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

Answer:

$T=92.16 \°C$

$y_{et}=0.433\\\\y_w=0.567$

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:

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

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:

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

Regards.