Chemical Funda

8/2/2019 Chemical Funda

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AN INTRODUCTION TO SATURATEDVAPOUR PRESSURE

This page looks at how the equilibrium between a liquid (or a solid)and its vapour leads to the idea of a saturated vapour pressure. It alsolooks at how saturated vapour pressure varies with temperature, andthe relationship between saturated vapour pressure and boiling point.

The origin of saturated vapour pressure

The evaporation of a liquid

The averageenergy of the particles in a liquid is governed by thetemperature. The higher the temperature, the higher the averageenergy. But within that average, some particles have energies higherthan the average, and others have energies lower than the average.

Some of the more energetic particles on the surface of the liquid canbe moving fast enough to escape from the attractive forces holding theliquid together. They evaporate.

The diagram shows a small region of a liquid near its surface.

Notice that evaporation only takes place on the surface of the liquid.That's quite different from boiling which happens when there is enoughenergy to disrupt the attractive forces throughout the liquid. That's why,if you look at boiling water, you see bubbles of gas being formed allthe way through the liquid.

If you look at water which is just evaporating in the sun, you don't see


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any bubbles. Water molecules are simply breaking away from thesurface layer.

Eventually, the water will all evaporate in this way. The energy which islost as the particles evaporate is replaced from the surroundings. As

the molecules in the water jostle with each other, new molecules willgain enough energy to escape from the surface.

The evaporation of a liquid in a closed container

Now imagine what happens if the liquid is in a closed container.Common sense tells you that water in a sealed bottle doesn't seem toevaporate - or at least, it doesn't disappear over time.

But there isconstant evaporation from the surface. Particles continueto break away from the surface of the liquid - but this time they aretrapped in the space above the liquid.

As the gaseous particles bounce around, some of them will hit thesurface of the liquid again, and be trapped there. There will rapidly bean equilibrium set up in which the number of particles leaving thesurface is exactly balanced by the number rejoining it.


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In this equilibrium, there will be a fixed number of the gaseousparticles in the space above the liquid.

When these particles hit the walls of the container, they exert apressure. This pressure is called the saturated vapour pressure(alsoknown as saturation vapour pressure) of the liquid.

Measuring the saturated vapour pressure

It isn't difficult to show the existence of this saturated vapour pressure(and to measure it) using a simple piece of apparatus.

Note: This experiment is much easier to talk about than do, given the safetyproblems in handling mercury because of its poisonous vapour. This isparticularly going to be a problem if you want to find the saturated vapourpressure of a liquid at a higher temperature. You would have to use a morecomplex bit of apparatus. That isn't a problem you need to worry about forUK A level purposes.

If you have a mercury barometer tube in a trough of mercury, at 1atmosphere pressure the column will be 760 mm tall. 1 atmosphere issometimes quoted as 760 mmHg ("millimetres of mercury").


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If you squirt a few drops of liquid into the tube, it will rise to form a thinlayer floating on top of the mercury. Some of the liquid will evaporateand you will get the equilibrium we've just been talking about -provided there is still some liquid on top of the mercury. It is only anequilibrium if both liquid and vapour are present.

The saturated vapour pressure of the liquid will force the mercury leveldown a bit. You can measure the drop - and this gives a value for thesaturated vapour pressure of the liquid at this temperature. In thiscase, the mercury has been forced down by a distance of 760 - 630mm. The saturated vapour pressure of this liquid at the temperature ofthe experiment is 130 mmHg.


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You could convert this into proper SI units (pascals) if you wanted to.760 mmHg is equivalent to 101325 Pa.

A value of 130 mmHg is quite a high vapour pressure if we are talkingabout room temperature. Water's saturated vapour pressure is about

20 mmHg at this temperature. A high vapour pressure means that theliquid must be volatile - molecules escape from its surface relativelyeasily, and aren't very good at sticking back on again either.

That will result in larger numbers of them in the gas state onceequilibrium is reached.

The liquid in the example must have significantly weakerintermolecular forces than water.

The variation of saturated vapour pressure withtemperature

The effect of temperature on the equilibrium between liquid andvapour

You can look at this in two ways.

There is a common sense way. If you increase the temperature, youare increasing the average energy of the particles present. That

means that more of them are likely to have enough energy to escapefrom the surface of the liquid. That will tend to increase the saturatedvapour pressure.

Or you can look at it in terms of Le Chatelier's Principle - which worksjust as well in this kind of physical situation as it does in the morefamiliar chemical examples.

Note: You could follow this link if you aren't sure aboutLe Chatelier'sPrinciple.

Use the BACK button on your browser to return to this page later.

When the space above the liquid is saturated with vapour particles,you have this equilibrium occurring on the surface of the liquid:
http://www.chemguide.co.uk/physical/equilibria/lechatelier.html#tophttp://www.chemguide.co.uk/physical/equilibria/lechatelier.html#tophttp://www.chemguide.co.uk/physical/equilibria/lechatelier.html#tophttp://www.chemguide.co.uk/physical/equilibria/lechatelier.html#tophttp://www.chemguide.co.uk/physical/equilibria/lechatelier.html#tophttp://www.chemguide.co.uk/physical/equilibria/lechatelier.html#top


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The forward change (liquid to vapour) is endothermic. It needs heat toconvert the liquid into the vapour.

According to Le Chatelier, increasing the temperature of a system in adynamic equilibrium favours the endothermic change. That means thatincreasing the temperature increases the amount of vapour present,and so increases the saturated vapour pressure.

The effect of temperature on the saturated vapour pressure ofwater

The graph shows how the saturated vapour pressure (svp) of water

varies from 0C to 100 C. The pressure scale (the vertical one) ismeasured in kilopascals (kPa). 1 atmosphere pressure is 101.325 kPa.

