CHAPTER 6 PROPERTIES OF GASES 6. The Ideal Gas Equation In 660, the Honorable Robert Boyle, Father of Chemistry and seventh son of the Earl of Cork, and one of the founders of the Royal Soiety of London, onduted ertain Experiments Physio- Mehanial Touhing the Spring of the Air. He held a quantity of air in the losed arm of a J- shaped glass tube by means of a olumn of merury and he measured the volume of the air as it was subjeted to greater and greater pressures. As a result of these experiments he established what is now known as Boyle's Law: The pressure of a fixed mass of gas held at onstant temperature (i.e. in an isothermal proess) is inversely proportional to its volume. That is, PV = onstant. 6.. Later experiments showed that the volume of a fixed mass of gas held at onstant pressure inreases linearly with temperature. In partiular, most gases have about the same volume oeffiient of expansion. At 0 o C this is about 0.0066 C o or /7 C o. If you extrapolate the volume of a fixed mass of gas held at onstant pressure to lower and lower temperatures, the extrapolated volume would fall to zero at 7 o C. This is not diretly the basis of our belief that no temperatures are possible below 7 o C. For one thing, a real gas would liquefy long before that temperature is reahed. Nevertheless, for reasons that will be disussed in a muh later hapter, we do believe that this is the absolute zero of temperature. In any ase: The volume of a fixed mass of gas held at onstant pressure (i.e. in an isobari proess) is diretly proportional to its Kelvin temperature. Lastly, The pressure of a fixed mass of gas held at onstant volume (i.e. in an isohori proess) is diretly proportional to its Kelvin temperature. If P, V and T are all allowed to vary, these three laws beome PV/T = onstant 6.. The value of the onstant depends on how muh gas there is; in partiular, it is proportional to how many moles (hene how many moleules) of gas there are. That is PV/T = RN, 6..
where N is the number of moles and R is a proportionality onstant, whih is found to be about the same for most gases. Of ourse real gases behave only approximately as desribed, and only provided experiments are performed over modest ranges of temperature, pressure and volume, and provided the gas is well above the temperature at whih it will liquefy. Nevertheless, provided these onditions are satisfied, most gases do onform quite well to equation 6.. with about the same proportionality onstant for eah. A gas that obeys the equation PV = NRT 6..4 exatly is alled an Ideal Gas, and equation 6..4 is alled the Equation of State for an Ideal Gas. In this equation, V is the total volume of the gas, N is the number of moles and R is the Universal Gas Constant. The equation an also be written PV = RT. 6..5 In this ase, V is the molar volume. Some authors use different symbols (suh as V, v and V m ) for total, speifi and molar volume. This is probably a good idea, and it is at some risk that I am not going to do this, and I am going to hope that the ontext will make it lear whih volume I am referring to when I use the simple symbol V in any partiular situation. Note that, while total volume is an extensive quantity, speifi and molar volumes are intensive. It is not impossible to go wrong by a fator of 0 when using equation 6..5. If you are using CGS units, P will be expressed in dynes per square m, V is the volume of a mole (i.e. the volume oupied by 6.0 0 7 moleules), and the value of the universal gas onstant is 8.45 0 erg mole K. If you are using SI units, P will be expressed in pasal (N m ), V will be the volume of a kilomole (i.e. the volume oupied by 6.0 % 0 6 moleules), and the value of the universal gas onstant is 8.45 0 J kilomole K. If you wish to express pressure in Torr, atm. or bars, and energy in alories, you're on your own. NNA You an write equation 6..4 (with V = total volume) as. RT P=, where N A is Avogadro's V N number, whih is 6.0 0 moleules per mole, or 6.0 0 moleules per kilomole. The first term on the right hand side is the total number of moleules divided by the volume; that is, it is the number of moleules per unit volume, n. In the seond term, R/N A is Boltzmann's onstant, k =.807 0 J K. Hene the equation of state for an ideal gas an be written A 6 P = nkt. 6..6 Divide both sides of equation 6..5 by the molar mass ("moleular weight ) µ. The density ρ of a sample of gas is equal to the molar mass divided by the molar volume, and hene the equation of state for an ideal gas an also be written
ρrt P =. 6..7 µ In summary, equations 6..4, 6..5, 6..6 and 6..7 are all ommonly-seen equivalent forms of the equation of state for an ideal gas. From this point on I shall use V to mean the molar volume, unless stated otherwise, so that I shall use equation 6..5 rather than 6..4 for the equation of state for an ideal gas. Note that the molar volume (unlike the total volume) is an intensive state variable. In September 007, the values given for the above-mentioned physial onstants on the Website of the National Institute of Siene and Tehnology (http://physis.nist.gov/uu/index.html) were: Molar Gas Constant R = 84.47 (5) J kmole K. Avogadro Constant N A = 6.0 4 79 (0) 0 6 partiles kmole. Boltzmann Constant k =.80 6504 (4) 0 J K per partile. The number in parentheses is the standard unertainty in the last two figures. [There is a proposal, likely to beome offiial in 05, to give defined exat numerial values to Avogadro s and Boltzmann s onstants, namely 6.0 4 % 0 partiles mole and.80 6 % 0 J K per partile. This may at first seem to be somewhat akin to defining π to be exatly, but it is not really like that at all. It is all part of a general shift in defining many of the units used in physis in terms of fundamental physial quantities (suh as the harge on the eletron) rather than in terms of rods or ylinders of platinum held in Paris.] 6. Real Gases How well do real gases onform to the equation of state for an ideal gas? The answer is quite well over a large range of P, V and T, provided that the temperature is well above the ritial temperature. We'll have to see shortly what is meant by the ritial temperature; for the moment we'll say the ideal gas equation is followed quite well provided that the temperature is well above the temperature at whih it an be liquefied merely by ompressing it. Air at room temperature obeys the law quite well. Gases in stellar atmospheres also obey the law well, beause there is no danger there of the gas liquefying. (In the ores of stars, however, where densities are very large, the gases obey a very different equation of state.) One measure of how well the law is obeyed by real gases is to measure P, V and T, and see how PV PV lose is to. The quantity is known as the ompression fator, and is often given the RT RT symbol Z. For most real gases at very high pressures (a few hundred atmospheres), it is found in fat that Z is rather greater than. As the pressure is lowered, Z beomes lower, and then, alas, it overshoots and is found to be a little less than. Then at yet lower pressures Z rises again. The exat shape of the Z : P urve is different from gas to gas, as is the pressure at whih Z is a minimum. Yet, for all gases, as the pressure approahes zero, PV/T approahes R exatly. For this reason R is sometimes alled the Universal Gas Constant as well as the Ideal Gas Constant. In the limit of very low pressures, all gases behave very losely to the behaviour of an ideal gas. In Setion 6. we shall be examining more losely how the ompression fator varies with pressure.
4 Another way to look at how losely real gases obey the ideal gas equation is to plot P versus V for a number of different temperatures. That is, we draw a set of isotherms. For an ideal gas, these isotherms, PV = onstant, are retangular hyperbolas. So they are for real gases at high temperatures. At lower temperatures, departures from the ideal gas equation are marked. Typial isotherms are skethed in figure VI.. Alas, my limited skills with this infernal omputer in front of me allow me only to sketh these isotherms rudely by hand. P Liquid Gas Liquid + Vapour Vapour CI V FIGURE VI. In the PV plane of figure VI., you will see several areas marked "gas", "liquid", "vapour", "liquid + vapour". You an follow the behaviour at a given temperature by starting at the right hand end of eah isotherm, and gradually moving to the left i.e. inrease the pressure and derease the volume. The hottest isotherm is nearly hyperboloidal. Nothing speial happens beyond the volume dereasing as the pressure is inreased, aording to Boyle's law. At slightly lower temperatures, a kink develops in the isotherm, and at the ritial temperature the kink develops a loal horizontal infletion point. The isotherm for the ritial temperature is the ritial isotherm, marked CI on the sketh. Still nothing speial happens other than V dereasing as P is inreased, though not now aording to Boyle's law. For temperatures below the ritial temperature, we refer to the gas as a vapour. As you derease the volume, the pressure gradually inreases until you reah the dashed urve. At this point, some of the vapour liquefies, and, as you ontinue to derease the volume, more and more of the vapour liquefies, the pressure remaining onstant while it does so. That s the horizontal portion of the isotherm. In that region (i.e. outlined by the dashed urve) we have liquid and vapour in equilibrium. Near the right hand end of the horizontal portion, there is just a small amount of liquid; at the left hand end, most of the substane is liquid, with only a small amount of vapour left.
5 After it is all liquid, further inrease of pressure barely dereases the volume, beause the liquid is hardly at all ompressible. The isotherm is then almost vertial. The temperature of the ritial isotherm is the ritial temperature. The pressure and molar volume at the horizontal infletion point of the ritial isotherm are the ritial pressure and ritial molar volume. The horizontal infletion point is the ritial point. 6. Van der Waals and Other Gases We have seen that real gases resemble an ideal gas only at low pressures and high temperatures. Various attempts have been made to find an equation that adequately represents the relation between P, V and T for a real gas i.e. to find an Equation of State for a real gas. Some of these attempts have been purely empirial attempts to fit a mathematial formula to real data. Others are the result of at least an attempt to desribe some physial model that would explain the behaviour of real gases. A sample of some of the simpler equations that have been proposed follows: van der Waals' equation: ( P + a / V )( V b) = RT. 6.. Berthelot s equation: ( P a /( TV ))( V b) = RT. + 6.. a Clausius's equation*: P + ( V b) = RT. T ( V ) 6.. + a /( RTV ) Dieterii's equation: P( V b) e = RT. 6..4 RT a Redlih-Kwong:. / P = 6..5 V b bt V V + b Virial equation: PV = A + BP + CP + DP +K. 6..6 In the virial equation in general the oeffiients A, B, C, are funtions of temperature. *In Clausius s equation, if we hoose = b, we get a fairly good agreement between the ritial ompression fator of a Clausius gas and of many real gases. The meaning of ritial ompression fator, and the alulation of its value for a Clausius gas is desribed a little later in this setion. There are many others, but by far the best known of these is van der Waals' equation, whih I shall desribe at some length. It is not possible for the voie-box of an English speaker orretly to pronoune the name van der Waals, although the W is pronouned more like a V than a W, and, to my ear, the v is somewhat intermediate between a v and an f. To hear it orretly pronouned espeially the vowels you must ask a native Duth speaker. The frequent spelling "van
6 der Waal's equation" is merely yet another symptom of the modern lamentable ignorane of the use of the apostrophe so muh regretted by Lynne Truss. The units in whih the onstants a and b should be expressed sometimes ause diffiulty, and they depend on whether the symbol V in the equation is intended to mean the speifi or molar volume. The following might be helpful. If V is intended to mean the speifi volume, van der Waals equation should be written ( P + a / V )( V b) = RT / µ, where µ is the molar mass ( moleular weight ). In this ase the dimensions and SI units of a are M L 5 T and Pa m 6 kg and the dimensions and SI units of b are M L and m kg If V is intended to mean the molar volume, van der Waals equation should be written in its familiar form ( P + a / V )( V b) = RT. In this ase the dimensions and SI units of a are ML 5 T mole and the dimensions and SI units of b are L mole and Pa m 6 kmole and m kmole The van der Waals onstants, referred to molar volume, of H O and CO are approximately: H O: a = 5.5 0 5 Pa m 6 kmole b =. 0 m kmole CO : a =.7 0 5 Pa m 6 kmole b = 4. 0 m kmole The van der Waals equation has its origin in at least some attempt to desribe a physial model of a real gas. The properties of an ideal gas an be modelled by supposing that a gas onsists of a olletion of moleules of zero effetive size and no fores between them, and pressure is the result of ollisions with the walls of the ontaining vessel. In the van der Waals model, there are supposed to be attrative fores between the moleules. These are known as van der Waals fores and are now understood to arise beause when one moleule approahes another, eah indues a dipole moment in the other, and the two indued dipoles then attrat eah other. This attrative fore redues the pressure at the walls, the redution being proportional to the number of moleules at the walls that are being attrated by the moleules beneath, and to the number of moleules beneath, whih are doing the attrating. Both are inversely proportional to V, so the pressure in the equation of state has to be replaed by the measured pressure P plus a term that is inversely proportional to V. Further, the moleules themselves oupy a finite volume. This is tantamount to saying that, at very lose range, there are repulsive fores (now understood to be Coulomb fores) that are greater than the attrative van der Waals fores. Thus the volume in whih the
7 moleules are free to roam has to be redued in the van der Waals equation. For more on the fores between moleules, see Setion 6.8. However onvining or otherwise you find these arguments, they are at least an attempt to desribe some physis, they do represent the behaviour of real gases better that the ideal gas equation, and, if nothing else, they give us an opportunity for a little mathematis pratie. We shall see shortly how it is possible to determine the onstants a and b from measurements of the ritial parameters. These onstants in turn give us some indiation of the strength of the van der Waals fores, and of the size of the moleules. Van der Waals' equation, equation 6.., an be written P RT a =. 6..7 V b V P P A horizontal infletion point ours where and V V RT ( V b) a V + = are both zero. That is 0 6..8 RT 6a and = 0. 4 ( V b) V 6..9 Eliminate RT/a from these to find the ritial molar volume of a van der Waals gas: V = b. 6..0 Substitute this into equation 6..8 or 6..9 (or both, as a hek on your algebra) to obtain the ritial temperature: T = 8a. 6.. 7Rb Substitute equations 6..0 and 6.. into equation 6..7 to obtain the ritial pressure: From these, we readily obtain a P = 7 b. 6.. PV RT 8 = = 0.75. 6..
