The problem of vapor pressure of liquid mixtures at atmospheric pressure
Thread Content
A small problem has been bothering me for a long time; today I finally mustered the courage to post a request for help on the forum. If there is a mixture of benzene and toluene, at atmospheric pressure and a temperature of 40°C, the phase state is liquid. However, in a storage tank at atmospheric pressure, there will definitely be a gas phase of benzene and toluene above the liquid; that is, there will be a vapor pressure. This can be calculated using Raoult’s law. At this temperature and pressure, equilibrium between the gas and liquid phases should exist. I’m quite confused about this – perhaps my foundation in these concepts is too weak, as I simply don’t know how to explain it. Does the pressure mentioned earlier refer to external pressure, that is, the pressure exerted from the outside environment other than benzene and toluene vapors?
I don’t understand what you mean. The sum of the partial pressures of benzene and toluene equals the total pressure; the specific values of these partial pressures depend on the composition of the liquid phase.
In a closed tank, if there are no other components, the gas and liquid phases remain in equilibrium once stability is achieved, which is what you described. In an open system, such as the atmospheric environment, the partial pressures do not reach equilibrium; therefore, the liquid phase that is present is not in a state of equilibrium. Once equilibrium is reached, everything has evaporated and turned into a gas phase
For example, if a atmospheric distillation column is used to separate these two substances, how is the pressure at the top of the column generated? Is this pressure the partial pressure of benzene and toluene combined? Is it still connected to the atmosphere, generated by atmospheric pressure?