Saturated vapour pressure and boiling point

A liquid boils when its saturated vapour pressure becomes equal to theexternal pressure on the liquid. When that happens, it enables bubblesof vapour to form throughout the liquid - those are the bubbles you seewhen a liquid boils.

If the external pressure is higher than the saturated vapour pressure,these bubbles are prevented from forming, and you just get


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evaporation at the surface of the liquid.

If the liquid is in an open container and exposed to normal atmosphericpressure, the liquid boils when its saturated vapour pressure becomesequal to 1 atmosphere (or 101325 Pa or 101.325 kPa or 760 mmHg).

This happens with water when the temperature reaches 100C.

But at different pressures, water will boil at different temperatures. Forexample, at the top of Mount Everest the pressure is so low that waterwill boil at about 70C. Depressions from the Atlantic can easily lowerthe atmospheric pressure in the UK enough so that water will boil at99C - even lower with very deep depressions.

Whenever we just talk about "the boiling point" of a liquid, we alwaysassume that it is being measured at exactly 1 atmosphere pressure. Inpractice, of course, that is rarely exactly true.

Saturated vapour pressure and solids

Sublimation

Solids can also lose particles from their surface to form a vapour,except that in this case we call the effect sublimationrather thanevaporation. Sublimation is the direct change from solid to vapour (orvice versa) without going through the liquid stage.

In most cases, at ordinary temperatures, the saturated vapourpressures of solids range from low to very, very, very low. The forcesof attraction in many solids are too high to allow much loss of particlesfrom the surface.

However, there are some which do easily form vapours. For example,naphthalene (used in old-fashioned "moth balls" to deter clothesmoths) has quite a strong smell. Molecules must be breaking awayfrom the surface as a vapour, because otherwise you wouldn't be ableto smell it.

Another fairly common example (discussed in detail on another page)is solid carbon dioxide - "dry ice". This never forms a liquid atatmospheric pressure and always converts directly from solid tovapour. That's why it is known as dry ice.

PHASE DIAGRAMS OF PURE


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SUBSTANCES

This page explains how to interpret the phase diagrams for simplepure substances - including a look at the special cases of the phasediagrams of water and carbon dioxide. This is going to be a long page,because I have tried to do the whole thing as gently as possible.

The basic phase diagram

What is a phase?

At its simplest, a phase can be just another term for solid, liquid or gas.

If you have some ice floating in water, you have a solid phase presentand a liquid phase. If there is air above the mixture, then that isanother phase.

But the term can be used more generally than this. For example, oilfloating on water also consists of two phases - in this case, two liquidphases. If the oil and water are contained in a bucket, then the solidbucket is yet another phase. In fact, there might be more than onesolid phase if the handle is attached separately to the bucket ratherthan moulded as a part of the bucket.

You can recognise the presence of the different phases because thereis an obvious boundary between them - a boundary between the solidice and the liquid water, for example, or the boundary between the twoliquids.

Phase diagrams

A phase diagram lets you work out exactly what phases are present atany given temperature and pressure. In the cases we'll be looking at

on this page, the phases will simply be the solid, liquid or vapour (gas)states of a pure substance.

This is the phase diagram for a typical pure substance.


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These diagrams (including this one) are nearly always drawn highlydistorted in order to see what is going on more easily. There areusually two major distortions. We'll discuss these when they becomerelevant.

If you look at the diagram, you will see that there are three lines, threeareas marked "solid", "liquid" and "vapour", and two special pointsmarked "C" and "T".

The three areas

These are easy! Suppose you have a pure substance at three differentsets of conditions of temperature and pressure corresponding to 1, 2and 3 in the next diagram.


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Under the set of conditions at 1 in the diagram, the substance wouldbe a solid because it falls into that area of the phase diagram. At 2, itwould be a liquid; and at 3, it would be a vapour (a gas).

Note: I'm using the terms vapour and gas as if they were interchangeable.There aresubtle differences between them that I'm not ready to explain for awhile yet. Be patient!

Moving from solid to liquid by changing the temperature:

Suppose you had a solid and increased the temperature while keepingthe pressure constant - as shown in the next diagram. As thetemperature increases to the point where it crosses the line, the solidwill turn to liquid. In other words, it melts.


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If you repeated this at a higher fixed pressure, the melting temperaturewould be higher because the line between the solid and liquid areasslopes slightly forward.

Note: This is one of the cases where we distort these diagrams to make themeasier to discuss. This line is much more vertical in practice than we normallydraw it. There would be very little change in melting point at a higher pressure.The diagram would be very difficult to follow if we didn't exaggerate it a bit.


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So what actually is this line separating the solid and liquid areas of thediagram?

It simply shows the effect of pressure on melting point.

Anywhere on this line, there is an equilibrium between solid and liquid.

You can apply Le Chatelier's Principle to this equilibrium just as if itwas a chemical equilibrium. If you increase the pressure, theequilibrium will move in such a way as to counter the change you have

just made.

If it converted from liquid to solid, the pressure would tend to decreaseagain because the solid takes up slightly less space for mostsubstances.

That means that increasing the pressure on the equilibrium mixture ofsolid and liquid at its original melting point will convert the mixture backinto the solid again. In other words, it will no longer melt at thistemperature.

To make it melt at this higher pressure, you will have to increase thetemperature a bit. Raising the pressure raises the melting point ofmost solids. That's why the melting point line slopes forward for mostsubstances.

Moving from solid to liquid by changing the pressure:

You can also play around with this by looking at what happens if youdecrease the pressure on a solid at constant temperature.

