Answer:
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
