Urine Formation

Created by Musango07/04/11

The kidney converts blood plasma to urine in three stages

1. Glomerular filtration
2. Tubular reabsorption and secretion
3. Water conservation 

 Glomerular Filtration

•As fluid travels through the nephron, its composition changes, thus its name

–Glomerular filtrate- fluid in the capsular space that is similar to blood plasma except that it has almost no protein

–Tubular fluid- fluid from the PCT through the DCT

•Differs from the glomerular filtrate in that substances are removed and added by the tubule cells

–Urine- fluid that enters the collecting ducts

• Components of Filtration

Filtration membrane- the barrier through which the fluid must pass to enter the capsular space. Consist of:

-          The fenestrated endothelium of the capillary-  the honeycombed structure of the capillary endothelium, containing large pores that allow various substances to pass through (not blood cells or large proteins)

-          The basement membrane- Its structure is like a kitchen sponge. While blood plasma is 7% protein, the glomerular filtrate is only 0.03% protein.  It has traces of albumin and smaller polypeptides, including some hormones.

-          Filtration slits- octopi like structure of the podocytes. Each arm has numerous little extensions called pedicles (foot processes) that wrap around the capillaries and interdigitate with each other. Pedicles have a negatively charged filtration slits; this restricts anions like albumins from being filtered.

–Water, electrolytes, glucose, fatty acids, nitrogenous wastes, and vitamins are filtered.

–Some substances that are of low molecular weights are retained in the blood plasma because they are bound to plasma proteins e.g. Calcium, iron, thyroid hormone and plasma fatty acids. The small fraction of the above that are unbound will pass through

•About 20% of the plasma that passes through the kidney gets filtered into the nephron

•Filtration  takes place in the glomerulus

•It is driven by the hydrostatic pressure of the blood (osmosis (oncotic pressure) opposes filtration, but the hydrostatic pressure is larger)

•Water and small molecules are filtered; blood cells and large molecules (most proteins) do not pass through the filter 

•  Kidney trauma and infections can damage the filtration membrane and allow albumin or blood cells to filter through

1. Proteinuria (albuminuria)- the presence of protein in the urine

1. Hematuria- the presence of blood in the urine

•Strenuous exercise can temporarily cause proteinuria or hematuria

–Strenuous exercise greatly reduces perfusion of the kidney causing glomerular damage due to prolonged hypoxia.

1. Albuminuria

•More than the normal amount of albumin in the urine. Albumin is the predominant protein in human blood and it is the key to the regulation of the osmotic pressure of blood.

•It is normal to have some albumin in urine. But too much albumin indicates that protein is leaking through the kidney.

•Albuminuria can mean many things. For example, albuminuria may be a sign of significant kidney disease or it may simply be a sequel of vigorous exercise. Albuminuria is a form of proteinuria.

1. Hematuria

•is a sign that something is causing bleeding in the genitourinary tract

•There are two types of hematuria,

–Microscopic- the amount of blood in the urine is so small that it can be seen only under a microscope.

–Gross (or macroscopic) - the urine is pink, red, or dark brown and may contain small blood clots.

1. Joggers hematuria

results from repeated jarring of the bladder during jogging or long-distance running.

•Reddish urine that is not caused by blood in the urine is called pseudohematuria. It may be caused by some drugs and some types of food. Find out from BNF

• Filtration Pressure

•Glomerular filtration follows the same principles that govern filtration in other blood capillaries but with some  significant differences in the magnitude of forces involved

–The blood hydrostatic pressure is much higher, about 60mm Hg compared with 10-15 mm Hg in most other capillaries

–Results from the fact that the afferent arteriole is substantially larger (than those in other parts of the body) than the efferent arteriole, giving the glomerulus a large inlet and small outlet.

– The osmotic pressure (Oncotic) which opposes the hydrostatic pressure is comparatively lower.

1. Nephrosclerosis 

High blood pressure in the glomeruli makes the kidneys especially vulnerable to hypertension, which can rupture glomerular capillaries and lead to scarring of the kidney (Nephrosclerosis) and atherosclerosis of renal blood vessels, leading to renal failure

Glomerular Filtration Rate (GFR)

Is the amount of filtrate formed per minute by the two kidneys combined. GFR depends upon;

–Permeability of the filtration barrier

–Surface area of the filtration barrier

GFR measurement 

It a way of measuring kidney function. It is not measured directly. We use a marker substance, which is neither reabsorbed nor secreted by the kidney. The reasoning is: the amount of the substance excreted per minute should be equal to the amount filtered.