Note: You have got to be a bit careful about this, because exactly what happensif you decrease the pressure depends on exactly what your starting conditionsare. We'll talk some more about this when we look at the line separating the solidregion from the vapour region.


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Moving from liquid to vapour:

In the same sort of way, you can do this either by changing thetemperature or the pressure.

The liquid will change to a vapour - it boils - when it crosses theboundary line between the two areas. If it is temperature that you arevarying, you can easily read off the boiling temperature from the phasediagram. In the diagram above, it is the temperature where the redarrow crosses the boundary line.

So, again, what is the significance of this line separating the two


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areas?

Anywhere along this line, there will be an equilibrium between theliquid and the vapour. The line is most easily seen as the effect ofpressure on the boiling point of the liquid.

As the pressure increases, so the boiling point increases.

Note: I don't want to make any very big deal over this, but this line is actuallyexactly the same as the graph for the effect of temperature on the saturatedvapour pressure of the liquid.Saturated vapour pressureis dealt with on aseparate page. A liquid will boil when its saturated vapour pressure is equal tothe external pressure.

Suppose you measured the saturated vapour pressure of a liquid at 50C, and itturned out to be 75 kPa. You could plot that as one point on a vapour pressurecurve, and then go on to measure other saturated vapour pressures at different

temperatures and plot those as well.

Now, suppose that you had the liquid exposed to a total external pressure of 75kPa, and gradually increased the temperature. The liquid would boil when itssaturated vapour pressure became equal to the external pressure - in this caseat 50C. If you have the complete vapour pressure curve, you could equally wellfind the boiling point corresponding to any other external pressure.

That means that the plot of saturated vapour pressure against temperature isexactly the same as the curve relating boiling point and external pressure - theyare just two ways of looking at the same thing.

If all you are interested in doing is interpreting one of these phase diagrams, youprobably don't have to worry too much about this.

The critical point

You will have noticed that this liquid-vapour equilibrium curve has atop limit that I have labelled as C in the phase diagram.

This is known as the critical point. The temperature and pressure

corresponding to this are known as the criticaltemperatureandcritical pressure.

If you increase the pressure on a gas (vapour) at a temperature lowerthan the critical temperature, you will eventually cross the liquid-vapourequilibrium line and the vapour will condense to give a liquid.
http://www.chemguide.co.uk/physical/phaseeqia/vapourpress.html#tophttp://www.chemguide.co.uk/physical/phaseeqia/vapourpress.html#tophttp://www.chemguide.co.uk/physical/phaseeqia/vapourpress.html#tophttp://www.chemguide.co.uk/physical/phaseeqia/vapourpress.html#top


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This works fine as long as the gas is below the critical temperature.What, though, if your temperature was above the critical temperature?There wouldn't be any line to cross!

That is because, above the critical temperature, it is impossible tocondense a gas into a liquid just by increasing the pressure. All youget is a highly compressed gas. The particles have too much energyfor the intermolecular attractions to hold them together as a liquid.

The critical temperature obviously varies from substance to substanceand depends on the strength of the attractions between the particles.

The stronger the intermolecular attractions, the higher the criticaltemperature.

Note: This is now a good point for a quick comment about the use of the words"gas" and "vapour". To a large extent you just use the term which feels right. Youdon't usually talk about "ethanol gas", although you wouldsay "ethanol vapour".Equally, you wouldn't talk about oxygen as being a vapour - you always call it agas.

There are various guide-lines that you can use if you want to. For example, if thesubstance is commonly a liquid at or around room temperature, you tend to callwhat comes away from it a vapour. A slightly wider use would be to call it a

vapour if the substance is below its critical point, and a gas if it is above it.Certainly it would be unusual to call anything a vapour if it was above its criticalpoint at room temperature - oxygen or nitrogen or hydrogen, for example. Thesewould all be described as gases.

This is absolutely NOT something that is at all worth getting worked up about!


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Moving from solid to vapour:

There's just one more line to look at on the phase diagram. This is the

line in the bottom left-hand corner between the solid and vapour areas.

That line represents solid-vapour equilibrium. If the conditions oftemperature and pressure fell exactly on that line, there would be solidand vapour in equilibrium with each other - the solid would besubliming. (Sublimation is the change directly from solid to vapour orvice versa without going through the liquid phase.)

Once again, you can cross that line by either increasing thetemperature of the solid, or decreasing the pressure.

The diagram shows the effect of increasing the temperature of a solidat a (probably very low) constant pressure. The pressure obviouslyhas to be low enough that a liquid can't form - in other words, it has tohappen below the point labelled as T.

You could read the sublimation temperature off the diagram. It will bethe temperature at which the line is crossed.

The triple point


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Point T on the diagram is called the triple point.

If you think about the three lines which meet at that point, theyrepresent conditions of:

solid-liquid equilibrium liquid-vapour equilibrium solid-vapour equilibrium

Where all three lines meet, you must have a unique combination oftemperature and pressure where all three phases are in equilibriumtogether. That's why it is called a triplepoint.

If you controlled the conditions of temperature and pressure in order toland on this point, you would see an equilibrium which involved thesolid melting and subliming, and the liquid in contact with it boiling to

produce a vapour - and all the reverse changes happening as well.

If you held the temperature and pressure at those values, and kept thesystem closed so that nothing escaped, that's how it would stay. Astrange set of affairs!

Normal melting and boiling points

The normal melting and boiling points are those when the pressure is

1 atmosphere. These can be found from the phase diagram bydrawing a line across at 1 atmosphere pressure.

AN INTRODUCTION TO KINETICTHEORY

This page takes a simple look at solids, liquids and gases, andchanges of state such as melting and boiling, in terms of the

behaviour of the particles present.

