Cooling water vapor in an airtight container? What would happen to gas water (100 degrees plus) when put in a strong airtight container? What would happen with the depressurization? Would it cool and condense into water anyway and leave a vacuum behind? Would it stay gaseous despite cooling? or would the vacuum being created by the depressurization cause the hydrogen and oxygen atoms to split, allowing some of the water to condense?
 A: Try this experiment: Get a metal can that has a screw-on cap.  Clean the can out thoroughly, and pour in about a half cup of water.  Put the can on your stove and heat the water until it is boiling.  Remove the can from the stove, quickly screw the lid on tightly, and set the can where it will cool.
The can will collapse on itself, dramatically.  
I think this should answer part of your question.  The water vapor condenses to form liquid water, leaving a near-vacuum behind.
A: If you keep the airtight strong container the same temperature (100degC plus), it will be stay gaseous. When you cool it as shown in the youtube, the pressure drops with temperature linearly until it reaches water vapor curve. After that point, condensation occurs, and it will be a mixture of liquid and vapor. Further cooling it, you will see more liquid, less vapor and lower pressure (vacuum) until you start to get ice. However, there's always some vapor in the container. It is the water molecular escaping from the surface of liquid and the ice. But hydrogen and oxygen would never split. It split when temperature is very high that the bond between the two can not hold the atoms. 