8 This quantity is often alled the ritial ompression fator or ritial ompressibility fator, and we shall denote it by the symbol Z. For many real gases Z is about 0.8; thus the van der Waals equation, while useful in disussing the properties of gases in a qualitative fashion, does not reprodue the observed ritial ompression fator partiularly well. Let us now substitute p = P/ P, v = V/ V, t = T/ T, and van der Waals' equation, in whih the pressure, volume and temperature are expressed in terms of their ritial values, beomes ( 8 p + / v )( v ) = t. 6..4 This an also be written p v ( p + 8t) v + 9v = 0. 6..5 For volumes less than a third of the ritial volume, this equation does not desribe the behaviour of a real gas at all well. Indeed, you an see that p = when v = /, whih means that you have to exert an infinite pressure to ompress a van der Waals gas to a third of its ritial volume. You might want to investigate for yourself the behaviour of equations 6..4 and 5 for volumes smaller than this. You will find that it goes to infinity at v = 0 and /, and it has a maximum between these two volumes. But the equation is of physial interest only for v > /, where the variation of pressure, volume and temperature bears at least some similarity to the behaviour of real gases, if by no means exat. In figure VI., I show the behaviour of a van der Waals gas for five temperatures one above the ritial temperature, one at the ritial temperature, and three below the ritial temperature. The lous of maxima and minima is found by eliminating t between equation 6..4 and p / v = 0. You should try this, and show that the lous of the maxima and minima (whih I have shown by a blue line in figure VI.) is given by p =. 6..6 v v
9.5 FIGURE VI. p 0.5 t.05.00 0.95 0.90 0.85 0 0 0.5.5.5.5 v Don t onfuse the blue urve in this figure (it shows the lous of maxima and minima) with the dashed urve in figure VI. (it shows the boundary between phases.). For the temperatures 0.85, 0.90 and 0.95 I have drawn the onstant pressure lines where liquid and vapour are in equilibrium in the real fluid. These are drawn so that they divide the van ver Waals urve into two equal areas, above and below. This means that the work done by the real fluid when it hanges from liquid to vapour at onstant pressure is equal to the work that would be done by its hypothetial van der Waals equivalent along its wiggly path. We shall later see that the plaement of the horizontal line is a onsequene of the fat that the Gibbs funtion (whih we have not yet met) is onstant while the liquid and vapour are in equilibrium. The dashed line of figure VI. would orrespond on figure VI. to the lous of the ends of the horizontal lines. I have drawn this lous, whih outlines the region where liquid and vapour are in equilibrium, in red in figure VI.. While the van der Waals equation is only a rough approximation to the behaviour of real gases, it is nevertheless true that, if pressures, temperatures and molar volumes are expressed in terms of the ritial pressures, temperatures and molar volumes, the atual equations of state of real gases are very similar. Two gases with the same values of p, v and t are said to be in orresponding states, and the observation that the p : v : t relation is approximately the same for all gases is alled the Law of Corresponding States. We may think of gases as being omposed of partiles (moleules) and the only differene between different gases is in the sizes of their moleules (i.e. their different van der Waals b onstants) and their dipole moments or their eletrial polarizabilities (i.e. their different van der Waals a onstants). In the dimensionless forms of the equation of state, these van der Waals onstants are removed from the equations, and it is not surprising that all gases then onform to the same equation of state.
0 I leave it to the reader to show that, for a Berthelot gas, the ritial molar volume, temperature and pressure and the ritial ompression fator are, respetively, b, 8a ar, 7bR b 6b and 0.75, that the equation of state in terms of the dimensionless variables is p 8t =, 6..7 v tv FIGURE VI.a (Berthelot).8.6.4. p 0.8 0.6 t =.0.00 0.98 0.96 0.94 0.4 0. 0 0 0. 0.4 0.6 0.8..4.6.8 v and that the lous of maxima and minima is p 4 6. / = v v v 6..8 These are shown in figure VI.a. It will be noted that the ritial ompression fator is the same as (and hene no better than) for a van der Waals gas. For a Clausius gas, the ritial molar volume, temperature and pressure and the ritial 8a ompression fator are, respetively,, ar b + b +, and. 7( b + ) R ( b + ) 6( b + ) 8( b + )
If = b. these beome, a ar 9, and 9R 5 = 0.85. I hoose = b beause that gives a good agreement with the ritial ompression fator for many real gases. In dimensionless units, the Clausius equation beomes t 48 p = 9v t(v + ). 6..9 The lous of maxima and minima is p = 80 44v 6(5 9v ) / v 6v 7v = + ( 9v )( + v ). 6..0 These are shown in figure VI.b FIGURE VI.b (Clausius).8.6.4 p. 0.8 0.6 t =.0.00 0.98 0.96 0.94 0.4 0. 0 0 0. 0.4 0.6 0.8..4.6.8 v The Clausius equation was hard work. Dieterii s is a little easier. The ritial molar volume, a a temperature and pressure and the ritial ompression fator are, respetively, b,, 4R b 4e b and /e = 0.7. Note that the ritial ompression fator is muh loser to that of many real gases. The dimensionless form of the Dieterii equation is t exp. p = 6.. v tv
The lous of maxima and minima is ( v ) p = exp. 6.. v v - These are shown in figure VI.. FIGURE VI. (Dieterii).8.6.4 p. 0.8 0.6 0.4 0. t =..0 0.9 0.8 0.7 0 0 0. 0.4 0.6 0.8..4.6.8 v The Redlih-Kwong equation of state, like those of van der Waals, Bethelot and Dieterii, has just two parameters (a and b). All of them are not too bad at temperatures appreiably above the ritial temperature, but, lose to the ritial temperature, the Redlih-Kwong empirial equation agrees a little better than the van der Waals equation does with what is observed for real gases. Obtaining the ritial onstants in terms of the parameters is done by exatly the same method as for the van der Waals and other equations, but requires perhaps a little more work and patiene. The reader might like (or might not like) to try it. For the ritial onstants I get V = xb, 6.. / a T = y 6..4 br a R and P = z, 5 6..5 b where x =.847 00, 6..6 y = 0.45 09 996 6..7 /
and z = 0.09 894 86. 6..8 The ritial ompression fator is xz /y, whih is exatly /. This is not as lose to the ompression fator of many real gases as the Dieterii ritial ompression fator is. We an invert these equations to obtain expressions for a and b in terms of the ritial temperature and pressure (or temperature and volume, or pressure and volume). Thus 5/ R T a = u 6..9 P RT and b = w, 6..0 P where u = 0.47 480 6.. and w = 0.086 640 50. 6.. (You an also do this for the other equations of state, of ourse.) In order to reprodue these results, you ll have to do a little work to see where all the onstants ome from. It turns out that the value of the onstant x is the positive real root of the equation x x x = 0. 6.. In the above analysis, I obtained all the onstants from a numerial solution of equation 6.., but the solution to this equation (and all subsequent onstants) an also be written in surds. Thus, with f =, g = 4, h = 6, the onstants an be alulated from 5g x =, y = g, z =, u = and w = f. 6..4 f g + h 9 f If we now introdue the dimensionless variables p = P/ P, v = V/ V, t = T/ T, and substitute these and equations 6..-5 into equation 6..6, we obtain the dimensionless Redlih-Kwong equation l t, / p = 6..5 xv mt xv xv + where l = y / z =.54 966 and m = gz = 0.07 559994. 6..6 The dimensionless Redlih-Kwong equation is illustrated in figure VI.d. I have not tried to find an expliit equation for the lous of maxima, but instead I alulated it numerially, illustrated by the dashed line in figure VI.d.