The arrangements of particles in solids, liquids andgases


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A simple view of the arrangement of the particles in solids,liquids and gases looks like this:

Solids

In the solid, the particles are touching, and the only motionallowed to them is vibration. The particles may be arrangedregularly (in which case, the solid is crystalline), or at random

(giving waxy solids like candles or some forms of polythene, forexample).

The particles are held in the solid by forces which depend on theactual substance - ionic bonds, covalent bonds, metallic bonds,hydrogen bonds or van der Waals attractions.

Liquids

In a liquid, the particles are mainly touching, but some gapshave appeared in the structure. These gaps allow the particles

to move, and so the particles are arranged randomly.The forcesthat held the solid particles together are also present in the liquid(unless melting has broken up a substance consisting only ofcovalent bonds - a giant covalent structure). However, theparticles in the liquid have enough energy to prevent the forcesholding them in a fixed arrangement.

For most liquids, the density of the liquid is slightly less than thatof the solid, but there isn't much difference. That means that theparticles in the liquid are almost as close together as they are ina solid. If you draw diagrams of liquids, make sure that most of

the particles are touching, but at random, with a few gaps.

Note: Water is an exception to this. Ice floats on water, and so the liquidmust be denser than the solid. You will find an explanation of this on apage aboutmolecular structures. If you choose to follow this link, use theBACK button on your browser to come back here afterwards.
http://www.chemguide.co.uk/atoms/structures/molecular.htmlhttp://www.chemguide.co.uk/atoms/structures/molecular.htmlhttp://www.chemguide.co.uk/atoms/structures/molecular.htmlhttp://www.chemguide.co.uk/atoms/structures/molecular.html


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Gases

In a gas, the particles are entirely free to move. At ordinarypressures, the distance between individual particles is of theorder of ten times the diameter of the particles. At that distance,any attractions between the particles are fairly negligible atordinary temperatures and pressures.

Changes of state

Melting and freezing

If energy is supplied by heating a solid, the heat energy causesstronger vibrations until the particles eventually have enough

energy to break away from the solid arrangement to form aliquid. The heat energy required to convert 1 mole of solid into aliquid at its melting point is called the enthalpy of fusion.

When a liquid freezes, the reverse happens. At sometemperature, the motion of the particles is slow enough for theforces of attraction to be able to hold the particles as a solid. Asthe new bonds are formed, heat energy is evolved.

Boiling and condensing

If more heat energy is supplied, the particles eventually movefast enough to break all the attractions between them, and theliquid boils. The heat energy required to convert 1 mole of liquidinto a gas at its boiling point is called the enthalpy ofvaporisation.

If the gas is cooled, at some temperature the gas particles willslow down enough for the attractions to become effectiveenough to condense it back into a liquid. Again, as those forcesare re-established, heat energy is released.

Remember: Breaking bonds needs energy; making bonds releases it.

The evaporation of a liquid

The averageenergy of the particles in a liquid is governed by


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the temperature. The higher the temperature, the higher theaverage energy. But within that average, some particles haveenergies higher than the average, and others have energieslower than the average.

Some of the more energetic particles on the surface of the liquidcan be moving fast enough to escape from the attractive forcesholding the liquid together. They evaporate.

The diagram shows a small region of a liquid near its surface.

Notice that evaporation only takes place on the surface of theliquid. That's quite different from boiling which happens whenthere is enough energy to disrupt the attractive forcesthroughout the liquid. That's why, if you look at boiling water, yousee bubbles of gas being formed all the way through the liquid.

If you look at water which is just evaporating in the sun, youdon't see any bubbles. Water molecules are simply breakingaway from the surface layer.

Eventually, the water will all evaporate in this way. The energywhich is lost as the particles evaporate is replaced from thesurroundings. As the molecules in the water jostle with eachother, new molecules will gain enough energy to escape fromthe surface.

The evaporation of a liquid in a closed container

Now imagine what happens if the liquid is in a closed container.Common sense tells you that water in a sealed bottle doesn'tseem to evaporate - or at least, it doesn't disappear over time.

But there isconstant evaporation from the surface. Particles


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continue to break away from the surface of the liquid - but thistime they are trapped in the space above the liquid.

As the gaseous particles bounce around, some of them will hitthe surface of the liquid again, and be trapped there. There willrapidly be an equilibrium set up in which the number of particlesleaving the surface is exactly balanced by the number rejoiningit.

In this equilibrium, there will be a fixed number of the gaseous

particles in the space above the liquid.

When these particles hit the walls of the container, they exert apressure. This pressure is called the saturated vapourpressure(also known as saturation vapour pressure) of theliquid.


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Sublimation

Solids can also lose particles from their surface to form avapour, except that in this case we call theeffect sublimationrather than evaporation. Sublimation is thedirect change from solid to vapour (or vice versa) without goingthrough the liquid stage.

In most cases, at ordinary temperatures, the saturated vapourpressures of solids range from low to very, very, very low. Theforces of attraction in many solids are too high to allow muchloss of particles from the surface.

However, there are some which do easily form vapours. Forexample, naphthalene (used in old-fashioned "moth balls" to

deter clothes moths) has quite a strong smell. Molecules mustbe breaking away from the surface as a vapour, becauseotherwise you wouldn't be able to smell it.

Another fairly common example (discussed in detail elsewhereon the site) is solid carbon dioxide - "dry ice". This never forms aliquid at atmospheric pressure and always converts directly fromsolid to vapour. That's why it is known as dry ice.


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The phase diagram for water

There is only one difference between this and the phase diagram thatwe've looked at up to now. The solid-liquid equilibrium line (the meltingpoint line) slopes backwards rather than forwards.

