MATHEMATICAL PROBLEM PAGE Directed by EDWARD L. HAENISCH Montana State College, Bozeman, Montana
A
SIMPLE working rule to find the maxima and minima of a function, f ( x ) ,is: (1) Find the first derivative, f'(x).
(2) Find the values of x (xl, x2, etc.), for which f'(x) vanishes. (3) Find the second derivative, f"(x). (4) Find the values of f"(xl), f"(x2), etc. If f"(x,) is positive then f(xl) is a minimum. Iff"(x,) is negative then f (xl) is a maximum. * Another test to see if X I , xm,etc., are points of minima or maxima is as follows. Obtain values for f(x) for x's which are slightly smaller and slightly larger than X I , xz, etc. If these values are larger than f(xl), X I is a minimum; if smaller, X I is a maximum. A point of inflection separates an arc concave downward from an arc concave upward. At such points f l ( x ) vanishes and f" ( x ) changes sign as it passes through the critical value. With a knowledge of the location of the points of maxima, minima, and inflection and with the aid of one or two points computed directly from the function, a rough plot of y = f(x) can be made.
Hint: Use the form P = -- (V-b) V2' 6 . Show that the probability curve, y = KedQ2 has a point of inflection for x = * 1/T-l2/h. SOLUTIONS TO APRIL PROBLEMS
1. (See Figure 1.) The tangent line at a temperature of 65'C. Ay 294 - 100 is drawn. The slope is - = = 8.62 mm. Ax 350.0 - 327.5
-
$
=
AH cal. molF X 187.5 mm. 1.987 cal. mol-' deg.-' X 33X2 deg.l AH = 10,440 cal. mol-'.
8.62 mm. deg.? =
REFERENCES
DANIELS: Chap. IX. MELLOR: 154-67, 172-5. PROBLEMS
1. Find the maxima and minima for (a) y = - 9x2 - 272 30 (b) y = 3x6 - 125xJ 21GOx 2. The density, d, of water a t any pmperature, t°C., is given by d = do(l at btz ct3) Where do is the density at O°C., a = 5.3 X lo-', b = -6.53 X 10W8,c = 1.4 X At what temperature does water have its maximum density? 3 . Trace the curve y = x3 - 9x2 242 - 7 . 4. The velocity of a certain antocatalytic reaction is dx given by, -" = K x ( a - x ) where x is the amount of material decomposed a t time, t; a is the amount of material originally present. Show that the velocity is greatest when x = 5 . Find the maxima and minima in Van der Waals'
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+
+ + +
+
2. (See Figure 2.) Notice that plotting log P against T changes thecurve of Figure 1 into a straight line. If we take the values for T = 313,353,373 we obtain the equations
.
1.7429 = 0.003195A
+ 2.4957B 4-
2.5504 = 0.002833A f 2.5479B 2.8808 = 0.002681A -t 2.5718B
+C +C
These may he solved by use of determinants, giving A = -2860.8, B = 4.37, C = 21.79. To find the slope:
-
l o g P =A~ + B l o g T + C
* I f i f ( x , ) also vanishes then x, may be a point of inflection. In such cases continue to differentiatef(x) untilf (n)(z) is obtained , which point f (z) = 0. If n 1s which d m not yanish at x ~ at even apply the tests in (4).
A B LnP ---+-InT+C 2.303 T 2.303
244
C
d(ln P ) - 2.303d (log P ) = - II,or d (lag P) AH, or R
2.303R'
I. e., if log P is plotted vs.
AY = From the graph, slope = A
-
1 - we gel a straight line of T
0 - 1x2 - -1.62 0.00251 - 0.002 0.00051
-AH = -2.303 X 1 9 8 i X 3176 = 14,530 cal. r n ~ l - ~ . (Note: Care must be taken to plot log P correctly. Thus. if P = 0.28 log P = -1 log 2.8 = -1 0.4472 = -0.5528.)
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4. (See Figure 4.) ing table:
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The data for the chord plot are in the follow-
dP This may he used with the values of the coefficients to get dT and AH calculated as in 1. dP P dP 3. (See Figure 3.) Rearranging - = AfLJ' - we get - = dT RTZ dT TZ
Moles CHsOH
%
Moles H.0
Au As
A sample calculation for - =
-
=
! ! As
34.0237 - 33.3615 = 55,96. 0.295148 - 0.283314
V2=as read from the curve for a 46% solution is 56.217.
