University of Connecticut OpenCommons@UConn Chemistry Education Materials Department of Chemistry 2015 The van der Waals Equation as a cubic Carl W. David University of Connecticut, [email protected] Follow this and additional works at: https://opencommons.uconn.edu/chem_educ Part of the Physical Chemistry Commons Recommended Citation David, Carl W., "The av n der Waals Equation as a cubic" (2015). Chemistry Education Materials. 88. https://opencommons.uconn.edu/chem_educ/88 The van der Waals Equation as a Cubic Carl W. David∗ Emeritus, Department of Chemistry University of Connecticut Storrs, Connecticut 06269-3060 (Dated: April 4, 2016) The van der Waals (from his thesis of 1873) equation is a cubic in the molar volume. Since this is the first equation of state studied in chemistry and physics more complicated than the ideal gas equation, it is noteworthy that the solution of this cubic is rarely if ever addressed. When (lowering the temperature) the two imaginary roots coalesce with the real root of the van der Waals equation we have the critical point. At Kelvin temperatures below Tc. the equation has three real roots; we explore obtaining those roots to aid in the Maxwell construction required to obtain the vapor pressure. I. INTRODUCTION The reduced van der Waals equations pr(vr;Tr) in the T p reduced variable scheme, i.e., Tr = , pr = and Tc pc v vr = , where the subscript c refers to the critical tem- It is well known that the van der Waals equation is vc a cubic equation in v whose solution is rarely addressed perature, pressure and/or molar volume and r, similarly, [1, 10, 11]. There no longer exists a reason for this lack of refers to the appropriate reduced variables, is familiarity, as computer assisted calculus programs have made it simpler than ever to manipulate the complicated 8Tr 3 pr(vr;Tr) = − 2 (1) expressions that constitute the traditional cubic polyno- 3vr − 1 vr mial solutions (vide supra). Rather than trusting a com- puter calculus program, however, this paper introduces We will derive the solution to the cubic, and then use the complete solution to the cubic using elementary al- that solution in the Maxwell construction searching for gebra. the vapor pressure. In an age in which algebra inclusion [4] in high school curricula is being questioned, it seems reasonable to ar- gue that the two years of calculus learning prior to study- III. THE GENERAL CUBIC EQUATION [7] ing physical chemistry is worth the effort. Therefore, avoiding the phrase "integration by parts" in Winn's pa- The general cubic equation is per [11] and avoiding the cubic solution which employs 3 2 only algebra (not counting DeMoivre's theorem) seems x + a1x + a2x + a3 = 0 (2) bad pedagogy. Instead, we should emphasize the useful- ness of what we've forced students to learn, validating which Cardan transformed using the prerequisite structure that differentiates STEM edu- a cation from other fields. x = y − 1 Since this solution is rather long-winded, this paper has 3 been written in sufficient detail that it should be able to i.e., be read without resorting to paper and pencil. 3 2 a1 a1 a1 y − + a1 a1y − + a2 y − + a3 = 0 II. PRELIMINARIES 3 3 3 which expands to Consider that the Maxwell construction [6] (vide infra) requires that one has the three roots (if T < T ) of the a2 2 a a c y3 + a − 1 y = − a3 + 1 2 − a (3) cubic van der Waals equation so that one can integrate 2 3 27 1 3 3 (pvdW − ptrial) dv from the first (smallest) to the second, and from the second to the third (largest) root (or vice which we write in the form: versa). Setting the sum of these two integrals equal to zero gives an equation whose only remaining variable is z3 + ηz + q = 0 (4) the pressure (ptrial); that solution pressure is the pre- dicted vapor pressure. This (Equation 4) is called a \reduced" cubic form (the mathematicians and chemists use the term \reduced" dif- ferently). We will solve the \reduced" equation for z ! y and then take the solutions and convert them back to x ∗ [email protected] at the end of the computation. 