1886.] certain geological phenomena, ete. 2 
where o’ —o is the increment of density over the surface density, 
due solely to the absorbed gas. The increment of density in 
nature is however thought to be partly due to the presence of 
heavier elements. A fortiwri then is ygf less than the hyperbolic 
logarithm of this ratio, such as it actually exists. Prof. Green 
has calculated the densities from Laplace’s formula, using 2°5 as 
the surface density*. This is probably too small. But since we 
are concerned with the ratio only, it is near enough for our pur- 
pose. He makes the density at 250 miles to be 3:1. Consequently 
at that depth, regarded as measured from the free surface, as 
Laplace’s law contemplates, we should have yg (+) =0:1914, 
which is less than ith. At a depth of 150 miles, it would be 
3ths, or about jth, and at a depth of 50 miles, it would be 1th 
of this, or less than jth. 
For the sake of illustration, suppose that, when 
€+k=150 miles = 6k, yg (€+k) = 0:115. 
Then ygk will be 0:019, which implies that the gas dissolved at 
the bottom of the crust is 0°019 of the mass of the liquid rock 
holding it in solution. Its mass will therefore be, in a unit mass 
of rock, «x 0-019, say 2°68 x 0:019, or 0°051 of the standard 
substance, which is liquid water. At this rate, one cubic foot of 
magma as it exists just beneath the crust, would yield 87 cubic 
inches of liquid water. For the reason given, this value of 
yg (€+k) is probably too large and the estimate excessive ; 
nevertheless it suffices to show that the theory will account for 
the emission of a very considerable quantity of steam from the 
lava during an eruption. 
Suppose that the pressure upon that portion of the magma 
where it was a becomes p, being less than w. Some of the gas 
will then be liberated; and by that means let the element df 
become dz. The volume of the gas which was wholly dissolved 
in d€ was mdé. In dz the total volume of the gas, which will be 
partly free and partly dissolved, being inversely as the pressure 
will be 
TY 
m — dt — md&. 
p (6 G 
This will be the volume by which the element is expanded by 
change of pressure from @ to p: so that . 
dz=df+ me dé —mdé. 
But since the amount of matter in the column is the same 
identically as before, p can differ from @ only by the difference of 
* Geology for Students, p. 482. 1876. 
