Question

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?

Answer 1

0.00833

Step-by-step 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