Two substances are used to measure GFR:

1. Inulin: a polysaccharide which is not metabolized by the body or reabsorbed from the urine. Inulin is not found in the body and must be injected. This substance gives the most accurate results and is used for research purposes.
2. Creatinine: a breakdown product from creatine phosphate, which is naturally found in the blood. Not quite as accurate as Inulin (about 10% is reabsorbed), but often used in medicine, since no injection is required.

Using Inulin in Measuring GFR

-          Is only filtered by the kidney; it is neither reabsorbed nor secreteted

-          Since no Inulin is reabsorbed from or secreted into the tubule, the amount filtered into the tubule at the glomerulus must equal the amount appearing in the urine

-          P X GFR = U X V ; where,

P = plasma concentration of Inulin, in mg/mL

GFR = glomerular filtration rate of plasma, in mL/min

U = urine concentration of Inulin, in mg/mL

V = rate of urine production, in mL/min

Solving the equation for GFR will give:

GFR = (U X V)/P

The normal value of GFR is 125ml/min

Renal Clearance

the volume of blood plasma from which a particular waste is completely removed in 1 minute.

Clearance= (U*V)/P

The clearance of Inulin= GFR and is taken as the standard, because it's neither secreted nor reabsorbed.

-          If a substance has a clearance greater than Inulin, then it must have been secreted into the tubular fluid by the nephron epithelium.

-          If it's lower, then either it was not filtered at the glomerulus or it must have been reabsorbed from the tubular fluid.

Regulation of Glomerular Filtration

GFR must be finely controlled:

If it is too high,

–Fluid flows through the renal tubules too rapidly for them to reabsorb the usual amount of water and solutes

–Urine output rises and creates a threat of dehydration and electrolyte depletion           

If it is too low

–Fluid flows sluggishly through the tubules and they reabsorb wastes that should be eliminated and azotemia may occur. GFR is adjusted by three homeostatic mechanisms:

1. Renal autoregulation- can even be observed in denervated kidneys e.g. transplanted kidneys!
2. Sympathetic control
3. Hormonal control

Renal Autoregulation

•The ability of the Nephrons to adjust their own blood flow and GFR without external (nervous or hormonal) control.

•Allows stable fluid and electrolyte balance in spite of alterations in mean arterial pressure.

•Therefore, a primary role of the renal autoregulatory mechanism is to regulate intrarenal haemodynamics and intrarenal pressures to levels that maintain an optimal balance with tubular metabolic functions.

•There are two mechanisms of autoregulation

• Myogenic mechanism

• Tubular glomerular feedback (TGF)

Myogenic Mechanism

Keeping mind that filtration pressure is a matter of blood pressure (or glomerular capillary pressure); Blood pressure changes are sensed through stretch receptors and respond accordingly through relaxation or constriction.

–afferent arteriolar vasoconstriction would serve to protect the glomerulus from uncontrolled systemic hypertension,

–while afferent arteriolar vasodilatation would allow for greater blood flow into the glomerulus in times of hypotension

The autoregulatory system accomplishes this by maintaining the glomerular capillary pressure around 60-70 mm Hg

Tubular glomerular Feedback

•The Juxtaglomerular apparatus (JGA) is made up of specialized cells in the wall of the afferent arteriole and granular cells in the wall of the distal tubule (the macula densa).

•This area is innervated by adrenergic fibers and the granular cells carry renin in intracellular granules

•The principle function of the JGA is adapting the GFR to early distal tubule fluid characteristics by modulating renin synthesis and release: this is known as the Tubular glomerular feedback (TGF) loop.  

•Afferent arteriolar caliber is principally controlled by TGF

•Besides altered sodium concentration at the macula densa of the distal tubule, release of renin can also be induced by changes in the blood flow patterns of the afferent arteriole, or by adrenergic stimulation.

Renin-Angiotensin Mechanism

•When blood pressure drops, the sympathetic nerves stimulate the JGA cells to secrete renin

•Renin acts on angiotensinogen to create angiotensin.

•Angiotensin is converted to angiotensin II by the action of angiotensin-converting enzyme (ACE) from the lungs and kidneys. Angiotensin II has multiple effects.