4 FIGURE VI.D (Redlih-Kwong).8 p.6.4. 0.8 0.6 t =.0.05.00 0.95 0.90 0.4 0. 0 0 0. 0.4 0.6 0.8..4.6.8 v Here is a summary of the results for the two-parameter equations of state: P V T Z a b a Van der Waals 7b b 8a 7Rb 8 7R T 64P RT 8P Berthelot ar b 6b b 8a 7bR 8 7R T 64P RT 8P a Dieterii 4e b b a 4Rb e RT e P 4R e T P Redlih-Kwong / a R z 5 xb b y a br / R T u P 5/ RT w P z = 0.099 x =.85 y = 0.45 u = 0.47 w = 0.0866
5 The reader an try to reprodue these (let me know (jtatum@uvi.a) if you find any mistakes!) or at least (a useful exerise) verify their dimensions. We mentioned in Setion 6. that a useful way of indiating how the behaviour of a real gas differs PV from that of an ideal gas is by plotting the ompression fator Z = versus pressure. As the RT pressure approahes zero, the ompression fator approahes. This is beause the moleules are then so far apart that there are no appreiable fores (attrative or repulsive) between them. As the pressure is inreased from zero, the ompression fator generally at first drops a little below, and then rises above as the pressure is inreased. It will be interesting to see how the ompression fator is expeted to vary with pressure for the various theoretial gases that we have been disussing. I ll do it just for a van der Waals gas, and I ll use the dimensionless form of van der Waals equation, whih was first given as equation 6..4: The ompression fator is 8 p + / v )( v ) = t. 6..7 ( PV Z = and the ritial ompression fator is RT PV Z =. From this, RT pv pv we see that Z = Z. For a van der Waals gas, Z = 8, so that Z =. Unfortunately, in t 8t order to plot Z versus p for a given t, we have to be able to express v in terms of p, whih means solving equation 6..7, whih is a ubi equation in v [p v ( p + 8t) v + 9v = 0]. I have done this numerially, and I show the resulting graphs of Z versus p for several temperatures, in figure VI.e. Notie that at p = and t =, (i.e. at the ritial point), the ompression fator is 0.75. The Z versus p urves for real gases have the same general shape, but the preise agreement in numerial detail is not quite so good. Where Z >, the pressure is greater than that of an ideal gas, the b (repulsive) part of the van der Waals equation being more important than the a (attrative) part. Where Z <, the pressure is less than that of an ideal gas, the a (attrative) part of the van der Waals equation being more important than the b (repulsive) part. I haven t investigated whether the other theoretial equations of state do appreiably better. Why not have a go yourself?!
6.8 FIGURE VI.e.6.4 Z. 0.8 t = 5.0.0.75.50.75 0.6 0.4.00.5 0. 0 4 5 6 7 8 9 0 p Figure VI.e. The ompression fator Z = PV / RT versus p (pressure in units of the ritial pressure for a van der Waals gas, for several values of t (temperature in units of the ritial temperature.) For a van der Waals gas the ompression fator is greater than for all temperatures greater than t = 7/8 =.75. At this temperature, the ompression fator is lose to up to p equals approximately, and this temperature is known as the Boyle temperature. At the Boyle temperature, the Z : p urve is flat and lose to for a fairly large range of pressures. Thus, at the Boyle temperature, even a non-ideal gas obeys Boyle s law fairly losely. For a van der Waals gas, the ritial temperature is 8a/(7Rb), so the Boyle temperature for a van der Waals is a/(rb). The reader should alulate this for H O and CO, using the values of the van der Waals onstants given in this Chapter. The dot on the t =.00 isotherm at p = and Z = 0.75 orresponds to the ritial point. Anyone who feels in need of more mental exerise might like to ask: For what value of p (other than zero) is Z =. For example, an you show that, for t =, Z = for p = 5 /7 = 5.60? The last proposed empirial equation of state that we mentioned is the virial equation, equation 6..6: PV = A + BP + CP + DP +K. This is sometimes written in the form B C D PV = A + + + K, but in these notes we ll use the form of equation 6..6. V V V The oeffiients A, B, et are alled the virial oeffiients and a funtions of temperature. The first oeffiient, A, is just RT. We an also write the virial equation as
Z 7 = + B' P + C' P + D' P +K. We ould measure the oeffiient B' for a real gas by plotting Z as a funtion of pressure in a Z similar manner to figure VI.e. The initial slope extrapolated to zero pressure gives P T the value of B '. At low temperatures B ' is negative; at high temperatures B ' is positive. At the Boyle temperature B ' is zero, and at that temperature the ompression fator is unity for a large range in pressures, and the gas aordingly losely onforms to Boyle s law. The oeffiient C' is small, so the term C' P omes into play only at higher pressures. At higher pressures, Z inreases, showing that C ' is a positive oeffiient. The oeffiient D ' is smaller still than C ' All the mathematially well-behaved equations of state below the ritial temperature have a maximum and a minimum i.e. the urve shows a wiggle. I illustrate this in figure VI.f. This is the van der Waals isotherm for t = 0.95 in dimensionless variables. It is the same as one of the urves shown in figure VI., drawn to a different sale so as to emphasize the wiggle. FIGURE VI.f 0.95 0.9 t = 0.95 p 0.85 a n 0.8 0.75 b d m e 0.7 0.6 0.8..4.6.8. v Using the little ylinder and piston to the right of the graph, try and imagine what happens to the enlosed liquid or vapour as you move the piston in and out at onstant temperature, moving from a to e and bak again on the graph. Start at e. The ylinder is filled with vapour. Move the piston inwards, going from e to d; the pressure inreases and the volume
