MATHEMATICAL PROBLEM PAGE Directed by PAUL C. CROSS The
University of Wisconsin, Madison, Wisconsin PROBLEMS
26-30
T
HE following problems illustrate the use of some of the tables and equations in volume I of the Iuternational Critical Tables. Problems 29 and 30 involve the use of the tables. Solutions will be given in the September issue.
(a) What is the value of the constant 6 in the Callendar equations?
26. The symbols denoting the fundamental diiensions, length, mass, time, and temperature are I, m, t, and T, respectively. (a) Energy, E, has the dimensions of m12t-2 and frequency, v, has the dimension of t-'. What are the dimensions of Planck's constant h, given by the relation hv = E? (b) What is the dimension of the quantity hv/c2, where G is the velocity of light? Velocity has the dimensions of It-'. (c) Stefan's law gives S = CT4. S is the energy emitted per unit area per unit time by a blackbody radiator a t the temperature T. C is a constant. What are the dimensions of C? Show that with proper choice of units C may be expressed in terms of erg c m . 3 see.-' deg.-'. (I.C.T., I, 1Gff.) 27. The unit cell of a crystal of MgO has been found to be a cube 4.20 Angstroms (4.2 X lo-' cm.1 On an edge. Each cell contains the equivalent of 4 atoms of Mg and 4 atoms of 0. What is the density of MgO crystals from this data? (I.C.T., I, 344.) 28. A platinum resistance thermometer had a resistance of 2.3675 ohms a t O°C., 3.2881 ohms a t 100°C. and 6.2543 ohms a t 444.G°C.
t-(pt)=6--1 (2) (10 (pt) is the uncorrected temperature or "platinum" temperature. These equations facilitate the determination of temperatures from resistances by successive approximations in the "correction" term; i. e., in the right side of equation (2). (b) What is the temperature to within O.l°C. of a thermostat in which the thermometer has a resistance of 5.4240 ohms? (I.C.T., I, 54.) 29. Physical measurements on different preparations of a certain organic compound were as follows: Boiling point 81-8Z°C.; Density 0.78-0.79 g./cc.; Refractive index 1.375-1.380. Identify the compound using the property-substance tables as directed on page 100 (I.C.T., I). 30. Molecular refraction, R,is given by the relation Z - 1 . - 7M R = n---n2+2 d where n is the refractive index, M the molecular weight, and d the density. Calculate the molecular refraction of phenyl ether (GHnOCeHs) from the data in the International Critical Tables.
21.
(a)
c
_
(pt)
I
= -2.456/2.303
= -1.067
(RIM Rt- ")lo0 Ro -
(1)
)A
= 273/(22.4 X 329) = 0.03705. = e-6.63 x 0.03705 x 10,
log
=
(c) 71.9% absorption is equivalent to 29.1% transmission. c = 82.7/(22.4 X 760) = 0.00486. 2.303 log 0.291 = -0.00486 X 10 X k. k = 2.303 X (-0.5361)/-0.0486 = 25.4.
= 2.933.
10
antilog 3.933 = 0.0857 = fraction transmitted. 1 - 0.0857 = 09143 = fraction absorbed or 91.43% absorbed. (h) 80% absorption is equivalent to 20% transmission. Thus 1/10= 0.2. c = (273 X P . d / ( 2 2 . 4 X 329 X 760) = 0.00004875 x Pam. 2.303 lag 0 . 2 = -6.63 X 20 X 0.00004875 X p,,. 2.303(-0 699) = -0.006464p,,. p,, = 249 mm.
22.
+ +
+
= ap/(l bp); I. bpz = ap. Divide by ax and transpose: p/x = bp/a l/a. Thus if p/x is plotted against 9, a straight line will result. The intercept on the p-axis is l/a; so a = llintercept. The slope is bra. Thus b = slope X e = slape/interceot. .. 23. The total heat capacity of the thermostat and water is 105 kg. cal. per degree. The amount of cooling per liter of water through the roil will be (t-20)/105 degrees. Thus, dtldn = -(t-20)/105. (The minus sign indicates that the temperature is being lowered.)
509
x
Separating the variables. dt/(t-20) = -du/105. Integrate between the limits t = 40 and v = 0 t o t = 25 and v = v. 2.303[1og(25-20) - log(40-20)l = -(v-0)/105. log = -0,602 = -u/241.8. v = 145.6 liters. At the rate of five liters per minute, this would correspond to 29.12 minutes of Bow. 24. 24 X 0.092 = 20 X N. where N is the normality of the acid solution. Thus N = 0.092 X 24/20 = 0.1104. M = ' / z N = 0.0552 mol per liter. 0.0552 X 126 = 6.955 grams per liter.
25. dc/c = -kdt. Integrating between limits, In
= -k(t-to) ce C
log - = (-k/2.303) LO
X 1. (Since to = 0.)
Thus log1/, = -0.301 = (-k/2.303) X $11,. When the substance is three-fourths decomposed, one-fourth remains, and c/co = I/,. log = -0.602 = (-k/2.303) X tv,. Therefore. t~/,/ta/, = 0.301/0.602 =
