Question

Using Raoult’s law, estimate the boiling temperature and mole fractions in the vapor phase that is in equilibrium with a liquid having 0.2608 mol fraction of ethanol, in a mixture ethanol/water at P = 1 atm.

If possible, solve non-numerically.

Answer 1

`T=92.16 \°C`

`y_(et)=0.433\\\\y_w=0.567`

Step-by-step explanation:

Hello,

In this case, the Raoult's law for this problem is:

`y_(et)P=x_(et)P_(et)^(sat)\\\\y_(w)P=x_(w)P_(w)^(sat)`

Which can be written as:

`P=x_(et)P_(et)^(sat)+x_(w)P_(w)^(sat)`

Thus, by using the Antoine equation, we can symbolically represent the the temperature at which such mixture boil:

`1atm=0.2608*10^{(8.13484-( 1662.48)/(238.131+T))}/760+0.7392*10^{(5.40221-(1838.675)/(-31.737+T-273.15))}/1.01325`

The solution, by numerical iteration process (there is not way to solve it analytically) is 92.16 °C considering the data extracted from NIST database. Next, vapor fractions are:

`y_(et)=x_(et)*10^{(8.13484-( 1662.48)/(238.131+T))}/760/P\\\\y_(et)=0.2608*10^{(8.13484-( 1662.48)/(238.131+92.16))}/760/1atm\\\\y_(et)=0.433\\\\y_w=1-y_(et)=1-0.433\\\\y_w=0.567`

Regards.