Renin-Angiotensin Mechanism

•Stimulates widespread vasoconstriction which raises the Mean Arterial Pressure throughout the body

•Constricts both the afferent and efferent arterioles (more prominent here because of greater concentration of Angiotensin II receptors) which reduces GFR and water loss

•Stimulates the secretion of antidiuretic hormone which promotes water reabsorption

•Stimulates the adrenal cortex to secrete Aldosterone, which in turn promotes sodium and water retention

•Stimulates the sense of thirst and encourages water intake

In summary, renal vascular resistance controlled by

•Intrinsic mechanisms

–Stretch receptors in wall of afferent and efferent arterioles

•Extrinsic mechanisms

–Hormones – renin-angiotensin II axis. Angiotensin II acts to increase efferent arteriolar tone

–Sympathetic innervation- in strenuous exercise or acute conditions such as circulatory shock. Norepinephrine directly increases afferent arteriolar tone causing a reduction in GFR and urine production, while redirecting blood from the kidneys to the heart, brain, and skeletal muscles.

Note: Angiotensin II release is dependent on renin release which results from

–blood flow changes in the afferent arteriole

–adrenergic stimulation

–solute changes at the macula densa 

Note: The normal GFR can only be maintained within systemic arterial pressures of 90-200 mmHg.

Circumstances under which GFR does not remain constant

1. Major heamorrhage- a large increase in sympathetic activity causes greater constriction of the afferent arterioles more than the efferent thus reducing GFR
2. Liver disease and malnutrition- reduces the plasma oncotic pressure considerably thus increasing GFR.

Assignment:

1. Define the terms; oliguria, polyuria, edema.
2. Describe the Pathophysiology of Glomerular nephritis.

Raoult's Law and Non-volatile Solutes

This page deals with Raoult's Law and how it applies to solutions in which the solute is non-volatile - for example, a solution of salt in water. A non-volatile solute (the salt, for example) hasn't got any tendency to form a vapour at the temperature of the solution. It goes on to explain how the resulting lowering of vapour pressure affects the boiling point and freezing point of the solution.
	Important:  If you haven't already read the page about saturated vapour pressure, you should follow this link before you go on. Use the BACK button on your browser to return to this page when you are ready.
Raoult's Law There are several ways of stating Raoult's Law, and you tend to use slightly different versions depending on the situation you are talking about. You can use the simplified definition in the box below in the case of a single volatile liquid (the solvent) and a non-volatile solute. The vapour pressure of a solution of a non-volatile solute is equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction. In equation form, this reads: In this equation, Po is the vapour pressure of the pure solvent at a particular temperature. xsolv is the mole fraction of the solvent. That is exactly what it says it is - the fraction of the total number of moles present which is solvent. You calculate this using: Suppose you had a solution containing 10 moles of water and 0.1 moles of sugar. The total number of moles is therefore 10.1 The mole fraction of the water is: A simple explanation of why Raoult's Law works There are two ways of explaining why Raoult's Law works - a simple visual way, and a more sophisticated way based on entropy. Because of the level I am aiming at, I'm just going to look at the simple way. Remember that saturated vapour pressure is what you get when a liquid is in a sealed container. An equilibrium is set up where the number of particles breaking away from the surface is exactly the same as the number sticking on to the surface again. Now suppose you added enough solute so that the solvent molecules only occupied 50% of the surface of the solution. A certain fraction of the solvent molecules will have enough energy to escape from the surface (say, 1 in 1000 or 1 in a million, or whatever). If you reduce the number of solvent molecules on the surface, you are going to reduce the number which can escape in any given time. But it won't make any difference to the ability of molecules in the vapour to stick to the surface again. If a solvent molecule in the vapour hits a bit of surface occupied by the solute particles, it may well stick. There are obviously attractions between solvent and solute otherwise you wouldn't have a solution in the first place. The net effect of this is that when equilibrium is established, there will be fewer solvent molecules in the vapour phase - it is less likely that they are going to break away, but there isn't any problem about them returning. If there are fewer particles in the vapour at equilibrium, the saturated vapour pressure is lower. Limitations on Raoult's Law Raoult's Law only works for ideal solutions. An ideal solution is defined as one which obeys Raoult's Law. Features of an ideal solution In practice, there's no such thing! However, very dilute solutions obey Raoult's Law to a reasonable approximation. The solution in the last diagram wouldn't actually obey Raoult's Law - it is far too concentrated. I had to draw it that concentrated to make the point more clearly. In an ideal solution, it takes exactly the same amount of energy for a solvent molecule to break away from the surface of the solution as it did in the pure solvent. The forces of attraction between solvent and solute are exactly the same as between the original solvent molecules - not a very likely event! Suppose that in the pure solvent, 1 in 1000 molecules had enough energy to overcome the intermolecular forces and break away from the surface in any given time. In an ideal solution, that would still be exactly the same proportion. Fewer would, of course, break away because there are now fewer solvent molecules on the surface - but of those that are on the surface, the same proportion still break away. If there were strong solvent-solute attractions, this proportion may be reduced to 1 in 2000, or 1 in 5000 or whatever. In any real solution of, say, a salt in water, there are strong attractions between the water molecules and the ions. That would tend to slow down the loss of water molecules from the surface. However, if the solution is sufficiently dilute, there will be good-sized regions on the surface where you still have water molecules on their own. The solution will then approach ideal behaviour. The nature of the solute There is another thing that you have to be careful of if you are going to do any calculations on Raoult's Law (beyond the scope of this site). You may have noticed in the little calculation about mole fraction further up the page, that I used sugar as a solute rather than salt. There was a good reason for that! What matters isn't actually the number of moles of substance that you put into the solution, but the number of moles of particles formed. For each mole of sodium chloride dissolved, you get 1 mole of sodium ions and 1 mole of chloride ions - in other words, you get twice the number of moles of particles as of original salt. So, if you added 0.1 moles of sodium chloride, there would actually be 0.2 moles of particles in the solution - and that's the figure you would have to use in the mole fraction calculation. Unless you think carefully about it, Raoult's Law only works for solutes which don't change their nature when they dissolve. For example, they mustn't ionise or associate (in other words, if you put in substance A, it mustn't form A2 in solution). If it does either of these things, you have to treat Raoult's law with great care.
	Note:  This isn't a problem you are likely to have to worry about if you are a UK A level student. Just be aware that the problem exists.
Raoult's Law and melting and boiling points The effect of Raoult's Law is that the saturated vapour pressure of a solution is going to be lower than that of the pure solvent at any particular temperature. That has important effects on the phase diagram of the solvent. The next diagram shows the phase diagram for pure water in the region around its normal melting and boiling points. The 1 atmosphere line shows the conditions for measuring the normal melting and boiling points.
	Note:  In common with most phase diagrams, this is drawn highly distorted in order to show more clearly what is going on. If you haven't already read my page about phase diagrams for pure substances, you should follow this link before you go on to make proper sense of what comes next. Use the BACK button on your browser to return to this page when you are ready.
The line separating the liquid and vapour regions is the set of conditions where liquid and vapour are in equilibrium. It can be thought of as the effect of pressure on the boiling point of the water, but it is also the curve showing the effect of temperature on the saturated vapour pressure of the water. These two ways of looking at the same line are discussed briefly in a note about half-way down the page about phase diagrams (follow the last link above). If you draw the saturated vapour pressure curve for a solution of a non-volatile solute in water, it will always be lower than the curve for the pure water.
	Note:  The curves for the pure water and for the solution are often drawn parallel to each other. That has got to be wrong! Suppose you have a solution where the mole fraction of the water is 0.99 and the vapour pressure of the pure water at that temperature is 100 kPa. The vapour pressure of the solution will be 99 kPa - a fall of 1 kPa. At a lower temperature, where the vapour pressure of the pure water is 10 kPa, the fall will only be 0.1 kPa. For the curves to be parallel the falls would have to be the same over the whole temperature range. They aren't!
If you look closely at the last diagram, you will see that the point at which the liquid-vapour equilibrium curve meets the solid-vapour curve has moved. That point is the triple point of the system - a unique set of temperature and pressure conditions at which it is possible to get solid, liquid and vapour all in equilibrium with each other at the same time. Since the triple point has solid-liquid equilibrium present (amongst other equilibria), it is also a melting point of the system - although not the normal melting point because the pressure isn't 1 atmosphere. That must mean that the phase diagram needs a new melting point line (a solid-liquid equilibrium line) passing through the new triple point. That is shown in the next diagram. Now we are finally in a position to see what effect a non-volatile solute has on the melting and freezing points of the solution. Look at what happens when you draw in the 1 atmosphere pressure line which lets you measure the melting and boiling points. The diagram also includes the melting and boiling points of the pure water from the original phase diagram for pure water (black lines). Because of the changes to the phase diagram, you can see that: the boiling point of the solvent in a solution is higher than that of the pure solvent; the freezing point (melting point) of the solvent in a solution is lower than that of the pure solvent. We have looked at this with water as the solvent, but using a different solvent would make no difference to the argument or the conclusions. The only difference is in the slope of the solid-liquid equilibrium lines. For most solvents, these slope forwards whereas the water line slopes backwards. You could prove to yourself that that doesn't affect what we have been looking at by re-drawing all these diagrams with the slope of that particular line changed. You will find it makes no difference whatsoever.

