The Redued van der Waals Equation of State The van der Waals equation of state is na + ( V nb) n (1) V where n is the mole number, a and b are onstants harateristi of a artiular gas, and R the gas onstant P, V, and T are as usual the ressure, volume, and temerature Here we are exressing the van der Waals equation in molar quantities; but as usual, we an relae nr by Nk and write it in terms of moleular quantities It turns out that if we examine the isotherms of a van der Waals gas on a P V lot, one sees a oint of infletion on the isotherm orresonding to the ritial oint of a gas In other words, we have 0 and 0 V V T T T T We set n 1 mole for onveniene and investigate these relations Our goal is to derive a redued form of the van der Waals equation that will not inlude the onstants a and b We first write the van der Waals equation in the form a () V b V Next we find the first and seond derivatives and set eah one equal to zero: a + 0 () V V b V Now we solve both Eq () and Eq (4) for T ( ) 6a 0 (4) V V T ( b) av (5) V ( b) av (6) 4 V Equating the right hand sides of these last two equations, we obtain 1
V b a a V V V b or finally, V b (7) We substitute this result bak into Eq (5) to obtain ( ) ( b) a b b 7b We substitute both of these results bak into Eq () to obtain a 7b a or V b V b 9b 7b Note that we have at this oint found the ritial quantities, V, and T in terms of the onstants a and b We ollet these three results as follows: V 7b b 7b or (8) (9) (10) Next, we define the following redued quantities: V T ; V ; T (11) V T Thus the molar van der Waals equation, a +, V beomes 8 a a 8 ( bv b) R T + a b b V R b 7 9 7 Colleting terms and simlifying, we at last obtain the redued van der Waals equation,
+ ( V 1) 8T (1) V Exerise: Plot the isotherms of the redued van der Waals equation and onfirm that there is a oint of infletion for the ritial temerature, T 1 This equation, whih imlies that the equation of state for any van der Waals gas takes exatly the same form, is sometimes alled the Law of Corresonding States In fat, that law is a good deal more general As the grah I will hand out in lass shows, exerimental data for a wide range of substanes fall on the same urves if P, V, and T are measured in terms of the redued quantities defined above (The grah is taken from Stanley s book on ritial henomena, ited below) PV It is of interest to onsider the omressibility ratio Z For an ideal gas, this quantity is of ourse one For a van der Waals gas, Z a ( b ) 7 b 075 (1) 8 7b This result should hold for any van der Waals gas If we omare this redition to the results of ritial oint measurements of real gases, we find something like the following table: Fluid PV water 00 arbon dioxide (CO ) 075 nitrogen (N ) 091 hydrogen (H ) 004 For a more omlete list (and disussion), see H Eugene Stanley, Introdution to Phase Transitions and Critial Phenomena, Oxford, 1971, age69ff The moral is that although the van der Waals gas is learly an imrovement on the ideal gas it at least redits a hase transition, and shows the saling features that an be found in redued lots of real gases it is by no means a highly aurate desrition Z
Inversion Curve Finally, onsider the inversion urve for a van der Waals gas We onsidered a Joule- Thomson (or Joule-Kelvin) roess, in whih a gas exands at onstant enthaly aross a Joule-Thomson valve We define the Joule-Thomson oeffiient µ as T µ (14) H To alulate this quantity, we onsider dh TdS + Vd (15) We substitute the seond TdS equation to obtain or, olleting terms, V dh CP dt T d + Vd T V dh CP dt + V T d (16) T Sine H is onstant, dh 0, whereuon Eq (16) redues to µ V V T T T V T C C H ( α 1) 1 V where α is the oeffiient of thermal exansion As we have seen, V T µ > 0 imlies the gas will beome ooler, µ < 0 imlies the gas will atually warm u Thus, µ 0, the equation of the inversion urve, is the urve that divides these two regions Refer to the grah of temerature vs ressure for hydrogen to understand how this situation lays out exerimentally Exerise: Show that for an ideal gas, µ is always zero Exerise: Various aroximation shemes lead to the result in for low densities, the inversion urve is given aroximately by a kti (18) b Confirm this result by working through G&T age 87 arefully and in detail Be sure you see how they introdue the low density aroximation Note that they imliitly define a density ρ N/ V See Reif for a similar aroah By omarison with Reif s exerimental grah for nitrogen, ersuade yourself that this result is not eseially desritive (17) 4
It is straightforward to find the inversion urve for a van der Waals gas We start with the molar van der Waals equation of state, a V b V and alulate the exansion oeffiient, as follows: We obtain after a little algebra the result R a V 0 + + T ( ) V b V b V T R V V b T a + V V b a T R V (19) Hene the exansion oeffiient is given by 1 V V b α V T a V b VT R V We seek the inversion urve: µ 0 αt 1 0 Using Eq (0), it is straightforward to show (0) a V b T VT R V, whih after a few lines of algebra redues to a V b b kt V This result for the inversion urve is onsiderably more lausible than the aroximate form ited above To ut it into a onvenient form, we use redued van der Waals quantities It follows from Equations (10) and (11) above that (1) 5
We substitute as follows into Eq (1): V bv T 7b a bv b b, 7 b T bv whih after a little algebra redues to 4T ( V 1) Equation () is the equation of the redued inversion urve It is onvenient to write it in terms of and T We will do so, in the roess disensing with the tilde symbol (~) for simliity We first note that, taking the square root of Eq (), ( V 1) V 4T 1 4T V V We use this result to substitute into the redued van der Waals equation [Eq (1) above], whih we write as we obtain 8T V V or 4T 8T V 1 ; V 8T 4T 4T 4T After several lines of algebra, we obtain at last the result ( T ) () 9 1 () whih gives us the inversion urve for the van der Waals equation in a useful form If we solve for T as a funtion of and make a grah, the resulting urve is qualitatively similar to the exerimental inversion urve for nitrogen in Reif s book! 6