DYNAMICAL PRINCIPLES TO PHYSICAL PHENOMENA. 
499 
If the alteration in the volume of tire saturated solution amounts to a cubic 
millimetre per gramme of salt dissolved, d(y + v')/dq is 10 -3 , so that 
n x 10~ 5 
Td v + v) 
so that with this amount of alteration in volume the terms proportional to the pressure 
will be the most important for pressures up to 100 atmospheres, while for pressures 
greater than this the term proportional to the square of the pressures will be the 
most important. In the case of a solution of sal-ammoniac the increase in volume is 
enormously greater than that stated, so that for such a salt the effect of pressure on 
the solubility ought to be very nearly proportional to the pressure. This agrees with 
Sobby’s experiments on the solubility of certain salts. # 
In this case the effect of pressure is expressed by the equation 
d0 cl (v + v') 
x -e=~i > —dT- 
We can test the formula by comparing it with the result of Sorby’s experiments 
on the effect of pressure upon the solubility. The two salts we shall take are potassium 
sulphate and sodium chloride, as we can get approximate values for these salts for all 
the quantities involved in equation (44). 
According to J. Thomsen, the heat absorbed when 174 grammes of K 3 S0 4 dissolve 
in water is 6380 gramme-degree units, so that the heat absorbed for one grain of 
K 3 S0 4 will be about 36 gramme-degrees or 36 X 4‘2 X 10 7 C.G.S. units. We may 
take this as an approximation to the value of A, though it must be remembered that 
A is the heat absorbed when one gramme of salt is dissolved in a nearly saturated 
solution. 
According to Sorby, when K 2 S0 4 crystallises out of a saturated solution, the 
volume increases from 100 to 134, so that, when a cubic centimetre of salt dissolves, 
the volume diminishes by about '25 c.c. The specific gravity of K 3 S0 4 is about 2'5, 
so that, when one gramme of salt dissolves, the volume diminishes by 'I c.c., and 
d(y v) dq = — 10 _1 , so that for 100 atmospheres the change in temperature 
required to produce the same change in solubility is at the temperature 15° C. given 
by the equation 
o 288 x 10 8 x 10- 1 
“ 36 x 4-2 x 10 7 
= 2 °. 
Sorby found that the K 3 S0 4 dissolved increased by about 3 per cent. According 
to Kopp, 100 parts of water dissolve 8'36 parts of K 3 S0 4 at 0°, and T741 part 
* Sorby, ‘Roy. Soc. Proc,,’ vol. 12, 1863, p. 538. 
3 s 2 
n/100, 
