Page:The American Cyclopædia (1879) Volume XI.djvu/730

712 MOLECULE hydrogen, expresses also the weight of a mole- cule of that substance with reference to the hydrogen molecule. It must be remembered moreover that as the molecule of hydrogen is a definite mass, its weight must be a definite quantity, however small, and may therefore be used as a standard of weight like a grain or a gramme. When therefore we determine the specific gravity of a gas with reference to hy- drogen, we thereby determine the weight of a molecule of that aeriform substance in terms of this molecular unit. For reasons based on chemical relations we have actually adopted as the unit of molecular weight one half of a hydrogen molecule, which we call a hydrogen atom ; and hence in the system in use the mo- lecular weight of a substance is equal to twice its specific gravity in the state of gas referred to hydrogen. As this unit of molecular weight, although a magnitude of a very different order, is as definite a mass of matter as a grain or a gramme, we shall aid our conceptions by giv- ing to it a definite name ; and since the mass of a cubic decimetre of hydrogen has been called a crith, the word microcrith will suggest both the nature of the molecular unit and the order of its magnitude. Kemembering then that the microcrith is the weight of the hydro- gen atom, and that this is one half of the hy- drogen molecule, we shall be understood when for the future we estimate molecular weights in microcriths. In the second place, it is evi- dent that the known pressure which a gas exerts against the sides of the containing ves- sel gives data from which we can calculate the mean velocity of the molecular motion under determinate conditions. Consider for exam- ple a cubic metre of hydrogen gas at the tem- perature of melting ice and under a pressure of one atmosphere. This aeriform mass weighs 0*08954 of a kilogramme, and exerts-a pressure of 10,332*96 kilogrammes against each face of the cubic enclosure. This pressure balances the molecular bombardment, and the momen- tum of the bombardment thus resisted during one second must be equal to that which the pressure would produce during the same period if acting on a mass of matter free to move. If the force of gravity at the place of observa- tion imparts to one kilogramme of matter a velocity of 9*8088 metres a second, then this momentum must be equal to 9*8088 x 10,332*96 =101,354. To find the momentum of the mo- lecular bombardment against the cube face, we must conceive of the mass of hydrogen divi- ded by planes parallel to the face in question into very thin sections, which are not thicker than the length of the mean path of a mole- cule, and between which the motion may be re- garded as uniform. Let V represent the mean velocity of the molecules, moving of course in all directions, and u one of the components of this velocity resolved perpendicularly to the face of the cube we are considering, and estimated of course as so many metres a sec- ond. Limiting our attention to this compo- nent, it is evident that we may regard the whole number of the hydrogen molecules with- in the enclosure as moving at any instant with the mean velocity u on lines normal to the face of the cube we are considering, and directed either toward this face or its opposite; and, since equilibrium is maintained, it is evident that the two opposite molecular volleys must be equal to each other. It it also further evident that the pressure against the face of the cube must be the sum of these two mo- lecular streams, that from the face as well as that toward it. If the molecules moved only one metre each second, it is manifest that each molecule would on the average move through the length of the cube in a second, were it not that the direction of its motion is continually being altered by collision with other molecules. But although the path of any one molecule may be very short, yet, as the molecules perfectly transmit their motion at each collision, and as the motion is always carried forward by a series of perfectly similar masses, the result is the same as if the same one had moved through the whole distance. There is therefore constantly passing between the small sections we have assumed, and also beating against the opposite faces of the cube, the same number of mole- cules as if the two streams were continuous. If the velocity were only a metre a second, there must pass every section in one or the other direction, and beat against one or the other of the two opposite faces of the cube during each second, a number of molecules, which we will represent by w, equal to the whole number of molecules in the cube ; and since the velocity we are considering is -M, the number of mole- cules thus passing on striking must be nu. If now in represents the weight of each mole- cule, then the total momentum resisted by the cube face each second must be mrm*, which is equal, as we have before seen, to 101,354. But in the expression 77mw 2 =101,354 the value mn is simply the weight of the cubic metre of hydrogen at the temperature of melting ice and at a pressure of one atmosphere, or 0*08954 of a kilogramme; so that w 2 =101,354 -4-0*08954=1,131,940. It must be remem- bered, however, that u is only one of the com- ponents of the molecular velocity, that per- pendicular to one pair of the cube faces ; and in order to determine the actual velocity, V, we must take into consideration the other two components, which are normal to the other two pairs of the cube faces; for V 2 =w 2 + 2 +w 2 . Now, although the values of these components for individual molecules may vary between the widest limits, yet their average values must be equal ; for otherwise the pres- sure of the gas could not be, as it is, equal in the directions of these components. If then our letters represent these average values, V a =3w 2 =3,395,820, and V=l,843 metres a sec- ond. The absolute velocity of the hydrogen molecules being now known at the tempera- ture of melting ice, we can readily calculate