In the case of water, the melting point gets lower at higher pressures.Why?

If you have this equilibrium and increase the pressure on it, accordingto Le Chatelier's Principle the equilibrium will move to reduce thepressure again. That means that it will move to the side with thesmaller volume. Liquid water is produced.

To make the liquid water freeze again at this higher pressure, you willhave to reduce the temperature. Higher pressures mean lower melting(freezing) points.

Now lets put some numbers on the diagram to show the exactpositions of the critical point and triple point for water.


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Notice that the triple point for water occurs at a very low pressure.Notice also that the critical temperature is 374C. It would beimpossible to convert water from a gas to a liquid by compressing itabove this temperature.

The normal melting and boiling points of water are found in exactly thesame way as we have already discussed - by seeing where the 1atmosphere pressure line crosses the solid-liquid and then the liquid-vapour equilibrium lines.

Just one final example of using this diagram (because it appeals tome). Imagine lowering the pressure on liquid water along the line in thediagram below.


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The phase diagram shows that the water would first freeze to form iceas it crossed into the solid area. When the pressure fell low enough,the ice would then sublime to give water vapour. In other words, thechange is from liquid to solid to vapour. I find that satisfyingly bizarre!

The phase diagram for carbon dioxide

The only thing special about this phase diagram is the position of thetriple point which is well above atmospheric pressure. It is impossibleto get any liquid carbon dioxide at pressures less than 5.11


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atmospheres.

That means that at 1 atmosphere pressure, carbon dioxide will sublimeat a temperature of -78C.

This is the reason that solid carbon dioxide is often known as "dry ice".You can't get liquid carbon dioxide under normal conditions - only thesolid or the vapour.

IDEAL GASES AND THE IDEAL GAS LAW

This page looks at the assumptions which are made in the KineticTheory about ideal gases, and takes an introductory look at the

Ideal Gas Law: pV = nRT. This is intended only as an introductionsuitable for chemistry students at about UK A level standard (for 16- 18 year olds), and so there is no attempt to derive the ideal gaslaw using physics-style calculations.

Kinetic Theory assumptions about ideal gases

There is no such thing as an ideal gas, of course, but many gasesbehave approximately as if they were ideal at ordinary working

temperatures and pressures. Real gases are dealt with in moredetail on another page.

The assumptions are:

Gases are made up of molecules which are in constantrandom motion in straight lines.

The molecules behave as rigid spheres. Pressure is due to collisions between the molecules and the

walls of the container. All collisions, both between the molecules themselves, and

between the molecules and the walls of the container, areperfectly elastic. (That means that there is no loss of kineticenergy during the collision.)

The temperature of the gas is proportional to the averagekinetic energy of the molecules.

And then two absolutely key assumptions, because these are the


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two most important ways in which real gases differ from idealgases:

There are no (or entirely negligible) intermolecular forcesbetween the gas molecules.

The volume occupied by the molecules themselves isentirely negligible relative to the volume of the container.

The Ideal Gas Equation

The ideal gas equation is:

pV = nRT

On the whole, this is an easy equation to remember and use. Theproblems lie almost entirely in the units. I am assuming below thatyou are working in strict SI units (as you will be if you are doing aUK-based exam, for example).

Exploring the various terms

Pressure, p

Pressure is measured in pascals, Pa - sometimes expressed asnewtons per square metre, N m-2. These mean exactly the samething.

Be careful if you are given pressures in kPa (kilopascals). Forexample, 150 kPa is 150,000 Pa. You must make that conversionbefore you use the ideal gas equation.

Should you want to convert from other pressure measurements:

1 atmosphere = 101,325 Pa 1 bar = 100 kPa = 100,000 Pa

Volume, V

This is the most likely place for you to go wrong when you use this


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equation. That's because the SI unit of volume is the cubic metre,m3 - notcm3 or dm3.

1 m3 = 1000 dm3 = 1,000,000 cm3

So if you are inserting values of volume into the equation, you firsthave to convert them into cubic metres.

You would have to divide a volume in dm3 by 1000, or in cm3 by amillion.

Similarly, if you are working out a volume using the equation,remember to covert the answer in cubic metres into dm3 or cm3 ifyou need to - this time by multiplying by a 1000 or a million.

If you get this wrong, you are going to end up with a silly answer,out by a factor of a thousand or a million. So it is usually fairlyobvious if you have done something wrong, and you can checkback again.

Number of moles, n

This is easy, of course - it is just a number. You already know thatyou work it out by dividing the mass in grams by the mass of onemole in grams.

You will most often use the ideal gas equation by first making thesubstitution to give:

I don't recommend that you remember the ideal gas equation in thisform, but you must be confident that you can convert it into thisform.

The gas constant, R

A value for R will be given you if you need it, or you can look it up ina data source. The SI value for R is 8.31441 J K -1 mol-1.


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Note: You may come across other values for this with different units. Acommonly used one in the past was 82.053 cm

3atm K

-1mol

-1. The units

tell you that the volume would be in cubic centimetres and the pressure inatmospheres. Unfortunately the units in the SI version aren't so obviouslyhelpful.

The temperature, T

The temperature has to be in kelvin. Don't forget to add 273 if youare given a temperature in degrees Celsius.

Using the ideal gas equation

Calculations using the ideal gas equation are included in mycalculations book (see the link at the very bottom of the page), andI can't repeat them here. There are, however, a couple ofcalculations that I haven't done in the book which give a reasonableidea of how the ideal gas equation works.

The molar volume at stp

If you have done simple calculations from equations, you haveprobably used the molar volume of a gas.