dereases. Now a real gas doesn t follow the van der Waals funtion all the way. At d, something different happens. Atually it is possible to take a vapour a little way past d towards (but not beyond) n. That would be a superooled vapour, suh as is used in a loud hamber. It will ondense immediately into a line of liquid droplets as soon as a harged partile flies through the vapour. However, what usually happens is that some of the vapour starts to ondense as liquid, and we move horizontally from d to b. As we move the piston down at onstant temperature, the volume of ourse dereases, and more and more liquid ondenses in suh a manner that the pressure remains onstant. In the portion db, we have liquid and vapour existing together in the piston, in thermodynami equilibrium. Near to the d end there is only a little liquid; near to the b end it is nearly all liquid, with only a little vapour left. Beyond b, towards a, the spae is ompletely filled with liquid. We an push and push, inreasing the pressure greatly, but there is very little hange in volume, beause the liquid is almost (though not quite) inompressible. The isotherm is very steep there. It is atually possible to take the liquid a little way from b towards (but not beyond) m without any of it vaporizing. This would be a superheated liquid, suh as is used in a bubble hamber. It will vaporize immediately into a line of bubbles as soon as a harged partile flies through the liquid. There will be further important material onerning hange of state in Chapters 9 and 4. At present, though, I want to ask: At what pressure does ondensation ommene? Putting it another way, what is the height of the line bd in figure VI.f? I have heard it argued that the height of bd, (the pressure at whih ondensation ours) must be suh that the area bm is equal to the area nd. I am not sure that I fully understand the arguments leading to this onlusion. After all, a real gas doesn t onform exatly to a van der Waals equation or any of the other theoretial/empirial equations that we have disussed. But perhaps it is not unreasonable to draw bd suh that the areas above and below it are equal, and in any ase it makes for an interesting (and hallenging) omputational exerise. The van der Waals equation, in dimensionless variables, is given as equation 6..4. Can you alulate the pressure suh that the area bm below bd is equal to the area nd above it? I make it p = 0.8, whih is the height where I have drawn it in the figure. I haven t done the alulation for the other equations. I leave that to you!. 6.4 Gas, Vapour, Liquid and Solid 8 Our desription of the behaviour of a real substane in setion 6. was inomplete in many ways, not least beause it made no mention of the solid state. At very low temperatures or at very high pressures, most substanes will solidify.
9 P O L S + S L CP L + V TL S + V G V CI V P O S L CP L + V TL S + V G V CI V FIGURE V. FIGURE V.4 In figures VI. and 4 I have skethed shematially, by hand, the several regions in the PV-plane in whih a substane exists in its several stages. Unlike in Figure VI. and VI., I have not drawn isotherms. The various lines are intended to represent the boundaries between phases, and are therefore more akin to the dashed urves in figures VI. and VI.. The one exeption is the ritial isotherm, CI, whih is indeed the urve that separates gas from vapour or liquid, but whih is also, of ourse, an isotherm. The differene between figures VI. and VI.4 is that figure VI. represents a substane that expands when it melts from solid to liquid, while figure VI.4 represents a substane that ontrats when it melts from solid to liquid (that is, the solid is less dense than the liquid, and will float upon it.) Most substanes expand upon melting, but we have to inlude those exeptional substanes that ontrat upon melting, beause one suh substane is one of the most important of all water. You an try to understand the figures a little by moving along a horizontal line (isobar) or along a vertial line (isohor) and notiing where phase hanges take plae. Can you see, for example, where a solid will hange to a vapour without going through a liquid phase (sublimation)? You will note, in the figures, the ritial isotherm CI, that separates gas from liquid or vapour, and you will note that, at temperatures above the ritial temperature, the only phase possible is gas, and the substane annot be liquefied merely by ompression. You will note also the ritial point CP. You will also see the triple line TL, along whih solid, liquid and vapour o-exist together. What of the region marked O? The substane annot exist here in solid, liquid or gaseous phase. To that extent, we see that the van der Waals equation may be a little bit better than we thought it was, beause you will remember that it went up to infinity at a third of the ritial volume. All that this means is that by then the moleules are so tightly jammed together that you simply annot ompress them any further. Although a substane annot exist in an ordinary solid, liquid or gas phase in the region marked O, if the matter is degenerate it will be in this region. The eletron struture of the atoms breaks down, so that it then does beame possible to jam the atoms loser together. This may mean something to those of you who are familiar with the onept of degenerate matter. If you have not heard of it, do not worry; you are unlikely to ome aross it unless you visit a white dwarf star, or the ore of a massive star, or have to take an examination in astrophysis. For the time being, we shall look the other way and pretend it doesn't exist.
0 We an get a little more insight by looking at the PT-plane. Figure VI.5 shows a substane that expands on melting, and figure VI.6 shows a substane (suh as water) that expands on freezing. In the PT-plane, the triple point (where solid, liquid and vapour) are in equilibrium with eah other, appears as the triple point, TP. (In PVT-spae it is a line, although the ritial point CP remains a genuine point in PVT-spae.) The line separating liquid from vapour terminates at the ritial point, and the line is often drawn as though it were somehow left hanging in mid-air, so that one is unertain whether a given point near the ritial point represents a gas, a vapour or a liquid. But in the PT-plane, the ritial isotherm is a vertial line (show as dashed in the figures), and the liquid/vapour boundary terminates at the ritial isotherm, and there is no question what phase is represented by a point near to the ritial point. To the right of the ritial isotherm, we have a gas. To the left, we have either a liquid or a vapour, depending on whether we are above or below the liquid/vapour boundary. As we ross the solid/vapour boundary, below the ritial temperature and below the ritial pressure (on Mars!) we have a phase hange diretly from solid to vapour or vapour to solid i.e. sublimation. (I have often heard that, below the triple point, a solid will "sublime". I think I prefer the verb "to sublimate".) P CI P CI L S TP L CP G S TP CP G V V T T FIGURE VI.5 FIGURE VI.6 Really to appreiate these diagrams you need to see and to handle a three-dimensional model in - spae. My skills at making drawings with my omputer are nowhere near good enough yet for me to attempt a three-dimensional drawing, but Mr Charles Card of the University of Vitoria was kind enough to photograph for me a model from the University s olletion, and I reprodue these below as figures VI 7,8 and 9.
