Proportional dependence

theorem at its face value and apply it, we can arrive at absurd results. For example: for ... Princeton University, Princeton, New Jersey. The questio...
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JOURNAL OF CHEMICAL EDUCATION

PROPORTIONAL DEPENDENCE LAWRENCE SLIFKIN Princeton University, Princeton, New Jersey

THIS article is intended to call to notice the too The question now arises: reconsidering the problem, frequent applications of nonrigorous logic in developing and recognizing the restrictions of constant third proportional dependences. It is often stated, especi- variable, how can we show that x is proportional to ally in elementary courses, that if a property x is pro- yz? This is easily seen from the following, due to L. G . portional to y and if x is proportional to z, then x is Peck, our premises are proportional to the product yz. If we accept this theorem a t its face value and apply it, we can arrive a t absurd results. For example: for an ideal gas and the volume is proportional to the temperature, and the pressure is proportional to the temperature. There- where k~ is really a function of z and k2 is a function of fore, the volume is proportional to the pressure. y Then we may say Our fallacy is that we have overlooked the fact that the first relation holds only when pressure is constant while the second is valid only when volume is held constant. We have used equations which are not and simultaneously applicable. In fact, by simple algebraic manipulation, we see that if x is proportional to y and z, then unless we specify that each relation holds only when the third variable js constant, we are led to Now, if x = f ( y , z), then from (3) f is linear in y and k4y the conclusion that'x is proportional to the square mot from (4)f is linear'in z. Thus f(y, z) = kayz fksy kc, where ka. . .k6 are really constants. Equaof yz.' tions (1) and (2) show that x = 0 when either y or z = If z = ay andz = bz, then z' = abyz and z = d d @ . 0: Therefore, kn = k6 = k6 = 0, and x = kayz.

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