A vapor composed of 0.60 mole fraction ethanol and 0.40 mole fraction acetic acid at 120.0 mmHg (absolute), is in equilibrium with a liquid phase also composed of ethanol and acetic acid. Assume that the liquid is an ideal solution and apply Raoult/'s law to find the following. What is the temperature of the liquid?

0.00833

Explanation:

Log p1sat (mm Hg) = 8.20417-1642.89 / (t+230.3), t in deg, c

Log p2sat (mm Hg) = 7.5596-1644.05 / (t+233.524), t in deg, c

Let y1 = mole fraction of ethanol in the vapor phase = 0.6, y2 = mole fraction of acetic acid in the vapor phase = 0.4

Let x1=mole fraction of ethanol in the liquid phase, x2 = mole fraction of acetic acid in the liquid phase.

From Raoult's law, y1P = x1P1sat (1), P1sat = saturation pressure of ethanol, P2sat = saturation pressure of acetic acid, P = total pressure, y2P=x2p2sat (2)

Eq. 1 can be written as x1 = y1P/P1sat and x2 = y2P//P2sat

Addition of x1 and x2 gives x1+x2 = y1P/P1sat + y2P/P2sat

Hence y1/P1sat + y2/P2sat = 1/P

0.6/P1sat + 0.4/P2sat = 1/120 = 0.00833 (1)

this proves by trial and error procedure. we can assume some temperature, Calculate p1sat (for ethanol and P2sat for acetic acid) and calculate the LHS of Eq. 3 and check whether it relates with RHS of the equation. The calculations are done in excel.

The iterative procedure gives temperature as 53.16 deg. c

T (deg. c) 53.16

P1sat (mm Hg) 256.0505949

P2sat (mm Hg) 66.81725201 0.6/P1sat 0.002343287

0.4/P2sat 0.005986478

Calc 0.008329765

Answer 0.00833

Hence y1P=x1P1sat, x1=y1P/P1sat = 0.6*120/256 = 0.28, x2 = 1-0.28=0.72