1 mole of any gas occupies 22.4 dm3 at stp (standard temperatureand pressure, taken as 0C and 1 atmosphere pressure). You mayalso have used a value of 24.0 dm3 at room temperature andpressure (taken as about 20C and 1 atmosphere).

These figures are actually only true for an ideal gas, and we'll havea look at where they come from.

We can use the ideal gas equation to calculate the volume of 1mole of an ideal gas at 0C and 1 atmosphere pressure.

First, we have to get the units right.

0C is 273 K. T = 273 K


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1 atmosphere = 101325 Pa. p = 101325 Pa

We know that n = 1, because we are trying to calculate the volumeof 1 mole of gas.

And, finally, R = 8.31441 J K -1 mol-1.

Slotting all of this into the ideal gas equation and then rearranging itgives:

And finally, because we are interested in the volume in cubicdecimetres, you have to remember to multiply this by 1000 toconvert from cubic metres into cubic decimetres.

The molar volume of an ideal gas is therefore 22.4 dm3 at stp.

And, of course, you could redo this calculation to find the volume of1 mole of an ideal gas at room temperature and pressure - or anyother temperature and pressure.

Finding the relative formula mass of a gas from its density

This is about as tricky as it gets using the ideal gas equation.

The density of ethane is 1.264 g dm-3 at 20C and 1 atmosphere.Calculate the relative formula mass of ethane.

The density value means that 1 dm3 of ethane weighs 1.264 g.

Again, before we do anything else, get the awkward units sorted

out.

A pressure of 1 atmosphere is 101325 Pa.

The volume of 1 dm3 has to be converted to cubic metres, bydividing by 1000. We have a volume of 0.001 m3.


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The temperature is 293 K.

Now put all the numbers into the form of the ideal gas equationwhich lets you work with masses, and rearrange it to work out themass of 1 mole.

The mass of 1 mole of anything is simply the relative formula massin grams.

So the relative formula mass of ethane is 30.4, to 3 sig figs.

Now, if you add up the relative formula mass of ethane, C2H6 usingaccurate values of relative atomic masses, you get an answer of30.07 to 4 significant figures. Which is different from our answer -so what's wrong?

There are two possibilities.

The density value I have used may not be correct. I did thesum again using a slightly different value quoted at adifferent temperature from another source. This time I got ananswer of 30.3. So the density values may not be entirelyaccurate, but they are both giving much the same sort ofanswer.

Ethane isn't an ideal gas. Well, of course it isn't an ideal gas- there's no such thing! However, assuming that the densityvalues are close to correct, the error is within 1% of what you

would expect. So although ethane isn't exactly behaving likean ideal gas, it isn't far off.

If you need to know about real gases, now is a good time to readabout them.


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REAL GASES

This page looks at how and why real gases differ from idealgases, and takes a brief look at the van der Waals equation. Ifyou have come straight to this page via a search engine, it mightbe a good idea to read the page aboutideal gasesfirst.

Real gases v ideal gases

Real gases and the molar volume

I want to use this to illustrate the slight differences between thenumerical properties of real and ideal gases at normaltemperatures and pressures.

If you have read the page about ideal gases, you will rememberthat we used the ideal gas equation to work out a value for themolar volume of an ideal gas at stp (standard temperature andpressure).

If you know the density of a gas at a particular temperature andpressure, it is very easy to work out its molar volume.

For example, at 273 K and 1 atmosphere pressure, the densityof helium is 0.1785 g dm-3.

That means that 0.1785 g of helium occupies 1 dm3 at stp. It is afairly simple sum to work out what 1 mole of helium, He, wouldoccupy.

1 mole of He weighs 4 g, and would occupy 4 / 0.1785 dm3 =22.4 dm3.

That's the same (at least to 3 significant figures) as the ideal gasvalue, suggesting that helium behaves as an ideal gas underthese conditions.

Note: If your maths isn't very good, and you can't understand why Idivided 4 by 0.1785 to get the answer, think of it like this:
http://www.chemguide.co.uk/physical/kt/idealgases.html#tophttp://www.chemguide.co.uk/physical/kt/idealgases.html#tophttp://www.chemguide.co.uk/physical/kt/idealgases.html#tophttp://www.chemguide.co.uk/physical/kt/idealgases.html#top


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Replace the awkward value of 0.1785 by something simple, like 2. If 2 g ofHe occupied 1 dm

3, what would 4 g occupy? It is obviously twice as much,

2 dm3

- but how did you get at that mathematically? You found out howmany times 2 would go into 4 - in other words, you divided 4 by 2. Do

exactly the same with the more complicated number.

If you do this for a random sample of other gases, you get thesevalues (to 3 significant figures) for the molar volume at stp (273K and 1 atmosphere pressure).

density (g dm-3) molar volume at stpHe 0.1785 22.4N2 1.2506 22.4O2 1.4290 22.4CH4 0.717 22.3CO2 1.977 22.3C2H4 1.260 22.2NH3 0.769 22.1SO2 2.926 21.9

So although for simple calculation purposes we use the value22.4 dm3 for all gases, you can see that it isn't exactly true. Evenat ordinary temperatures and pressures, real gases can deviate

slightly from the ideal value. The effect is much greater undermore extreme conditions, as we will see next.


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Compression factors

For an ideal gas, pV = nRT. If pV and nRT are the same, and

you divide one by the other, then the answer will, of course, be1. For real gases, pV doesn't equal nRT, and so the value will besomething different.

The term pV / nRT is called the compression factor. Thegraphs below show how this varies for nitrogen as you changethe temperature and the pressure.