FIGURE VI.7
FIGURE VI.8
FIGURE VI.9 I now give some numerial values for the ritial temperature and pressure, the ompression fator, and the temperature and pressure of the triple point for H O and for CO. These are not intended as definitive values. I looked them up in a number of soures and I found a surprisingly wide range of the numbers quoted. They are given here merely to give the reader a rough idea of what the values are for these two substanes. The temperature quoted for the triple point of H O is, of ourse, exat, being one of the fixed points of the Kelvin sale. Reall that one atmospheri pressure is about.0 0 5 Pa. T (K) P (Pa) Z T (K) P (Pa) H O 647. 0 7 0.4 7.6 6. 0 CO 04 7. 0 6 0.76 7 5. 0 5
4 The reader might like to see whether these numbers are ompatible with the numbers I gave for the van der Waals onstants in Setion 6.. Exat agreement is not to be expeted, beause the figures I quote are only approximate and are gleaned from a variety of soures and also, of ourse, neither gas an be expeted to obey van der Waals equation exatly. If the numbers seem to be wildly disrepant, please let me know. We who live on the surfae of Earth are familiar with water in its solid, liquid and vapour forms, and this might suggest that the onditions on the surfae of Earth, the temperature and pressure, must be lose to the triple point of water. We see from the above table that the triple point of water (whih is defined to be 7.6 K = 0.0 ºC in the International Temperature Sale), is indeed near our typial ambient temperatures, but the triple point pressure of water is 6.7 Pa, whih is only about 0.006 atm. However, we are near the triple point if the partial pressure of water vapour in the atmosphere is lose to 0.006 atm, whih it often is. So we are indeed lose to the triple point, whih is why we so often see water in its three phases. Inidentally, the P : T diagram for the water system is a good deal more ompliated that the ideal diagram of figure VI.6, partiularly in the solid region, sine there are apparently many (about 5) different forms, or phases, of water ie. Some idle thoughts on vapours. There is a question of how to spell vapour. In the United States, vapor is usual, and in the United Kingdom vapour is usual. Vaporize is a bit trikier. The spelling vaporize is usual in the United States, but what to do in the United Kingdom? Is it vapourize, vapourise, vaporize or vaporise? Is there a u or no u? Is it z or s? To answer the first question: In the United Kingdom, the u, as in the United States, is omitted. Only weak spellers and those who would try to be more English than the English would try to insert a u. As for s or z, either seems to be used in the United Kingdom. Etymologially, z would be the better hoie, so the spelling vaporize is perfetly aeptable on both sides of the Atlanti Oean. More idle thoughts on vapours. Is a vapour a gas? What is a fluid? And is glass a liquid? Some authors treat gas and vapour as though they were quite different things: a gas is not a vapour, and a vapour is not a gas. Others regard a vapour as being a sort of gas namely a gas whose temperature is below the ritial temperature and whih an be liquefied by inreasing the pressure. In that ase, what do you all a gas that is above the ritial temperature? The term permanent gas is often used. Thus a vapour is a gas below its ritial temperature, and a permanent gas is a gas above its ritial temperature. A fluid is something that flows. Thus liquids and gases (inluding vapours) are fluids. There is, you would imagine, always a lear distintion between a liquid and a gas. But is the distintion always so lear? I admit that I have never atually seen the phenomenon that I am about to desribe, but it is desribed so often that I presume someone has seen it! Consider a losed ontainer with a liquid in equilibrium with its vapour. The liquid and vapour are separated by a sharp, horizontal boundary. That is to say, the system is on the line separating liquid and vapour in figures VI.5 and 6. This line an be regarded, if you like, as a graph of boiling point versus temperature, or equally of vapour pressure versus temperature. If you raise the pressure, the boiling point inreases; or if you inrease the temperature, the vapour pressure inreases. More liquid will enter the vapour state, and, as the pressure of the vapour inreases, so does its density. The liquid, on the other hand, is almost inompressible, and, beause of thermal expansion, its density dereases. As we move up the line separating liquid form vapour in the P:T plane, the density of the vapour inreases and the density of the liquid dereases. Their densities beome more and more equal until, as we approah the ritial point, the boundary between liquid and vapour beomes less and less distint, and less onstrained by gravity to be horizontal, until eventually, at the ritial point, the distintion between liquid and vapour blurs and ultimately disappears. So what have you got then? It is ertainly a fluid, but are you going to all it a gas, a vapour or a liquid? Sine none of these words would seem to have a stronger laim than either of the others, some authors refer to the substane when a little above and to the right of the ritial point in the P:T plane as a superritial fluid. There is also the question as to whether glass is a solid or a liquid. A famous radio personality many years ago, on a Brains Trust programme broadast by the British Broadasting Corporation, Professor C. E. M. Joad, was famous for his sentene: It all depends on what you mean by So I suppose the question as to whether glass is a liquid or a solid depends on what you mean by a liquid or a solid. The moment when I drop a tumbler and it shatters into many
5 viiously sharp fragments is not a good moment to onvine me that glass is a liquid. Those who assert that glass is a liquid say that it has not got a solid rystalline struture, and that it flows, albeit very slowly. It has a very large visosity. We are told that windows in anient mediaeval athedrals are thiker at the bottom than at the top, as a result of the visous liquid flow over the enturies. I don t know if any of the many people who have told me that have atually personally measured the thikness of a athedral window. At any rate, before you started this hapter, you had a very lear idea in your mind about the differenes between a solid, liquid and a gas. Now that I have painstakingly explained it all, you are ompletely onfused, and are no longer at all sure that you know the differene. 6.5 Kineti Theory of Gases: Pressure There will be more about marosopi PVT relations for gases when we go further into thermodynamis. In this setion, we deal with mirosopi properties, and how pressure and temperature are related to the number density of moleules and their speed. We shall onsider an ideal gas, ontaining n moleules per unit volume, eah of mass m, held in a ubial box of side l. The veloity of a partiular moleule is to be denoted by = ui + v j + wk. Here u, v, w are the omponents of the veloity parallel to the sides of the box. As ever, I shall use the word veloity to mean "veloity" and the word speed to mean "speed". Thus the veloity of the moleule is and its speed is. We are going to start by alulating the pressure on the walls, assumed to be aused by the ollisions of millions of moleules repeatedly olliding with the walls. ("Why do you keep banging your head against the wall?" "Beause it feels so good when I stop.") Consider the x-motion. Assuming that ollisions are elasti, we note that the hange of the x- omponent of momentum when a moleule bounes off a yz-wall is mu. The time taken to ross to the other side of the ube and bak again is l/u. The number of ollisions that this moleule makes with one yz-wall per unit time is u/(l). The rate of hange of momentum of that moleule at that wall is therefore mu % u/(l) = mu /l. The rate of hange of the x-omponent of the momentum at that wall of all the nl moleules in the box is nl mu / l = nml u. That is, the fore on that wall is nml u, and so the pressure on the wall is nm u. But u = v = w (that s assuming that the veloities are isotropi and there s no wind) and Pythagoras s theorem), and therefore u =. So the pressure is u + v + w = (that s P = nm = ρ. 6.5. Here ρ is the density = mass volume = molar mass molar volume = µ/v, (here V = molar volume) and therefore PV = µ. 6.5.
