Does a water and kerosene mixture have the vapor pressure of kerosene alone?

[Srijit Sen]

When we mix water and kerosene, what is the vapor pressure of the mixture? Is it just that of kerosene, or an average of vapor pressure of both liquids? (Kerosene floats on top of water which is why I felt like asking this question)

 

[Prannoy Mehta]

Let us take a scenario.

Initially in a completely isolated close container. Water is kept (A significant amount) at 1 Atm, 273 K. Constant temperature, constant pressure, constant number of moles. And well the volume of the container is fixed.

Some amount of water evaporates and there is an equilibrium established between water vapor and the liquid water. Now, magically Kerosene is brought inside the closed container. Yes, it is to avoid the escaping of water vapor. (Though it should not affect the final result) If kerosene was added slowly. Now the water vapor and water liquid equilibrium is disturbed.

Because there is no pressure by the water applied on the water vapor. So equilibrium is disturbed and the water vapor condenses to form liquid water. Now, the liquid vapor will sink down after it is formed as it is denser than kerosene. Now even kerosene evaporates and the liquid and vapor phase of Kerosene establish an equilibrium between each other. So at the end, Kerosene vapors would exist. Now if suppose a group of highly energetic water even manages to escape the liquid water. It will condense back to liquid water and give its energy to the Kerosene molecules.

So at the end, we will end up with Kerosene Vapors in the container.

I think this would be the answer.

Best Wishes

 

[Srijit Sen]

So can it be concluded that by mixing kerosene and water at room temperature the change in entropy of the system is negative? (Vapor pressure of water is greater than vapor pressure of kerosene at 25 degree celsius)

 

[Chestermiller]

When we mix water and kerosene, what is the vapor pressure of the mixture? Is it just that of kerosene, or an average of vapor pressure of both liquids? (Kerosene floats on top of water which is why I felt like asking this question)

The equilibrium pressure will be the sum of the vapor pressures of water and kerosene: http://www.separationprocesses.com/Distillation/DT_Chp01m.htm

 

[Kevin McHugh]

I didn't know you could mix kerosene and water.

 

[Prannoy Mehta]

The equilibrium pressure will be the sum of the vapor pressures of water and kerosene: http://www.separationprocesses.com/Distillation/DT_Chp01m.htm

Chestermiller is right. Sorry for the error. I will dig into vapor pressure and then reply to it

 

[Srijit Sen]

The equilibrium pressure will be the sum of the vapor pressures of water and kerosene: http://www.separationprocesses.com/Distillation/DT_Chp01m.htm

Thanks for the answer Chestermiller. But I find it hard to believe that water can penetrate a liquid of lower density and then evaporate (establish equilibrium) to exert vapor pressure. How is this possible? Suppose I have 1 m of water on top of which i add 100 m of kerosene in a closed vessel of height 102 m. How can water molecules travel 100 m upward to exert pressure? I mean the surface is so so far away... (just keen to know what exactly happens)

 

[Chestermiller]

Thanks for the answer Chestermiller. But I find it hard to believe that water can penetrate a liquid of lower density and then evaporate (establish equilibrium) to exert vapor pressure. How is this possible? Suppose I have 1 m of water on top of which i add 100 m of kerosene in a closed vessel of height 102 m. How can water molecules travel 100 m upward to exert pressure? I mean the surface is so so far away... (just keen to know what exactly happens)

The water is always going to have a small solubility in the kerosine. At equilibrium, the chemical potential of water in the underlying water phase, in the kerosine, and in the gas phase must all be equal. This guarantees that the partial pressure of the water in the gas phase is equal to the equilibrium vapor pressure.

The water molecules travel through the kerosine by molecular diffusion.

It doesn't matter how far away the surface is. That just determines how long it takes for the system to reach equilibrium. The molecules of water in the gas phase are exchanging with the molecules dissolved in the kerosine, and the molecules of water in the kerosene are also exchanging with the liquid water below.

 

[Srijit Sen]

The water is always going to have a small solubility in the kerosine. At equilibrium, the chemical potential of water in the underlying water phase, in the kerosine, and in the gas phase must all be equal. This guarantees that the partial pressure of the water in the gas phase is equal to the equilibrium vapor pressure.

The water molecules travel through the kerosine by molecular diffusion.

It doesn't matter how far away the surface is. That just determines how long it takes for the system to reach equilibrium. The molecules of water in the gas phase are exchanging with the molecules dissolved in the kerosene, and the molecules of water in the kerosene are also exchanging with the liquid water below.

Thanks a lot, this makes sense somewhat. Now suppose i have an osmosis set up. In one side I have salt water (0.5 M NaCl in 1L water) and on the other side I have 1L of pure water. To the side having salt water I add benzene. According to what I have studied, vapor pressure of salt water is slightly lesser than that of pure water. But now having added benzene to salt water vapor pressure of salt water + benzene > water according to data provided by you earlier (Experiment conducted at 100 degree F). What will the initial movement of water be? From pure water to salt water or from salt water to pure water (the normal case).

 

[Kevin McHugh]

When you increase the ionic strength of the water, you will effectively reduce the solubility of benzene in the salt water. Ksp is already very small for hydrocarbons in water and vice versa.