Note: These diagrams were generated using data produced fromPatrickBarrie's program, and converted into graphs using Excel. The figures arederived from the van der Waals equation - not because it is the bestsource, but because it is the only one you are likely to come across at thislevel, and I shall mention it below. If you wanted to play around with some

of the other equations, you would find that the results produce similarlyshaped curves, but the absolute sizes of the deviations would vary.

If nitrogen was an ideal gas under all conditions of temperatureand pressure, every one of these curves would be a horizontal
http://www.ceb.cam.ac.uk/thermo/http://www.ceb.cam.ac.uk/thermo/http://www.ceb.cam.ac.uk/thermo/http://www.ceb.cam.ac.uk/thermo/http://www.ceb.cam.ac.uk/thermo/http://www.ceb.cam.ac.uk/thermo/


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straight line showing a compression factor of 1. That's obviouslynot true!

Things to notice

At low pressures of about 1 bar (100 kPa - just a bit lessthan 1 atmosphere), the compression factor is close to 1.Nitrogen approximates to ideal behaviour at ordinarypressures.

The non-ideal behaviour gets worse at lowertemperatures. For temperatures of 300 or 400 K, thecompression factor is close to 1 over quite a largepressure range. The nitrogen becomes more ideal over agreater pressure range as the temperature rises.

The non-ideal behaviour gets worse at higher pressures. There must be at least two different effects causing these

deviations. There must be at least one effect causing thepV / nRT ratio to be too low, especially at lowtemperatures. And there must be at least one effectcausing it to get too high as pressure increases. We willexplore those effects in a while.

Other gases

Is the same behaviour shown by other gases? The next diagramshows how the compression factors vary with pressure for avariety of gases at a fixed temperature.


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If you were to redo the set of original nitrogen graphs (at varyingtemperatures) for any of these other gases, you would find thateach of them will produce a set of curves similar to the nitrogenones. What varies is the temperature at which the differentgraph shapes occur.

For example, if you look at the carbon dioxide graph at 273 K, itlooks similar to the nitrogen one at 100 K from the first set ofcurves, although it doesn't increase so steeply at higherpressures.

It is easy to say that gases become less ideal at lowtemperatures, but what counts as a low temperature varies fromgas to gas. The closer you get to to the temperature at which thegas would turn into a liquid (or, in the case of carbon dioxide, asolid), the more non-ideal the gas becomes.

What causes non-ideal behaviour?

In the compression factor expression, pV / nRT, everything onthe bottom of the expression is either known or can be


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measured accurately. But that's not true of pressure andvolume.

In the assumptions we make about ideal gases, there are two

statements which say things which can't be true of a real gas,and these have an effect on both pressure and volume.

The volume problem

The kinetic theory assumes that, for an ideal gas, the volumetaken up by the molecules themselves is entirely negligiblecompared with the volume of the container.

For a real gas, that assumption isn't true. The moleculesthemselves do take up a proportion of the space in thecontainer. The space in the container available for things tomove around in is less than the measured volume of thecontainer.

This problem gets worse the more the gas is compressed. If thepressure is low, the volume taken up by the actualmolecules isinsignificant compared with the total volume of thecontainer.

But as the gas gets more compressed, the proportion of the totalvolume that the molecules themselves take up gets higher andhigher. You could imagine compressing it so much that themolecules were actually all touching each other. At that point the


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volume available for them to move around in is zero!

Suppose at some high pressure, you measure the volume of thecontainer as, say, 1000 cm3, but suppose the molecules

themselves occupy as much as 100 cm

3

of it.

The ideal gas equation was worked out by doing calculationsbased on Kinetic Theory assumptions. The V in pV is assumedto be the volume which the molecules are free to move aroundin - but in this case, it would only be 900 cm3, not 1000 cm3.

If you worked out the compression factor, pV / nRT, by puttingthe total volume of the container into the formula, the answer isbound to be higher than it ought to be. It doesn't allow for thevolume taken up by the molecules themselves.

Let's just repeat one of the earlier diagrams so that you can seethis effect in operation.

For an ideal gas, the compression factor would be 1 over thewhole pressure range. For a real gas like nitrogen, notice howthe compression factor tends to increase with pressure.


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The value of the compression factor is too high at highpressures for a real gas. The reason for that is that themeasured volume that you put into the expression is too highbecause you aren't allowing for the volume taken up by the

molecules. That error gets relatively worse the morecompressed the gas becomes.

The pressure problem

Another key assumption of the Kinetic Theory for ideal gases isthat there are no intermolecular forces between the molecules.That is wrong for every real gas.

If there weren't any intermolecular forces then it would beimpossible to condense the gas as a liquid. Even helium, withthe weakest of all intermolecular forces, can be turned to a liquidif the temperature is low enough.

So what effect do intermolecular forces have?

For a gas molecule in the middle of the gas, there is no neteffect. It will be attracted to some extent to all the othermolecules around it, but, on average, those attractions willcancel each other out. Attractions from behind a molecule,tending to slow it down, will be cancelled out by attractions fromin front of it, tending to speed it up.

Despite all the intermolecular forces it is experiencing, themolecule picked out in green will just continue to move in the


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same direction at the same speed.

That's different if the molecule is just about to hit the wall of thecontainer.

Now there aren't any gas molecules in front of it, and the net pullis backwards into the body of the gas. The molecule will slowdown just before it hits the wall.

If it slows down, it will hit the wall less hard, and so exert lesspressure.

The overall effect of this is to make the measured pressure less

than it would be if the gas was ideal. That means that if you putthe measured pressure into the expression pV / nRT, the valueof the compression factor will be less than it would be if the gaswas ideal.

This is why, under some conditions, graphs of compressionfactors drop below the ideal value of 1.