6 But µ is of the translational kineti energy of a mole of gas, and we already know that PV = RT, so that we dedue that the translational kineti energy of the moleules in a mole of gas is equal to RT. That is to say the mean translational kineti energy per moleule is kt, where k is Boltzmann's onstant (see Setion 6.). 6.6 Collisions In this setion, we are going to ask: What is the mean time between intermoleular ollisions? What is the mean free path between ollisions? How many intermoleular ollisions are there per unit volume per unit time? How many ollisions with the walls of a ontaining vessel are there per unit area per unit time? Sine I know little hemistry, I shall assume that moleules are hard spheres of diameter d. This may not be too bad for monatomi gases suh as the rare gases. For others, the assumption is tantamount to assuming that moleules repel eah other when their entres of mass approah within a distane d. In any ase, we shall assume that the ollision ross-setion is of area πd. Notie, from the sketh below, that two equal spheres ollide when their entres are separated by their diameter d, and onsequently the ollision ross setion (shown as a dashed irle) is of area πd. In fat in what follows, I m just going to all the area of the ollision ross setion σ; in doing that, I don t even have to assume that its shape is irular. In time t, a moleule moving with speed sweeps out a ylinder of volume σ t. If there are n moleules per unit volume, the number of ollisions that that partiular moleule will experiene in time t would appear to be σ tn, whih is to say that the number of ollisions it experienes in unit
7 time is σ n. Thus the mean time τ between ollisions would appear to be τ = /( σn), and the mean free path λ between ollisions is λ = /( σn). But this isn't quite right, beause we have not taken into aount the fat that all the moleules in the above-mentioned ylinder are moving. It is not as though our hero moleule were olliding with a set of stationary moleules. The relevant speed to use in this analysis is the mean relative speed between moleules, and this is a little greater than the speed of eah. Let s see if we an do a little better. Let s start by supposing that all of the moleules are moving with speed. There are two extreme sorts of ollision: The head on ollision: For suh a ollision, the relative speed between the moleules is. Then there is the sort of ollision in whih one moleule barely athes up with another one: In that ase the relative speed is zero. The average relative speed is evidently somewhere between 0 and. These are extreme ases. The average situation is somewhat in between. We may argue that the average situation is for the two moleules to be travelling in perpendiular paths:
8 If we think of this as the average situation, then we may argue that the average relative speed between two moleules is. In that ase, we may onlude that the mean time between ollisions is τ = / ( σn), and the mean free path is λ = / ( σn). This argument may or may not be ompletely onvining, but it is probably loser to the mark than our previous effort. Let s see if we an make a further improvement. As before, we ll suppose that eah moleule is moving with speed. Suppose our hero moleule to be moving upward with speed, and another moleule approahes at an angle θ, as in the sketh below. θ By vetor addition of veloities, it will be seen (a little thought will be needed) that the relative / speed of approah between the two moleules is ( + os θ). Now the fration of moleules approahing from angles between θ and θ + d θ is sin θd θ. This is beause the area of an elemental zone of a sphere of unit radius between θ and θ + d θ is πsin θdθ, and the total area of the sphere is 4π - see the sketh below: πsin θdθ 4π
9 / Thus the mean relative speed of all the moleules is ( + os θ) sin θdθ, whih works out to be 4. In this model, then, the mean time between ollisions would be τ =, and the mean free 4 σn path would be λ =. 4 σn However, we have still assumed that all the moleules are moving at the same speed. I am told (but I have not verified it myself) that, if you take aount of the Maxwell-Boltzmann distribution of speeds (see Setion 6.7), the mean relative speed of ollision is, where is the mean speed of the Maxwell-Boltzmann distribution (equal to τ = σn and λ =. σn π 0 8kT.) If that is so, then we obtain πm In any ase, sine moleules are not hard spheres (they are neither spheres nor hard) and the details of a ollision depend on the shape of the moleules and the fore law between them, it may not be meaningful to try to obtain an extremely preise formula for the mean free path, but instead settle for τ = and λ =, and if you wish to take b, you won t be far out. bσn bσn Of more interest would be to alulate the mean time between ollisions for various pressures and temperatures, and ask how does this ompare, for example, with the mean lifetime of an atom in an exited atomi level, or a metastable level. Or to ompare the mean free path between ollisions with the mean nearest-neighbour distane between moleules in a gas. I think under typial familiar onditions, you ll find that the mean free path is rather longer than the mean nearestneighbour distane. Also of interest is the number of ollisions per unit volume per unit time. If we suppose that a single moleule experienes bσn ollisions per unit time, and there are n moleules per unit volume, then the number of ollisions per unit volume per unit time is Z = bσ n The fator of is neessary so that we don t ount ollisions of A with B and of B with A as two different ollisions. Another useful result is that the number of moleules striking the walls of a ontaining vessel per unit area per unit time is n. 4.
