edited by JOHN J. ALEXANDER University of Cincinnati Cincinnati. OH 45221
exam que~tionexchange On a Relation between Fugacity and Pressure
it leads to an untidy solution (cf. reference9).Therefore, we convert the integral I over p to an integral over u , yielding
Salem Jagennathan University of Nonh Florida Jacksonville,FL 32216
,-
problem shows that a simple relation ~h~ following exists between fugacity, f , and pressure, p, even for a gas satisfying a nonideal equation of state. Problems on fugacity are usually of the kind requiring the determination of a relation between f and p for gases satisfying: (1) a one-parameter van der Wads equation of state (14), (2) a virial equation of state in powers of p (5,6),or, (3) a Beattie and Bridgeman equation of state (7).
p = RTIu
+ RTBlu2
(9)
Evaluating aplau from eq 9 and substituting it into eq 8 yields
The right-hand side of eq 10 simplifies to
Unfortunately, on the one hand, (1)and (2), being easy, do not challenge serious students, while, on the other hand, (3) yields a complicated expression that fails to show that a simple relation often exists between f and p, even for a gas satisfying a nonideal equation of state. The following exam problem has been posed elsewhere (8)for a slightly different equation of state. Its stated solution (9), however, does not exhibit the simplicity of our solution (cf. eq 2 below), nor does it appear to have been obtained using the method described in this note. Ouesllon Show that for a gas satisfying the approximate virial equation of state
pul(RT) = 1 + Blu
-
where, in passing from eq 7 to eq 8, we have replaced dp with (dpl Ju)dv and the limits 0 5 p 5 p' with 5 u 5 u'. The limits of u are obtained from (1) by rewriting it as
where the limits of the integrals are reversed. The two integrals on the right-hand side of eq 11 can be evaluated by decomposing their integrands into partial fractions. Thus,
= In [(u'
+ B)lu']lB
and
(1)
the fugacity, f , is related to the pressure, p, by flp = [ul(u + B)] exp (2Blu)
(2)
where u, T,R, and B are the volume, temperature, universal gas constant, and the second virial coefficient, respectively. Acceptable Solullon Now the general expression (10) for f is
f = P' eap (fop' KZ - 1)lpldp)
Substituting eqs 12 and 13 into eq 11yields
I (3)
Define
+ In ["'I("'+ B)]
(14)
Substituting eq 14 into eq 6 yields flp'
where Z is the compressibility factor, Z = pul(RT)
= 2Blu'
Finally, settingp' eq 2.
= [u'/(o'
+ B)] exp (2Blu')
(15)
= p and u' = u in eq 15yields the desired solution,
Acknowledgment I thank J. S. Huehner for comments.
Literature Cited
Therefore, eq 3 becomes
f
= P'
exp (1)
Substitutingeqs 1and 4 into eq 5 yields
I. Denhigh. K. The P~inriplasof Chemieol Equilibrium: Cambridge University: Great Britain, 1968: Chapter 3, Equation 3.53. 2. Atkin.. P. W. PhvsicolChemisliu.. 3rd ed.: W. H.Freeman: NewYork. 1966:Cha~ter 6. ~ ~ ~6.3.k ~ l e 3. M O O P ~ .w J. B.Z& Phvsieal Chemiatry; Prentice-Hall: E n g i e w d Cliffs. NI. 1983; Chapter lo. Example 10.1. 4. G1asstone.S. Thsrmodynomirr for Chemists:Krieger: New York. 1972: Chapter XII, p
.
~~
.
916 . .~.
Although1can he evaluated by expressing u in terms of p, using eq 1, that is,
5. Reference I . Chapter 3. Pmhlcm8. 6. Refercnce4, Chapter XII, Exercise 4. 7. TunelLG.J. Phyr. Chem. 1931,35,2885 8. Reference?. Chanter 6. Problem 6.32. 9. Referenee 2, p 840. lo. Reference 7, Equation X.
Volume 64
Number 8
August 1987
677
Counting Halomethanes 'atlhasarathy Narnbi Veterans Administration Medical Center Kansas City, M O 64128 Undergraduate students learning organic chemistry are often asked t o enumerate t h e possible isomers of simple s aiven t o illustrate alkanes. However.. verv. little e m ~ h a s i is t h e range of molecular species t h a t can b e i e r i v e d b y substituting t h e hvdroaens . - of a n alkane, for example, b y halogens. T h e simple case is t o substitute t h e hydrogens of methane with t h e halogens ~uestion How m a n y halomethanes are there? Acceptable Solutions The ohvious solution is to m i t e out all the possible halomethanes. One does arrive at the correct answer of 70 halomethanes, including methane. However, there isan easier way, which isalso e simple example of the application of romhinatories m chemistry. We consider the four hydroaenr of mrrhnne as "boxea". We ha\,e five "obiects" (H, F. CI, Br, and I) t o u a ~ a n dplace in there boxes. \Ve need to r~memberthat repcrltimu arc possible, that is, two boxes can haw ohjectd oithe samr kind. Caw I: \Vc ran hnw all the hoxes conrainrng the same
678
Journal of Chemical Education
kind of objects, that is halomethanes of the type CXn.Since we have five different objects it is ohvious we have 5 halomethanes of this type. Case 2: Three boxes can have objects of the same kind, and the fourthcan he different, that is, CXaY. In this case for X we have five choices and for Y four choices. Therefore, there are 5 X 4 = 20 halomethanes of the type CX3Y. Case 3: In this case two hoxes have objects of the same kind, and the other two hoxes have different ohjects. Now, we have two possibilities with the halomethanes of the type CX2Y2andCX2YZ.Let us first consider the case of CX2Y2.For X we have five choices and for Y four choices, but we have to remember that CX2Y2= CY2Xz.Hence, there are (5 X 4)/2 = 10 halomethanes of the type CXZYZ.For the case of halamethanes of the type CX2YZ,notethat far X we have fii.e choices and for Y and Z there are four and three choices, respectively. Since CXzYZ = CX2ZY, we must divide by 2; that is, 5 X (4 X 3/2) = 30 halomethanes of the type CX2YZ. Case 4: In this case all the hoxes contain different objects, that is, halomethanes of the type CWXYZ. I t is a simple matter to find out that there are five halamethanes of this type. Adding all the above cases we find that there are 70 halomethanes. Moreover. there is an muation that can be used to arrive a t the ansa,crinonestep"l'hp nwntwr ufcomhinarims oin ohjerrs taken h at a time with the poss~hilnyof rrperltim i i given bv the equation: (n
C".k = & 8
+ k - I)! - I)!
In our case n = 5 and k = 4. Substituting these values forn and k we find that C , a = 70.
