1384 THE COLLIERY GUARDIAN, June 27, 1913. i- flight between the planes, then the actually transmitted momentum will be i n V G x l and this, on account 6 K of the elastic rebound of the molecules, will produce a transmitted pressure of 2 x m V Gr X y. But, as has been said, experiment shows that in oxygen gas for two planes of unit area = 1 cm.2, when V = 1 cm. per second, and the distance apart of the planes = 1 cm., = 0’000187 dynes, whence then 0000187 dynes = 2 x j nm V G X where V = 1, nm = p = 000143, G = 46,100 cm. and k is the average number of collisions during an average flight of a molecule through 1 cm. of space, i) was measured by means of a torsion disc, suspended with its surface near to a swiftly-moving disc. It varies, of course, with different gases. But if k be the number of collisions during a flight of 1 cm., and 0 be the number of collisions per second, and the mean average speed be Q, and V the velocity of the moving plane be 1 cm. per second, then q __q ___n m Q G_____ pQ G- 3 y 3^ ’ p being, as previously said, the density of the gas at 0 deg. Cent, and 760 mm. of mercury. mi, * 0*00143 x 42,500 x 46,100 Therefore 0= ----------3 x 0.000187---- = 5,000,000,000. This huge number, usually written 5 X 109, represents the number of times an average molecule of oxygen collides in a second. Its average, or mean free path, is obviously found by dividing the mean velocity by the number of collisions, and thus X = ?_ = 0’0000085 cm. Certain corrections C have, however, to be applied to this formula due to the fact that when we are calculating the mean free path it is not O, the absolute velocity of the molecules, that must be taken into consideration, but the relative velocity with regard to one another. And as they are all in motion it can be shown that the true value of the free path is three-fourths of what it would be at rest. On the other hand, there are corrections to be put in on account of the law of the distribution of velocities. Whence it results that the proper value of X = 1-17 x ® = 117 C 5 x 109 = 000000995 cm. Having thus determined the mean length of the free path of a molecule, it is next to be seen whether this value can be used to calculate its diameter. If a molecule A of diameter a moves towards a molecule B of the same diameter, it is obvious that it will strike it, if the centre of A goes within a distance =