0 To avoid repetition, I don't derive this here, but you will find a derivation in Chapter Setion.7 of Stellar Atmospheres, where I do the derivation with photons rather than with moleules. The only differene is that, in the ase of the photons, all are moving at the same speed (the speed of light), whereas here we have a distribution of speeds, and we use, the mean speed of the moleules. 6.7 Distribution of Speeds I am tempted to start by saying "Let f(u)du be the fration of moleules of whih the x-omponent of their veloities is between u and u + du." But we an go a little further than this with the realization that this distribution must be symmetri about u = 0, and therefore, whatever the funtion is, it must ontain only even powers of u. So we an start with: Let f(u )du be the fration of moleules of whih the x-omponent of their veloities is between u and u + du. Then, unless there is a systemati flow on the x-diretion or the x-diretion is somehow speial, the fration of moleules with y veloity omponents between v and v + dv is f(v )dv, and the fration of moleules with z veloity omponents between w and w + dw is f(w )dw. The fration of moleules in a box du dv dw of veloity spae is f ( u ) f ( v ) f ( w ) du dv dw. Sine the distribution of veloity omponents is independent of diretion, this produt must be of the form f(u ) f(v ) f(w ) = F( ), 6.7. or f(u ) f(v ) f(w ) = F(u + v + w ). 6.7. (Question: Dimensions of f? Of F?) It is easy to see that this is satisfied by ± f ( u ) Ae u / = m, 6.7. where A and m are onstants to be determined. It should also be lear that, of the two possible solutions represented by equation 6.7., we must hoose the one with the minus sign. Sine we must have f ( u ) du =, 6.7.4 it follows that A = π. 6.7.8 m
= (To see this, you have to know that ax π. e dx a ) 0 u m Thus we now have ( ) / f u = e. 6.7.9 π m This is the gaussian distribution of a veloity omponent. interpretation for the onstant m. We shall shortly find a physial The area under the urve represented by equation 6.7.9 is, of ourse, unity; the maximum value of f ( u ) is /( π). m Figure VI.0 illustrates this distribution. In this figure, the unit of speed is m. The area under the urve is. The maximum (at u = 0) is / π = 0.564. Exerise: Show that the FWHM (full width at half maximum) is ln m =.665m. This gives one physial interpretation of m ; we shall soon give another one, whih will explain the use of m as a subsript. 0.6 Figure VI.0 0.5 0.4 f(u ) 0. 0. 0. 0 - -.5 - -0.5 0 0.5.5 u The gaussian distribution deals with veloity omponents. We deal now with speeds. The fration of moleules having speeds between and + d is F( ) times the volume of a spherial shell in veloity spae of radii and + d. (Some readers may reall a similar argument in the Shrödinger equation for the hydrogen atom, in whih the probability of the eletron's being at a distane between r and r + dr is the probability density ψψ* times the volume of a spherial shell.
You'll notie that physis beomes easier and easier, beause you have seen it all before in different ontexts. In the present ontext, F is akin to the ψψ* of wave mehanis, and it ould be onsidered to be a "speed density".) Thus the fration of moleules having speeds between and + d is 4 / Φ( ) d = e m d. π m 6.7.0 I shall leave it to those who are skilled at alulus to show that Φ( ) d =, and also to show that the maximum of this distribution ours for a speed of = m and that the maximum value of Φ( ) is 4 /( me π).. This provides another interpretation of the onstant m. The speed at whih the maximum of the distribution ours is alled the mode of the distribution, or the modal speed hene the subsript m. Equation 6.7.0 is the Maxwell-Boltzmann distribution of speeds. It is shown in figure VI.7, in whih the unit of speed is m. The area is, and the maximum is 4 /( e π) = 0.80. 0 Figure VI. 0.9 0.8 0.7 Φ( ) 0.6 0.5 0.4 0. 0. 0. 0 0 0.5.5.5.5
The mean speed is found from Φ( ) d and the root mean square speed RMS is found from 0 RMS = Φ( ) d. If you have not enountered integrals of this type before, you may find that 0 the first of them is easier than the seond. If you an do these integrals, you will find that = and =. 6.7. π m RMS m The root mean square (RMS) speed, for whih I am here using the symbol RMS, is of ourse the square root of. We have seen from Setion 6.5 that the mean kineti energy per moleule, m, is equal to kt, so now let's bring it all together: π kt kt = = 0. 886 = = 0. 86 = =. 44 6.7. m RMS RMS 8 8kT kt = =. 8 = = 0. 9 = =. 596 6.7. π m m π RMS RMS πm π kt kt RMS = m =. 5m = =. 085 = =. 7 6.7.4 8 m m m m m Gauss: mu m f u kt e kt ( ) =. π 6.7.5 Maxwell-Boltzmann: m kt Φ ( ) = e. 6.7.6 π kt / m One last thing ours to me before we leave this setion. Can we alulate the median speed / of the Maxwell-Boltzmann distribution? This is the speed suh that half of the moleules are moving slower than /, and half are moving faster. It is the speed that divides the area under the urve in half. If we express speeds in units of m, we have to find / suh that 4 / π 0 e d =, 6.7.7 or / π e d = = 0.556 74. 0 8 6.7.8 That should keep your omputer busy for a while. Mine made the answer =.087 65. / m
6.8 Fores Between Moleules 4 We desribed in a qualitative manner in Setion 6. the fores between moleules the long-range attrative van der Waals fores aused by indued-dipole/indued-dipole interation, and the shortrange repulsive Coulomb fores as the moleules approah eah other losely, and how these intermoleular fores give rise to deviations from the Boyle s Law expetations for the equation of state for an ideal gas. Presumably, if we knew the exat equation for the fore law as a funtion of intermoleular distane, we ould in priniple alulate the equation of state; onversely, if we knew, through measurement, the form of the equation of state, we ould dedue the form of the intermoleular fores. I have not atually done this myself; an early referene worthwhile to look up would be Lennard-Jones, Pro. Roy, So. A, 4, (96). Qualitatively, the fore law for the interation between moleules would show a repulsive fore rapidly falling off with distane when the moleules are very lose (the moleules are hard ) and a longer-range attrative fore at larger distanes. Two of the simpler equations that have been used to desribe this are the Lennard-Jones potential: V = D 6 r e re + 6.8. r r and the Morse potential: V ) / ( ) ( r re e a. = D 6.8. Eah of these goes to V D as r, and V = 0 when r = r e. The Lennard-Jones potential (but not the Morse potential) goes to as r 0. These expressions annot be derived in the usual sense; they are merely expressions that are useful for disussion in that they desribe qualitatively the shape of the potential funtion that you would expet. The Lennard-Jones expression is often used in disussions of the van der Waals fore: if the van der Waals attrative fore is due mostly to indued-dipole/indued-dipole interation, an r 6 term is about right. The Morse potential is used more often in disussion of the fore between atoms in a bound moleule. If the Morse potential is put into the Shrödinger equation for an anharmoni osillating diatomi moleule, it results in a simple solution for the eigenfuntions and eigenvalues, with the energy levels being given as quadrati (and no higher) in v +.
5 FIGURE VI..8.6.4 Potential. 0.8 0.6 0.4 0. a/r e = 0. 0.77 L-J 0. 0.4 0 0.6 0.8..4.6.8 r The parameter a in the Morse expression determines how narrow or how broad the minimum is. It is left as an exerise for the reader to show that the FWHm (full width at half minimum) of the Morse expression is the same as for the Lennard-Jones potential for / 6 ( + ) ( ) ln( + 8) / 6 a = = 0.77 908. 6.8. In figure VI., I show, as ontinuous urves, the Morse potentials (in order of inreasing width) for a / r e = 0., 0.77, 0. and 0.4, and the Lennard-Jones potential as a dashed urve. Further omparisons between these two potential funtions an be found in T.-C. Lim, Z. Naturforshung, 58a, 65, (00).