 7s far as you are concerned, all the gasses we discuss will $ehae asideal gasses. So, if you hae a samle of gas containing 3.466 " 546Agas article, this samle of gas will

hae a olume of 66.?5? L, or a$out 1 gallons. Thin' a$out an o$>ect that is 1 gallons. - icture an office

water cooler $ottle, or a one of those L7(GE $uc'ets of aint.

Lets lug our num$ers into the equation:

95.444 atm; 966.?5? L; H 95.444 mol; 9(; 962A.51 K;

Notice how atmosheres were used as well as the e"act alue for standard temerature.

Soling for ( gies 4.4643 L atm 8 mol K, when rounded to four significant figures. This is usually

enough. (emem$er the alue. )oull need it for ro$lem soling.

Notice the weird unit on (. Say out loud liter atmosheres er mole Kelin.

This is not the only alue of ( that can e"ist. -t deends on which units you select. Those of you that ta'e

more chemistry or hysics will most li'ely meet u with .A5?1 Doules er mole Kelin, $ut thats foranother course. *e will only use the 4.4643 alue in gas=related ro$lems.

E"amle: 7 samle of gas with a mass of 6.5461 grams is found to occuy a olume of 6.14 L at 66.4J#at a ressure of 2?4.4 mm%g. %ow many moles of the gas are resent

Notice that the units for ressure MST $e in atm., so the 2?4.4 mm %g must $e conerted first.

2?4.4 mm %g 234.4 mm %g8atm H 4.B2A2 atm

%oweer, the unrounded=off alue should $e used in the calculation >ust $elow.

Now, lug into the equation:

94.B2A2 atm; 96.14 L; H 9n; 94.4643 L atm 8 mol K; 96B1.4 K;

and sole for n H 4.551 mol

E"amle: sing the ro$lem a$oe, what is the molar mass of the gas

This is a ery common use of this law and the odds are ery good you will see this tye of question on a

test.

The 'ey is to remem$er the units on molar mass: grams er mole.

*e 'now from the ro$lem statement that 6.5461 grams of the gas is inoled and we also 'now how

many moles that is.

*e 'now that from doing the calculation a$oe and getting 4.55?3 mol.


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So all we hae to do is diide the grams of gas $y how many moles it is:

6.5461 g 4.55?3 mol H 5.A? g8mol

*ith a molar mass of 5.A? g er mol can you ma'e an educated guess as to what gas this might $e

Lets go oer those stes for using the -deal Gas Law to calculate the molar mass of the gas:

5. )ou hae to 'now the grams of gas inoled. sually the ro$lem will >ust gie you the alue, $utnot always. )ou might hae to calculate it.

6. )ou are going to hae to calculate the moles of gas. se 0 H n(T and sole for n. Ma'e sure to use

L, atm and K.

A. !iide grams $y moles and theres your answer.

Dalton's La of Partial Pressures:/or any ure gas 9lets use helium; 0 H n(T holds true. Therefore, 0 is directly roortional to n if and T remain constant. 7s n goes u, so would 0.


10/17

7 container holds three gases: o"ygen, car$on dio"ide, and helium. The artial ressures of the three

gases are 6.44 atm, A.44 atm, and ?.44 atm, resectiely. *hat is the total ressure inside the container

0TH 0


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)ntermolecular +orces:

-ntermolecular forces are the forces of attractions that e"ist $etween molecules in a comound. Thesecause the comound to e"ist in a certain state of matter, solid, liquid or gas, and affect the melting and

$oiling oints of comounds as well as the solu$ilities of one su$stance in another.

The melting oint of a comound is the temerature at which a comound turns from a solid to a liquid or

a liquid to a solid.

The $oiling oint of a comound is the temerature at which a comound turns from a liquid to a gas or agas to a liquid. This temerature is a true measure of the forces of attractions $etween molecules as

molecules searate from one another when they turn from a liquid to a gas.