Look yet again at the nitrogen curves:


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This effect is most important at low temperatures. Why is that?

At lower temperatures, the molecules are moving more slowlyon average. Any pull they feel back into the gas will haverelatively more effect on a slow moving particle than a faster

one.

At higher temperatures, where the molecules are moving a lotfaster, any small pull back into the body of the gas is hardlygoing to be noticeable. At high temperatures, the effect ofintermolecular forces is indeed negligible.

And there is one final effect concerning intermolecular forceswhich is slightly more hidden away.

As pressure increases, the molecules are forced more closely

together. If they are closer, the intermolecular forces will becomemore important. So, as pressure increases, you would expectmore lowering of the compression factor relative to the idealcase. The molecules which slow down the one just about to hitthe wall will be closer to it, and so more effective.


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Is that what happens? Yes, up to a point.

Look again at the nitrogen curve at 100 K. As the pressureincreases, at first the value of the compression factor falls.

But it soon starts to rise again. Why? Because at this point, theeffect of the size of the molecules starts to become moreimportant - and as the pressure is increased even more, thisother effect becomes dominant.

Which is the most ideal gas?

You are looking for a gas with the smallest possible molecules,and the lowest possible intermolecular forces. That is helium.

A helium molecule consists of a single small atom, and the vander Waals dispersion forces are as low as it is possible for themto be.

Note: If you aren't happy about the factors which affect the size ofvander Waals dispersion forces, follow this link. Use the BACK button tocome back to this page afterwards.

Other gas laws - Boyle's Law andCharles' Law

This page takes a simple look at Boyle's Law and Charles'Law, and is suitable for 16 - 18 year old chemistry studentsdoing a course the equivalent of UK A level. The aim issimply to show how these laws relate to Kinetic Theory (in anon-mathematical way), and to the ideal gas equation.

Before you waste time on this, be sure that you actually needto know about it. Certainly in the UK exam system, it is prettyrare for chemistry students to be expected to know either ofthese laws these days. They have been almost completely
http://www.chemguide.co.uk/atoms/bonding/vdw.html#tophttp://www.chemguide.co.uk/atoms/bonding/vdw.html#tophttp://www.chemguide.co.uk/atoms/bonding/vdw.html#tophttp://www.chemguide.co.uk/atoms/bonding/vdw.html#tophttp://www.chemguide.co.uk/atoms/bonding/vdw.html#tophttp://www.chemguide.co.uk/atoms/bonding/vdw.html#top


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replaced by the ideal gas equation.

Boyle's LawStatement

For a fixed mass of gas at constant temperature,the volume is inversely proportional to thepressure.

That means that, for example, if you double the pressure, you will

halve the volume. If you increase the pressure 10 times, the

volume will decrease 10 times.

You can express this mathematically as

pV = constant

Is this consistent with pV = nRT ?

You have a fixed mass of gas, so n (the number ofmoles) is constant.

R is always constant - it is called the gas constant.

Boyle's Law demands that temperature is constant aswell.

That means that everything on the right-hand side of pV =nRT is constant, and so pV is constant - which is what wehave just said is a result of Boyle's Law.

Simple Kinetic Theory explanation

I'm not going to try to prove the relationship between pressureand volume mathematically - I'm just showing that it isreasonable.

This is easiest to see if you think about the effect ofdecreasing the volume of a fixed mass of gas at constant


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temperature.

Pressure is caused by gas molecules hitting the walls of thecontainer.

With a smaller volume, the gas molecules will hit the wallsmore frequently, and so the pressure increases.

You might argue that this isn't actually what Boyle's Law says- it wants you to increase the pressure first and see whateffect that has on the volume. But, in fact, it amounts to thesame thing.

If you want to increase the pressure of a fixed mass of gaswithout changing the temperature, the only way you can do it

is to squeeze it into a smaller volume. That causes themolecules to hit the walls more often, and so the pressureincreases.

Charles' Law

Statement

For a fixed mass of gas at constant pressure, the

volume is directly proportional to the kelvintemperature.

That means, for example, that if you double the kelvintemperature from, say to 300 K to 600 K, at constantpressure, the volume of a fixed mass of the gas will double aswell.

You can express this mathematically as

V = constant x T

Is this consistent with pV = nRT ?

You have a fixed mass of gas, so n (the number ofmoles) is constant.


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R is the gas constant. Charles' Law demands that pressure is constant as

well.

If you rearrange the pV = nRT equation by dividing both sidesby p, you will get

V = nR/p x T

But everything in the nR/p part of this is constant.

That means that V = constant x T, which is Charles' Law.

Simple Kinetic Theory explanation

Again, I'm not trying to prove the relationship betweenpressure and volume mathematically - just that it isreasonable.

Suppose you have a fixed mass of gas in a container with amoveable barrier - something like a gas syringe, for example.The barrier can move without any sort of resistance.

The barrier will settle so that the pressure inside and outsideis identical.

Now suppose you heat the gas, but notthe air outside.

The gas molecules will now be moving faster, and so will hit


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the barrier more frequently, and harder. Meanwhile, the airmolecules on the outside are hitting it exactly as before.

Obviously, the barrier will be forced to the right, and the

volume of the gas will increase. That will go on until thepressure inside and outside is the same. In other words, thepressure of the gas will be back to the same as the air again.

So we have fulfilled what Charles' Law says. We have a fixedmass of gas (nothing has been added, and nothing hasescaped). The pressure is the same before and after (in eachcase, the same as the external air pressure). And the volumeincreases when you increase the temperature of the gas.

What we haven't shown, of course, is that there is a "directly

proportional" relationship. It can be done, but it needs somemaths.

Chemical Funda

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