JOHN J. ALEXANDER University of Cincinnati Cincinnati. 45221
Collision Theory
volume = 2.5 mi3 X (1.6 X 103m = 1.0 X
Richard S. Myers Department of Physical Sciences Delta State University Cleveland, Mississippi 38733
total number = (2 X
Definine an observation as the "collision" of an obiect with an ohservation device permits an interesting application of collision t h e w to a ~ r o b l e mof current interest. Deoendine on the numhe; of hi& supplied by the instructor,-the fo< lowing question could be employed in general or physical chemistry courses. Ouesfion Given the following information, what is the maximum number of "monsters" which could inhabit Loch Ness?
lo-"
101° m3
creatures m+) X (1.0 X 101°m3)
= 2 X 102creatures Biomass calculations,based an assumptions about the diet of h e creatures, indicate a population of lo2*' in agreement with this calculation. Literature Cited (1) Swan. C..Notwe 294,497 (1976).
Temperature Dependence of K, M. J. Steffel
1) The loeh is 24 mi long and 1mi wide with an average depth of 540 ft. 2) A rough estimate of the swimming velocity is 3 m sec-l. 3) The frequency of sightings by underwater strobcecopic pbotography in 1975was once every 1@-lo5see. 4 ) Because of the loch's turbidity, the visibility range is only about 10 m. Hints 1) Consider a sighting to be a collision. 2) A moving object collides with a stationary and dissimilar target with a mean time of collision t = ( n m - ' where
n = number of density of objecta o = collision cross-section u = velocity
3) Consider o = s r 2where r is the range of visibility around the target. Satisfactory Answer 1
The mean time for collision is t = not' 1
The number density of creatures is n = to" t is given as 10C105sec. Assume t = 5 X lo4 see o = sr2 = (3.14)(10m)2 = 3.1 X 102m2
n
=
1 (5 X 10'sec)(3.1 X 102m2)(3m sec-') =2X
Ohio State University Marion Campus Marion, Ohio 43302 First year studentn nometimes fail to connect general featuresof equilibrium theorv with the studv of ioniceauilibria in aqueous solution. This question requires the resofution of a conflict between a common sense notion (How can pure water be anything but neutral?) and an isolated fact (The p H of pure water is 7.). The dilemma is resolved by the realization that K,, like other equilibrium constants, depends on T. Answering this question requires students to synthesize material which usually has been presented in separated fashion. Ouesfion At 100°C the pH of pure water is 6.12. (a) Calculate the [HsO+]in water at 100°C. (h) Calculate the [OH-] in water at I W C . (c) Is this water acidic, basic, or neutral? What is the basis for your answer? (d) Why is the pH not 7.W? Satisfactory Answer (a) pH = 6.12 = -log [H30+] [H30+] = 10-"I2 = 7.6 X 10-7 M (b) In pure H20, [OH-] = [H30f], so [OH-] = 7.6 X M (c) Neutral, since [H30+]= [OH-] (d) Water has apH of 7.0 at 25'C since [H@] = M. Apparently the disswiation is more extensive at 100eCoroducine hieher ion concentrations. At this temperature the vaiue of ~ , - w i ihe different from that at 25%
monsters m-3
total number = number density X volume = n X volume 540 ft volume = 24 mi X 1mi X 5280 ft mi-'
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Volume 55, Number 4, April 1978 1 243
