“…Pu + (2AIa)/(3p) y/xyf + vli'T 111 a (13) When x is effectively large, the second term in the left hand side of (13) vanishes and the saturation activity a assumes its normal value a0 given by pu = yu' + vRT In ao (14) Finally, from (13) and ( 14) we obtain Log a/au = (2/'3Alccy)/(2.303v RT px) (15) If the geometry of the crystal is known a can easily be calculated since, as will be seen from ( 4), ( 5) and (11), it is the ratio of the total surface area to the volume of a crystal of unit x. For instance, for a cube, if x is the length of one edge, a = 6.…”