# Calculating gas pressure following water transformation to vapor (in contact with boiling oil)

When water falls into a pot of boiling oil, it turns into vapor, and in an enclosed space this will create a higher gas pressure. Is there a way to figure out how high of a gas pressure there will be given amount of water and oil, temperature of water and oil, and size of the enclosed space?

I have looked around but found nothing that I could understand well

• Do you want just the total gas pressure (air, including oil vapor and water vapor)? Jul 20, 2018 at 1:15
• The calculation is absolutely possible. You're looking for steam tables. I'd assume that oil is incompressible and contained no water, so pressure is just a function of temperature, mass of water, and volume of headspace.
– MaxW
Jul 20, 2018 at 1:43
• Depending on relative volumes and temperatures you can get a "steam explosion" which will also spray oil droplets out of your container. Jul 20, 2018 at 2:37
• @blacksmith37 This is more theoretical... wouldn't experiment with that in the kitchen ;) Jul 20, 2018 at 2:38

If you assume that the system equilibrates with the air, oil, and water all at the same temperature T in the liquid and gas phases and also assume that the water and oil are immiscible, the "partial pressure" of the air in the gas phase will be given by the ideal gas law: $$P_{air}=\frac{nRT}{V}$$ where n is the number of moles of air in the room V is the volume of the room minus the liquid (head space). The partial pressure of water vapor in the gas phase will be equal to the equilibrium vapor pressure of water at the temperature T: $$P_{water}=p_{equ\ water}(T)$$And the partial pressure of oil in the gas phase will be equal to the equilibrium vapor pressure of oil at the temperature T: $$P_{oil}=p_{equ\ oil}(T)$$The total pressure in the enclosure will be equal to the sum of the partial pressures: $$P=P_{air}+P_{water}+P_{oil}$$