2 IV. THE VAN DER WAALS EQUATION AS A which suggests writing CUBIC 4η 2 = −3 The van der Waals gas equation can be expanded to u 2 so that 8Tr v 9vr 1 v3 − + 1 r + − = 0 r r p 3 3p p 4η r r r u = −3 which is of the form becomes the choice we have to make. Then, 3 2 vr + a1vr + a2vr + a3 = 0 (5) 3 4η 4q 4cos # + 2 cos# − 3 = 0 In terms of the original cubic equation (Equation 2), we u q 4η have −3 8Tr + 1 which is a mimic of the equation we are trying to solve. a = − pr (6) 1 3 3 3q 4cos # − 3cos# + q = 0 η 4η 3 −3 a2 = (7) pr so, we obtain and finally 3q cos3# = − q 4η 1 η −3 a3 = − (8) pr or In terms of the reduced cubic Equation 4, we have 3q 3q r−3 cos3# = − = − 2 q η a1 2η η η = a − (9) 2η −3 2 3 and which leads us to values of # for which this equation will be true. 2 a a We have q = − + a3 − 1 2 + a (10) 27 1 3 3 1 3q r−3 # = cos−1 − Consider the known equation 3 2η η 4cos3# − 3cos# − cos3# = 0 (11) which means, back-substituting r 4η 1 3q r−3 (DeMoivre's theorem gave us Equation 11) and the afore- ucos# = cos cos−1 − mentioned prototype \reduced" cubic: −3 3 2η η z3 + ηz = q (12) or, in terms of the reduced cubic equation we have r r where η and q are constants (see Equation 4) . Their 4η 1 3q −3 z = y = cos cos−1 − ≡ z commonality suggests a procedure, e.g., defining −3 3 2η η 1 z = ucos# Dividing this z −z1 into the original reduced cubic would yield a quadratic which then itself would be soluble, i.e., where u is the be chosen to force the two forms into conformance. To carry out this agenda, we cube our z3 + ηz − q z3 + ηz − q = expression for z and substitute the result into our reduced q h q i (z − z1) z − 4η cos 1 cos−1 − 3q −3 cubic, i.e., −3 3 2η η u3cos3# + ηucos# − q = 0 Synthetic division yields 2 and divide both sides by u3. Multiplying by 4, we obtain 2 2 ( η + z1 z1 − q z + z1z + (η + z1 ) − z − z1 4η 4q 4cos3# + cos# − = 0 u2 u3 where the numerator of the residual is zero! 3 The other two roots of V. THE MAXWELL CONSTRUCTION [6] 2 2 z + z1z + (η + z1 ) = 0 To achieve the Maxwell construction (see Figure 1) one are needs to integrate prdvr from the lowest root to the cen- tral root, and from the central root to the largest root, p 2 2 −z1 + z − 4(η + z ) and set these two integrals equal to each other (except z = 1 1 2 2 for the sign, i. e., using the absolute magnitude of the integrals) thereby obtaining an equation for ptrial (to be and vp called pr after solution) which we declare to be the va- p 2 2 por pressure for this particular isotherm. Alternatively, −z1 − z1 − 4(η + z1 ) R z3 = one could write (pvdW −ptrial)dv = 0 with the rule that 2 vp ptrial ! p when the equation is true. which lead to the second and third roots of the \reduced" The integral itself is trivial, i.e., cubic. Z v` Z v` 8Tr 3 prdvr = − 2 dvr 1. back substituting v v 3vr − 1 vr g8T (3v g − 1) 3 3 r `n r` + − (13) We thus have for the original cubic we started this 3 (3vrg − 1 vr` vrg 3 2 discussion with: x + a1x + a2x + a3 = 0, Equation 2, (with v rather than x): r This must be set equal to the equivalent path integral from "g" to "`" using the trial vapor pressure, i.e., r 4η 1 3q r−3 a v = cos cos−1 − 1 r` −3 3 4η η 3 Z ` p dv = pvp (v − v ) (14) (corresponding to the z1 root of the \reduced" cubic) trial r ` g where we have employed Equation 5 and its similar form g Equation 2 to return from the reduced form to the ulti- mately desired v form of the cubic van der Waals equation so the equation whose solution yields the vapor pressure we are trying to solve. is 2 a1 2 3 We had η = a2 − 3 (Equation 9) and q = − 27 a1 + a1a2 −a3 (Equation 10). It is clear that substituting these 8T (3v − 1) 3 3 3 r `n r` + − = pvp v − v (15) into our equation for v would lead to a fearful mess. And 3 (3v − 1 v v r r` rg then the substitutions which was Equation 6, Equation rg r` rg 7 and finally Equation 8 make evaluation possible. Un- fortunately, the resultant equation is enormous. This equation depends on a trial pressure which itself is vp A similar discussion for each of the other two roots adjusted until the equality holds (ptrial = pr ). However, leads to explicit solutions for all three roots. as noted by Lekner[5] [8], this can be written as 8T (3v − 1) 3 3 8T 3 r `n r` + − = p v − v = r − v − v (16) 3 (3v − 1 v v trial r` rg 3v − 1 v2 r` rg rg r` rg r` r` or ! 8T (3v − 1) 3 3 8T 3 r `n r` + − = r − v − v (17) 3 (3v − 1 v v 3v − 1 v2 r` rg rg r` rg rg rg since the trial pressure is the same at \`" and \g", or ! 8T (3v − 1) 3 3 8T 3 8T 3 r `n r` + − = r − v − r + v (18) 3 (3v − 1 v v 3v − 1 v2 r` 3v − 1 v2 rg rg r` rg r` r` rg rg or \vice versa".