42 Prof. J. H. Poynting on 



Taking a and V as the same for ice and water as an approxi- 

 mation, then equation (9) gives us 



^ = ot + P-=ot+ — , 



(10) 



,17— -l/ 



Subtracting, we have 



P -p'= OT - OT '-p^p (ii) 



But by equation (3) we have 



o>— or=Jr y » 



then 



p-p'=0 (12) 



Or, under the pressure P at the melting-point, the vapour- 

 tension of ice equals that of water, and there is an equal inter- 

 change of molecules taking place. According to this, then, 

 we may thus regard the alteration of melting-point by pres- 

 sure. The pressure increases the number of molecules given 

 off from the surfaces in contact with each other in both states; 

 but the increase is greater in the case of the less dense state. 

 Now, in the case of ice-water, ice is the less dense state, and 

 below 0° it has the less vapour-tension. Hence a sufficiently 

 great increase of pressure, while increasing both vapour-ten- 

 sions,, can make that of ice overtake that of water, or can lower 

 the melting-point. For paraffin, the liquid is the less dense 

 state. Then, increase of pressure can only render the two 

 vapour-tensions equal above the normal melting-point when 

 the liquid vapour-tension is less than that of the solid. 



Suppose now only one of the two states (the ice) to be 

 subjected to increase of pressure. For instance, let the ice be 

 compressed on a porous plate through which the water can 

 circulate freely. Then the pressure increases the rate at which 

 molecules escape from the ice into the water, but does not 

 affect the rate of escape of the water-molecules into the ice, 

 and a much less pressure will suffice to produce equilibrium 

 of exchange for a given temperature below 0° than when both 

 ice and water are subjected to the pressure. 



To calculate the fall in melting-point produced by a pres- 

 sure P' on the ice alone, we have, instead of (10), 



P =™ + -v- J 



