What is the pressure in a 10L sealed container containing 8L of alcohol?

Antoine’s equation for alcohol: if there is gas present before adding alcohol, then use Antoine’s equation and Dalton’s law of partial pressures

The original poster didn’t explain this issue clearly; at least they should have specified the initial state. Take a look at this post: How to simulate pressure changes due to temperature increases in a sealed container? ? http://bbs.hcbbs.com/thread-1299794-1-1.html (Source: Haichuan Chemical Forum)

You are calculating the pressure after 10 L of air at atmospheric pressure is compressed to a gas volume of 2 L.

The data for this can be found at http://www.doc88.com/p-546619948850.html: the pressure is 26.66 kPa at 48.4℃. As mentioned above, what is the initial state of this tank? If it is evacuated to a pressure of -0.1 Mpa and then alcohol is added and the mixture is heated to 50℃, the pressure will be -74 kPa (gauge pressure). If the tank initially contains air at atmospheric pressure and alcohol is added followed by heating to 50℃, the pressure will range between 0 and 27 kPa (gauge pressure)

To add, I’m referring to the situation under normal atmospheric pressure: 8 liters of alcohol are poured into an open 10L container, which is then sealed, and the container is heated to 50°C. How should the pressure inside this sealed container be calculated at this point? Thank you all for your guidance. Please help analyze it again based on the additional conditions I’ve provided. Thank you~~~

First of all, thank you for your reply. To clarify, what I’m referring to is a situation where, under normal atmospheric pressure, 8 liters of alcohol are poured into an open 10-liter container, which is then sealed. The container is subsequently heated to 50°C. How should the pressure inside this sealed container be calculated at that point?

First of all, thank you for your reply. To clarify, what I’m referring to is a situation where, under normal atmospheric pressure, 8 liters of alcohol are poured into an open 10-liter container, which is then sealed. The container is subsequently heated to 50°C. How should the pressure inside this sealed container be calculated at that point?

This post was last edited by Qingchengjun on 2017-5-25 at 10:02. Initially, the container is open and in contact with the atmosphere; the pressure inside it equals the atmospheric pressure (gauge pressure of 0, absolute pressure of about 100 kPa). If the container is sealed immediately after alcohol is added, it will essentially be filled with air. If it takes a long time before sealing, the vapor resulting from the evaporation of alcohol will occupy some of the space, displacing air. In this case, the gauge pressure remains 0. For the first calculation: after sealing and heating, the pressure of the air increases by approximately 323/298*100 = 108 kPa. The saturated pressure due to alcohol evaporation is around 27 kPa; thus, the total pressure is 108 + 27 = 135 kPa, with a gauge pressure of 35 kPa. For the second calculation: some of the air is displaced before sealing, resulting in a partial pressure of approximately 92 kPa. 323/298*92 = 100; therefore, the total pressure is 100 + 27 = 127 kPa, with a gauge pressure of 27 kPa