Question

Find the compositon and temperature of a vapor in equilibrium with a liquid that is 0.400 mol acetic acid/mol and 0.600 mole water/mol at 1 atm.

y_a=

y_w=

T_bp=

Find the composition and temperature of a liquid in equilibrium with a gas that is at 1 atm and is composed of 0.500 mol acetic acid/mol, 0.300 mol water/mol, and the rest nitrogen.

x_a=

x_w=

T_dp=

A gas mixture of 0.230 mol acetic acid/mol, 0.200 mol water/mol, and 0.570 mol inerts/mol at 93.0°C is compressed isothermally until condensation occurs. At what pressure does the condensation first form, and what is the composition of this condensate?

P_dp=

x_a=

x_w=

Answer 1

The liquid mixture of water / acetic acid is in equilibrium with its vapor. The composition
of water and acid in the liquid (Xw = 0.6 and Xa = 0.4, respectively) is different from the composition of
water and acid in the steam (Yw and Ya, respectively). Under these conditions, the fractions
molars Yw and Ya can be related to the partial vapor pressure of each component,
by Dalton's law:
Yw= Pw/P [1] , Ya= Pa/P [2]

Where Pw and Pa are the partial pressures of water and acid, respectively. And P is the total pressure
of the system (1 atm = 760 mmHg). On the other hand, partial pressures can be related to the
composition of the liquid, according to Raoult's law:

Pw= Xw.Pw* [3] , Pa = Xa. Pa* [4]

Where Pw * and Pa * are the vapor pressures of pure liquids. 
These vapor pressures,depends on the temperature, which according to Antoine's law you have to for water and acid
acetic is given by

Pw*= e~[18.3036 - ( 3816.44/(-46.13 + T)) ] [5]
Pa* = e~[16.8080 - ( 3405.57/(-56.34 + T)) ] [6]

Combining the equations [1], [3], [5] and the equations [2], [4], [6] we obtain that  
Yw= 0.6* e~[18.3036 - ( 3816.44/(-46.13 + T)) ] [7]
Ya = 0.4 e~[16.8080 - ( 3405.57/(-56.34 + T)) [8]

We also know that, because they are molar fractions, 
Yw + Ya = 1;---> so that, when solving
the previous system of equations (using excel), we obtain that:
T= 200 K ; Yw = 0.685 ; Ya= 0.305 

For this case, we apply again equations [1] to [6], however, in this case, they give us as data Yw = 0.30 and Ya = 0.50 (the rest are 0.20 of Nitrogen), so that we calculate the composition in the liquid (Xw and Xa), so that by combining equations [1], [3]. [5] and [2], [4], [6] and clearing Xw and Xa, we have that:

Xw*= 0.30e~-[18.3036 - ( 3816.44/(-46.13 + T)) ] [8]
Xa* = 0.50e~-[16.8080 - ( 3405.57/(-56.34 + T)) ] [9] 

So that when solving the system of equations 
(noticing that ----> Xa + Xb = 1 -Xnitrogen = 0.8
 then we have
T= 385 K ; Xw = 0.198 ; Xa= 0.602