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27 August, 04:10

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?

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  1. 27 August, 05:26
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    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
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