Find Mass Fractions of Benzene in Liquid and Vapour

[annastewert]

Homework Statement

A vessel contains a mixture of benzene and o-xylene. At 473K the pressure in the vessel is 1000 kPa (abs). Using the data provided below, and any others necessary, what are the mass fractions of benzene in the liquid and vapour? Assume the liquid phase is an idea liquid mixture and the gas phase behaves like and ideal gas mixture.

Homework Equations

Raoult's equation P=xP*(T)
Antoine's equation P*=exp[A-B/(T+C)]

The Attempt at a Solution

I used Antoine's equation to find P*(473). And I know P+P=1000 kPa but I don't know where to go from here. ANy help would be appreciated.

 

[Charles Link]

I think you need to provide the data that was furnished with the problem. Did they give you the composition of the liquid? Did they supply vapor pressure tables? In any case, 1000 kPa is very nearly P=10 atm.

 

[Chestermiller]

If $P_B$ and $P_0$ are the equilibrium vapor pressures of benzene and o-xylene, and x is the mole fraction of benzene in the liquid, in terms of x, what are the partial pressures of benzene and o-xylene in the vapor?

 

[Charles Link]

It does appear that at 200 degrees centigrade, one of the vapor pressures will be greater than 10 atm, and the other less than 10 atm, and a little algebra with Raoult's law that @Chestermiller is hinting at, should provide the complete solution. As I mentioned in post #2, it would help to have the data sheet to see this though. $\\$ editing... A google of the vapor pressure of Benzene gives a program that uses Antoine's equation and computes it for you. It gives P=14.5 atm at T=200 C. It did not have o-xylene but had m and p xylene and those had vapor pressures of about 4.2 atm. (Another website had a graph and shows o-xylene has a vapor pressure of about 3.0 atm at =200 C.) The problem then is to find the liquid mixture that will result in a total pressure P=10 atm, etc.