An Alternative Method for Calculating the Critical Compressibility Factor for the Redlich-Kwong Equation of State Carlos Castillo S. Universidad del Valle. Apartado ABreo 25360. Cali. Colombia A recent article' in this Journal describes the calculation of the critical compressibility factor for the Redlich-Kwong equation of state
Substituting RT, from eq 7 into eq 6 allows us to solve for the parenthesis explicity, i.e.,
(=IT,'"
- 3P,Vcb - P,bz) = 3P,V:
(8)
By similar substitutions of eqs 7-and 8 into eq 5, one has where V is the molar volume of the gas and a and b are the usual parameters in equations of state of real gases. Although the method uses two relationships that stem from the above equation and its first and second derivatives (aPl a V T a n d (d2PlaV)~equatedto zero a t the critical point, the algebraic simplifications are very tedious for an undergraduate physical chemistry student. I propose an alternative method using calculus to find the critical comprensibility factor for this same equation of state, which leads to its value (ZJRK= 113 directly without any additional artifice and can be used also for other cubic equations of state of gases. I start by writing the Redlich-Kwong equation as a cubic equation in V:
P V - R T V +[ ( d ~ '-"R)~ T -b 2 p ] v - ( a / ~ l ~= )o b (2) Differentiating eq 2 implicitly to obtain ( ~ P I ~ V = P, T the following equation holds: 3 V P + V F - 2RTV+ [ ( a l ~ ' " ) RbT
- b2P]- b2VF = 0
3V:b
+ 3Vcb2+ b3 = V,3
(10)
Which is solved by a simple artifice: we add the term VC3in both sides of eq 10 and therefore we obtain: V,3+3V,2b+3Vcb2+ b 3 = 2V:
(11)
This equation is also equal to ( V , + bP, which immediately yields b = (zLI3- 1)V"
(12)
Substituting eq 12 into eq 9 let us solve for the value for a: a = P,V;T,'"/(~'/~- 1)
(13)
We find also a and bin terms of PCinstead of V , by solving eq 7 for V, and substituting it into eqs 12 and 13. The results are: b = (2'"
- l)RT,/3Pc
(14)
and
6VP+3VP'+VP"+3VP'-2RT-bzP'-bzVP"-bZF=O (4)
a t the critical point P = PC;V = V,; T = T,; P = 0 and P" = 0, and therefore eqs 2,3, and 4 become, respectively,
+ ( a / ~ : / '- RbT, - b2PJV,- ( a / ~ : ' ~ =) b0
(9)
Solving the last two equations for a/T,L'2, the final result is:
(3)
Again differentiating eq 3 implicitly I obtain
P,V,3 - RT,V,Z
(alT,'")b = P,V,3
(5)
a = R2T,"2/9(21'3- 1)P,
(15)
Another simple method for calculating 2, for the R-K equation of state (which does not require calculus) was presented by M. E. Cardinali, et aL2 Acknowledgment The author wishes to thank R. Paredes and A. Jaramillo for review of the manuscript and for helpful suggestions.
Equation 7 provides us with
' Hakala, R. W. J. Chem. Educ. 1985, 62,110-1 11.
Cardinali, M. E.; Giomini, C. J. Chem. Educ. 1989, 66, 402.
Volume 68 Number 1 January 1991
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