Ideal Gases and the Ideal Gas Law

This page looks at the assumptions which are made in the Kinetic Theory about ideal gases, and takes an introductory look at the Ideal Gas Law: pV = nRT. This is intended only as an introduction suitable for chemistry students at about UK A level standard (for 16 - 18 year olds), and so there is no attempt to derive the ideal gas law using physics-style calculations. Kinetic Theory assumptions about ideal gases There is no such thing as an ideal gas, of course, but many gases behave approximately as if they were ideal at ordinary working temperatures and pressures. Real gases are dealt with in more detail on another page. The assumptions are: Gases are made up of molecules which are in constant random 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, are perfectly elastic. (That means that there is no loss of kinetic energy during the collision.) The temperature of the gas is proportional to the average kinetic energy of the molecules. And then two absolutely key assumptions, because these are the two most important ways in which real gases differ from ideal gases: There are no (or entirely negligible) intermolecular forces between the gas molecules. The volume occupied by the molecules themselves is entirely 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. The problems lie almost entirely in the units. I am assuming below that you are working in strict SI units (as you will be if you are doing a UK-based exam, for example). Exploring the various terms Pressure, p Pressure is measured in pascals, Pa - sometimes expressed as newtons per square metre, N m-2. These mean exactly the same thing. Be careful if you are given pressures in kPa (kilopascals). For example, 150 kPa is 150,000 Pa. You must make that conversion before 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 equation. That's because the SI unit of volume is the cubic metre, m3 - not cm3 or dm3. 1 m3 = 1000 dm3 = 1,000,000 cm3 So if you are inserting values of volume into the equation, you first have to convert them into cubic metres. You would have to divide a volume in dm3 by 1000, or in cm3 by a million. Similarly, if you are working out a volume using the equation, remember to covert the answer in cubic metres into dm3 or cm3 if you 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 fairly obvious if you have done something wrong, and you can check back again. Number of moles, n This is easy, of course - it is just a number. You already know that you work it out by dividing the mass in grams by the mass of one mole in grams. You will most often use the ideal gas equation by first making the substitution to give: I don't recommend that you remember the ideal gas equation in this form, but you must be confident that you can convert it into this form. The gas constant, R A value for R will be given you if you need it, or you can look it up in a data source. The SI value for R is 8.31441 J K-1 mol-1.
	Note:  You may come across other values for this with different units. A commonly used one in the past was 82.053 cm3 atm K-1 mol-1. The units tell you that the volume would be in cubic centimetres and the pressure in atmospheres. Unfortunately the units in the SI version aren't so obviously helpful.
The temperature, T The temperature has to be in kelvin. Don't forget to add 273 if you are given a temperature in degrees Celsius. Using the ideal gas equation Calculations using the ideal gas equation are included in my calculations book (see the link at the very bottom of the page), and I can't repeat them here. There are, however, a couple of calculations that I haven't done in the book which give a reasonable idea of how the ideal gas equation works. The molar volume at stp If you have done simple calculations from equations, you have probably used the molar volume of a gas. 1 mole of any gas occupies 22.4 dm3 at stp (standard temperature and pressure, taken as 0°C and 1 atmosphere pressure). You may also have used a value of 24.0 dm3 at room temperature and pressure (taken as about 20°C and 1 atmosphere). These figures are actually only true for an ideal gas, and we'll have a look at where they come from. We can use the ideal gas equation to calculate the volume of 1 mole of an ideal gas at 0°C and 1 atmosphere pressure. First, we have to get the units right. 0°C is 273 K. T = 273 K 1 atmosphere = 101325 Pa. p = 101325 Pa We know that n = 1, because we are trying to calculate the volume of 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 it gives: And finally, because we are interested in the volume in cubic decimetres, you have to remember to multiply this by 1000 to convert 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 of 1 mole of an ideal gas at room temperature and pressure - or any other 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 20°C 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, by dividing by 1000. We have a volume of 0.001 m3. The temperature is 293 K. Now put all the numbers into the form of the ideal gas equation which lets you work with masses, and rearrange it to work out the mass of 1 mole. The mass of 1 mole of anything is simply the relative formula mass in 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 using accurate values of relative atomic masses, you get an answer of 30.07 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 the sum again using a slightly different value quoted at a different temperature from another source. This time I got an answer of 30.3. So the density values may not be entirely accurate, but they are both giving much the same sort of answer. Ethane isn't an ideal gas. Well, of course it isn't an ideal gas - there's no such thing! However, assuming that the density values are close to correct, the error is within 1% of what you would expect. So although ethane isn't exactly behaving like an ideal gas, it isn't far off. If you need to know about real gases, now is a good time to read about them.

