Re: How to calculate the internal pressure of a closed container as its heated?

A “closed container” means that the total volume – volume of the liquid + 
volume of the gas above it in constant.  There are several assumptions 
that one can make to simplify the situation.  Assume that no air is in the 
gas space – only the inert gas.  Air was not mentioned – only the inert 
gas.  Assume that very little of the inert gas dissolves in the water.  
The inert gasses have negligible solubilities in water.  Also, one would 
need to know which inert gas is used to take this into account.  Assume 
that the volume of the water remains constant.  This is not absolutely 
true, BUT as temperature rises, the volume of the water would increase 
slightly due to the variation of density with temperature, AND as 
temperature rises more water vaporizes so this would lead to a slight 
decrease in the volume of the liquid. These two effects might just offset 
each other.  So the pressure in the gas phase is the sum of two factors: 
1) the pressure of the inert gas given by P=nRT/V (need to know n, T and 
V) or P2 = P1xT2/T1 and 2) the vapor pressure of water as it changes with 
temperature.   This is an example of Dalton’s Law of Partial Pressures.  
See http://intro.chem.okstate.edu/1515SP01/Database/VPWater.html for the 
values of the vapor pressure of water as temperature changes.  I hope this 
helps.  If the above assumptions cannot be made, then the problem becomes 
much more complicated.