The stronger the attractions $etween articles, the more difficult it will $e to searate the articles. *hensu$stances melt, the articles are still close to one another $ut the forces of attraction that held the

articles rigidly together in the solid state hae $een sufficiently oercome to allow the articles to moe.

*hen su$stances $oil, the articles are comletely searated from one another and the attractions $etween

molecules are comletely oercome and the articles float away as a gas. The energy required to causesu$stances to melt and to $oil, and thus disrut the forces of attraction, comes from the enironment

surrounding the material. -f you lace a iece of ice in your hand, the ice will melt more quic'ly than if itis laced on a cold counter to. The energy required to melt the ice comes from your hand, your handgets colder and the ice gets warmer.

Loo' at the ta$le of melting oints and $oiling oints for the halogens, shown $elow.

Melting 0oints and @oiling 0oints of Similar Su$stances with -ncreasing /ormula *eights

S@ST7N#E /* 9g8mole; m 9o#; $ 9o#;

/6 A =664 =5

#l6 25 =544.B =A?.3

@r6 534 =2.6 1.2

-6 61? 55A.1 5?.A1

7s the sie of the halogens increases, the melting and $oiling oints increase. The energy required tomoe and searate the molecules from one another increases as the sie of the molecules increases. More

massie molecules hae more inertia, which must first $e oercome $efore the molecules can $e

searated.

-f it ta'es more energy to searate the molecules, the attractions $etween molecules must $e greater. The

tyes of intermolecular forces resonsi$le for the increase in melting oints and $oiling oints of these

non=olar coalent comounds are called disersion forces also named London forces.

Now loo' at the ta$le $elow:

Melting 0oints and @oiling 0oints of Su$stances with Similar /ormula *eights

S@ST7N#E /* 9g8mole; m 9o#; $ 9o#;

/6 A =664 =5

N< A4 =53? =516

#%A


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#a ?4 BA 5??

Na/ ?6 BBA 53B1

7ll the su$stances in this ta$le hae similar formula masses, thus they hae similar disersion forces. -fthe only attractions $etween su$stances hae to do with sie, then they should hae similar melting oints

and $oiling oints. They do not. Let+s loo' more closely at the nature of the su$stance to see if we can

relate the structure of the material with its roerties.

+luorine an$ Nitrogen ,ono-i$e:

/luorine and nitrogen mono"ide are similar in sie and therefore hae similar disersion forces. /luorine

is a non=olar coalent molecule while nitrogen mono"ide is a olar coalent moleculeF it has an oeralldiole. Since nitrogen mono"ide has the higher melting oint and $oiling oint, it must hae the stronger

intermolecular forces. Gien the same sie, olar coalent molecules must hae stronger forces of

attraction than non=olar coalent molecules. These forces of attractions are called $i*ole"$i*ole forces.

Nitrogen ,ono-i$e an$ ,ethanol:

Nitrogen mono"ide and methanol are similar in sie and thus hae similar disersion forces. Nitrogenmono"ide and methanol are olar coalent molecules and thus hae diole=diole forces. Since methanol

has the higher melting oint and $oiling oint, it must hae the stronger intermolecular forces. The

difference in these molecules is the resence of a certain e"tremely olar $ond resent in methanol that is

not resent in nitrogen mono"ide. This is the hydrogen $ond or %=$ond, formed $etween the o"ygen andhydrogen.


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The tyes of interactions resonsi$le for the e"tremely high melting and $oiling oints of metals and ionic

comound are called electrostatic forces and are the strongest of all the intermolecular forces.

-ntermolecular /orces


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6. Now, - in>ect some liquid into the tan', $ut - do not fill the tan' with liquid. *hat will haen to the

liquid

A. Thats right. Some, may$e all, of the liquid will eaorate into gas, filling the emty sace. Now if allthe liquid eaorates, we >ust hae a tan' of gas. Thats not what we want. So lets suose that only

some of the liquid eaorated and that there are $oth the liquid state and the gas state resent in the

tan'.

