Two Further Simple Methods for Calculating the Critical Compressibility Factor for the Redlich-Kwong Equation of State Mario Emilio Cardinali and Claudio Giomini iCMMPM Dept., "La Sapienza" University, 00161 Rome, Italy Inarecent paper,' R. W. Hakalapresented both anumerical and an analytical method to calculate the compressibility ' factor a t the critical point
according to the Redlich-Kwong (RK) equation of state of gases:
where u is the molar volume of the gas and a and b have their usual meanings for the two-parameter equations of state of real gases. Both of Hakala's methods lead to the unequivocal result ( Z ~ ) R K= and have, as common starting point, the cubic equations
Second Method The curve described by eq 1,written for T = T,(critical isotherm), shows a point of horizontal inflection, whose abscissa and ordinate are u = u, and P = PC,respectively. In other words, this curve has a third-order (or triple) contact point with the isobar P = PCin correspondence with u = u,. As a consequence, the following cubic equation in u:
obtained from eq 1 after setting T = T,and P = PC,has a triple root u = u,. Hence, it can be put in the equivalent form
and which stem from eq. 1and its first and second derivatives, ( a P l a u ) and ~ ( ~ 2 P l a u 2equated )~, to zero a t the critical point. In addition to Hakala's brilliant ao~roaches,here we would suggest two further methods to cakulate ( z r ) ~ f iBoth of them are analytical. The first leads to ( Z < ) H K = %Idirectly from eqs 2 and 3 by means of a simple artifice; the second does not require calculus and can be used also for van der Waals's and Berthelot's equations.
or, alternatively, can he written as
Flrst Method Let us write the left-hand member of eq 3 as F. Since F = 0, then also k.F = 0,where k is any numerical factor. Therefore, we can add a term -1.F to the numerator, and a term -2.F t o the denominator, of the ratio a t the right-hand member of eq 2, without altering its value. Consequently, we obtain:
hence,
u: (Zc'RK
= 2":
- 2b.v: - b2.u, - F - 3b.": + b3 -
P,.u3
- 3Pc.uc.u2+ 3P,.",2."
-p;u:
=0
Since eqs 4 and 5 are different forms of the same equation, the coefficients of the Dowers with the same exoonent must be the same; in particular, for the terms in u2,
-R.T, = -3P,w,
The authors believe that, owing to its simplicity and straightforwardness, the method here suggested can profitably replace, at a didactic level, the conventional approach, which involves the first and second derivatives of the equation of state for a real gas (see e.g., Atkins2).
' Hakala. ~~.R. W. J. Chem. Educ. 1985.62. 110-111. ~~
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2Alkins, P. W. Physical Oxford, 1988: p 31.
402
Journal of Chemical Education
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he mist^, 3rd ed.; Oxford University:
