Page:Popular Science Monthly Volume 80.djvu/442

whether the molecule is large or small. Hence, the energy of agitation of our oil drop ought to be exactly the same as that of one of the molecules of the gas which surrounds it. But this is the quantity which we have just determined experimentally, and which, furthermore, can be computed with great precision from the kinetic theory.^[1]

Hence, we may consider that this quantity is known. The second factor, however, is not known with certainty, except under conditions which may or may not be fulfilled in any experimental work, and herein lies the uncertainty in all preceding attempts like those of Perrin to subject the kinetic theory of Brownian movements to any rigorous experimental test. Fortunately for the present work, however, this factor does not need to be known at all. For obviously the resistance which the medium offers to the motion at a given speed of this particular drop though it must be the same whether it is an electrical force, a gravitational force, or a force arising from molecular bombardments which is causing the motion. Consequently all that was necessary for us to do in order to eliminate this resistance factor entirely was first to observe the successive displacements of the balanced drop as indicated above and then to destroy the balance and measure how fast the drop moved on the average, both under gravity and under an electrical field of known strength, in precisely the way we had done when we were determining the successive values of the charge carried by the oil drops. From the results of the two experiments we could then eliminate the resistance factor and obtain the average displacement in terms of quantities every one of which was measurable with the greatest precision.^[2] Indeed the experimental error in measuring the aver-

1. The kinetic theory equation is $\scriptstyle E=3/2.RT/N$ in which E is the mean energy of molecular agitation, R an accurately known gas constant, T the absolute temperature, and N the number of molecules in 2 grams of hydrogen. Although N is not accurately known save through experiments of this sort, it fortunately does not need to be known, as will be shown in the next footnote, for the quantitative test here sought. When the above value of E is substituted in the equation of the last footnote it becomes

   $\scriptstyle D={\sqrt {\frac {4}{\pi }}}{\frac {RT}{NK}}t$.

2. When the drop is moving down through the medium under the force of gravity, mg, alone, its average velocity v is given by $\scriptstyle mg=Kv_{1}$. The substitution of this value of $\scriptstyle mg/v_{1}$ in the equation of the footnote on page —— gives $\scriptstyle e=K/F.(v_{1}+v_{2})$ and the elimination of K between this equation and that given in the preceding footnote gives

   $\scriptstyle D={\sqrt {\frac {4}{\pi }}}{\frac {RT(r_{1}+v_{2})t}{FE(Ne)}}$.

   Since D was of course different for different drops instead of making the comparison between the observed and calculated values of D it was thought preferable to make the comparison in every case between the value of Ne obtained