The gas that is a$oe the liquid is called its aor and it creates a ressure called aor ressure.

*hat - mean is, suose you attached a ressure gauge to the tan', would a gas ressure $e recordedThe answer is )ESQ

%oweer, here is a 'ey oint. The aor must $e in contact with the liquid at all times. (emoe the liquid

and you >ust hae a $o" of gasF you do not hae aor or aor ressure.

Lets emhasie this ointQQQQ /or aor ressure to e"ist, the aor 9gas hase; MST $e in hysicalcontact with the liquid 9or solid; it came from. )ou #7NT hae aor ressure without two hases $eing

resent and in contactQQQQQQ

%ow is aor ressure created 7nother way to ut it = how do molecules of the liquid $ecome molecules

of gas

Each molecule in the liquid has energy, $ut not the same amount. The energy is distri$uted according tothe Ma"well=@oltmann distri$ution. Een if you dont 'now what that is, the oint is that some

molecules hae a fairly large amount of energy comared to the aerage. Those are the ones we are

interested in.

*e are ES0E#-7LL) interested if one of these high energy molecules haens to $e sitting right at thesurface of the water. Now, all the molecules are in motion $ecause of their energy, $ut none hae

sufficient energy to $rea' the mutual attractie force molecules hae for each other. Suose that oursurface molecule was moing u away from the surface 7N! had enough energy to $rea' away from theattractie forces of the molecules around it.

*here would that molecule go -t would continue to moe away from the liquid surface 7N! -T

@E#


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@ut waitQ The aor ressure stos going u and winds u staying at some fi"ed alue. *hats going on

%eres the answer, - hoe you can handle itQ

7s more and more molecules LE7E the surface, what do some start to do Thats right, some (ET(Nto the surface and resume their former life as a liquid molecule. Soon the num$er of molecules in the

aor hase is constant $ecause the rate of returning equals the rate of leaing and so the ressure stays

constant.

This image is my attemt to summarie this rocess. -ll ut it here without comment:

/actors that will affect aor ressure:

5. Temerature of the su$stance

6. The secific su$stance you are studying

5. -f you the temerature you R the 'inetic energy of the gas articles R seed at which these

articles are moing, then the article has a $etter chance of $rea'ing free of the liquid.

6. !ifferent su$stances hae different intensities of self=attraction. The name for these attraction forces

are intermolecular forces.

/eating Diagram:


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This diagram deicts the cure of temerature erses time of a gien su$stance. Starting with a solid that

is melted into a liquid that is $oiled to a aor. 7s heat is added this is the shae of the cure generated./lat lines are due to intermolecular forces $eing $ro'en, solid ice turning to liquid water, liquid water

$eing turned into steam.


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amount of heat needed for melting is the same as that gien off in freeing. Li'ewise, the amount of heat

needed for aoriation is the same as that gien off in condensation.The equation will ta'e the form of one of the following:

H m %a heat H mass heat of aoriation

H m %cond heat H mass heat of condensation

H m %fusheat H mass

heat of fusion H m %sol heat H mass heat of solidification

*ould it ta'e more heat to melt 54 grams of solid water, ice or aorie 54 grams of liquid water Loo'

at the lengths of the flat lines on the a$oe lot.

-t ta'es more heat to $oil something than it does to melt it. Thin' a$out water on the stoeto and an ice

cu$e on the counter. 0lus, loo' at the %eat of aoriation & %eat of /usionF the %eat of /usion is 4

cal8gram and the %eat of aoriation is 1?4 cal8g.

E"amle:

%ow much heat would it ta'e to melt 61 grams of solid water H 61 grams 4 cal8gram H 6444 cal

E"amle:

%ow much heat would it ta'e to aorie 61 grams of liquid water

H 61 grams 1?4 cal8gram H 5A144 cal

Longer 0ro$lem: Start from ice go to steam.

States of Matter Gasses

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