Phase Diagrams of Pure Substances

This page explains how to interpret the phase diagrams for simple pure substances - including a look at the special cases of the phase diagrams 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 present and a liquid phase. If there is air above the mixture, then that is another phase. But the term can be used more generally than this. For example, oil floating on water also consists of two phases - in this case, two liquid phases. If the oil and water are contained in a bucket, then the solid bucket is yet another phase. In fact, there might be more than one solid phase if the handle is attached separately to the bucket rather than moulded as a part of the bucket. You can recognise the presence of the different phases because there is an obvious boundary between them - a boundary between the solid ice and the liquid water, for example, or the boundary between the two liquids. Phase diagrams A phase diagram lets you work out exactly what phases are present at any 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. These diagrams (including this one) are nearly always drawn highly distorted in order to see what is going on more easily. There are usually two major distortions. We'll discuss these when they become relevant. If you look at the diagram, you will see that there are three lines, three areas marked "solid", "liquid" and "vapour", and two special points marked "C" and "T". The three areas These are easy! Suppose you have a pure substance at three different sets of conditions of temperature and pressure corresponding to 1, 2 and 3 in the next diagram. Under the set of conditions at 1 in the diagram, the substance would be a solid because it falls into that area of the phase diagram. At 2, it would 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 are subtle differences between them that I'm not ready to explain for a while yet. Be patient!
Moving from solid to liquid by changing the temperature: Suppose you had a solid and increased the temperature while keeping the pressure constant - as shown in the next diagram. As the temperature increases to the point where it crosses the line, the solid will turn to liquid. In other words, it melts. If you repeated this at a higher fixed pressure, the melting temperature would be higher because the line between the solid and liquid areas slopes slightly forward.
	Note:  This is one of the cases where we distort these diagrams to make them easier to discuss. This line is much more vertical in practice than we normally draw 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.
So what actually is this line separating the solid and liquid areas of the diagram? 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 it was a chemical equilibrium. If you increase the pressure, the equilibrium 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 decrease again because the solid takes up slightly less space for most substances. That means that increasing the pressure on the equilibrium mixture of solid and liquid at its original melting point will convert the mixture back into the solid again. In other words, it will no longer melt at this temperature. To make it melt at this higher pressure, you will have to increase the temperature a bit. Raising the pressure raises the melting point of most solids. That's why the melting point line slopes forward for most substances. Moving from solid to liquid by changing the pressure: You can also play around with this by looking at what happens if you decrease the pressure on a solid at constant temperature.
	Note:  You have got to be a bit careful about this, because exactly what happens if you decrease the pressure depends on exactly what your starting conditions are. We'll talk some more about this when we look at the line separating the solid region from the vapour region.
Moving from liquid to vapour: In the same sort of way, you can do this either by changing the temperature or the pressure. The liquid will change to a vapour - it boils - when it crosses the boundary line between the two areas. If it is temperature that you are varying, you can easily read off the boiling temperature from the phase diagram. In the diagram above, it is the temperature where the red arrow crosses the boundary line. So, again, what is the significance of this line separating the two areas? Anywhere along this line, there will be an equilibrium between the liquid and the vapour. The line is most easily seen as the effect of pressure 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 actually exactly the same as the graph for the effect of temperature on the saturated vapour pressure of the liquid. Saturated vapour pressure is dealt with on a separate page. A liquid will boil when its saturated vapour pressure is equal to the external pressure. Suppose you measured the saturated vapour pressure of a liquid at 50°C, and it turned out to be 75 kPa. You could plot that as one point on a vapour pressure curve, 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 75 kPa, and gradually increased the temperature. The liquid would boil when its saturated vapour pressure became equal to the external pressure - in this case at 50°C. If you have the complete vapour pressure curve, you could equally well find the boiling point corresponding to any other external pressure. That means that the plot of saturated vapour pressure against temperature is exactly the same as the curve relating boiling point and external pressure - they are just two ways of looking at the same thing. If all you are interested in doing is interpreting one of these phase diagrams, you probably don't have to worry too much about this.
The critical point You will have noticed that this liquid-vapour equilibrium curve has a top 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 critical temperature and critical pressure. If you increase the pressure on a gas (vapour) at a temperature lower than the critical temperature, you will eventually cross the liquid-vapour equilibrium line and the vapour will condense to give a liquid. 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 to condense a gas into a liquid just by increasing the pressure. All you get is a highly compressed gas. The particles have too much energy for the intermolecular attractions to hold them together as a liquid. The critical temperature obviously varies from substance to substance and depends on the strength of the attractions between the particles. The stronger the intermolecular attractions, the higher the critical temperature.
	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. You don't usually talk about "ethanol gas", although you would say "ethanol vapour". Equally, you wouldn't talk about oxygen as being a vapour - you always call it a gas. There are various guide-lines that you can use if you want to. For example, if the substance is commonly a liquid at or around room temperature, you tend to call what 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 critical point at room temperature - oxygen or nitrogen or hydrogen, for example. These would all be described as gases. This is absolutely NOT something that is at all worth getting worked up about!
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 of temperature and pressure fell exactly on that line, there would be solid and vapour in equilibrium with each other - the solid would be subliming. (Sublimation is the change directly from solid to vapour or vice versa without going through the liquid phase.) Once again, you can cross that line by either increasing the temperature of the solid, or decreasing the pressure. The diagram shows the effect of increasing the temperature of a solid at a (probably very low) constant pressure. The pressure obviously has to be low enough that a liquid can't form - in other words, it has to happen below the point labelled as T. You could read the sublimation temperature off the diagram. It will be the temperature at which the line is crossed. The triple point Point T on the diagram is called the triple point. If you think about the three lines which meet at that point, they represent conditions of: solid-liquid equilibrium liquid-vapour equilibrium solid-vapour equilibrium Where all three lines meet, you must have a unique combination of temperature and pressure where all three phases are in equilibrium together. That's why it is called a triple point. If you controlled the conditions of temperature and pressure in order to land on this point, you would see an equilibrium which involved the solid 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 the system closed so that nothing escaped, that's how it would stay. A strange 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 by drawing a line across at 1 atmosphere pressure. The phase diagram for water There is only one difference between this and the phase diagram that we've looked at up to now. The solid-liquid equilibrium line (the melting point 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, according to Le Chatelier's Principle the equilibrium will move to reduce the pressure again. That means that it will move to the side with the smaller volume. Liquid water is produced. To make the liquid water freeze again at this higher pressure, you will have to reduce the temperature. Higher pressures mean lower melting (freezing) points. Now lets put some numbers on the diagram to show the exact positions of the critical point and triple point for water. Notice that the triple point for water occurs at a very low pressure. Notice also that the critical temperature is 374°C. It would be impossible to convert water from a gas to a liquid by compressing it above this temperature. The normal melting and boiling points of water are found in exactly the same way as we have already discussed - by seeing where the 1 atmosphere pressure line crosses the solid-liquid and then the liquid-vapour equilibrium lines.
	Note:  Further up the page I mentioned two ways in which these diagrams are distorted to make them easier to follow. I have already pointed out that the solid-liquid equilibrium line should really be much more vertical. This last diagram illustrates the other major distortion - which is to the scales of both pressure and temperature. Look, for example, at the gaps between the various quoted pressure figures and then imagine that you had to plot those on a bit of graph paper! The temperature scale is equally haphazard.
Just one final example of using this diagram (because it appeals to me). Imagine lowering the pressure on liquid water along the line in the diagram below. The phase diagram shows that the water would first freeze to form ice as it crossed into the solid area. When the pressure fell low enough, the ice would then sublime to give water vapour. In other words, the change 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 the triple point which is well above atmospheric pressure. It is impossible to get any liquid carbon dioxide at pressures less than 5.11 atmospheres. That means that at 1 atmosphere pressure, carbon dioxide will sublime at a temperature of -78°C. 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 the solid or the vapour.