Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN: 9781259696527
Author: J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher: McGraw-Hill Education
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Chapter 11, Problem 11.1P
Interpretation Introduction

Interpretation:

The partial molar volume V¯1 and V¯2 in a mixture containing 40 mole % of species 1 at the given conditions should be calculated.

Concept Introduction:

ΔV will be calculated after putting x1 and x2 value in the given expression. Since x1 is given, x2 will be calculated from formula : x1+x2=1 for binary mixture.

  VE=VxiVi and VE=ΔV

Since ΔV, V1, V2, x1 and x2 are known, V can be calculated.

  V1¯=V+x2dVdx1 and V2¯=V-x1dVdx1

Absolute value of V is known and it can be expressed as function of variable x1 using the above two equations. Therefore, absolute value of dVdx1 can be calculated by differentiation with respect to x1 . So, we can get value of V¯1 and V¯2 .

Expert Solution & Answer
Check Mark

Answer to Problem 11.1P

Partial molar volume V¯1 =124.76 cm3/mol

Partial molar volume V¯2 =93.36 cm3/mol

Explanation of Solution

Given Information:

At 250C and atmospheric pressure the volume change of mixing of binary liquid mixtures of species 1 and 2 is given by the following equation:

  ΔV=x1x2(45x1+25x2)

Where ΔV is in cm3.mol-1. At this condition V1 = 110 and V2 = 90 cm3.mol-1.

Mole fraction of species 1 ( x1 ) is 0.4.

Calculation:

  ΔV calculation:

  x1=0.4 (given)

  x2=10.4=0.6

  ΔV=0.4×0.6(45×0.4+25×0.6)=7.92 cm3/mol

  V (total molar volume of solution) calculation:

  V=ΔV+x1V1+x2V2=7.92 cm3/mol + 0.4×110 cm3/mol + 0.6×90 cm3/mol=105.92 cm3/mol

Calculating V1¯

V1¯=V+x2dVdx1=105.92 cm3/mol + x2×d(x1V1+x2V2+x1x2×(45x1+25x2))dx1cm3/mol=105.92 cm3/mol + 0.6×d(110x1+90(1x1)+x(1x1)×(45x1+25(1x1)))/dx1 cm3/mol=105.92 cm3/mol + 0.6×(4510x160x12)cm3/mol=105.92 cm3/mol + 0.6×(4510×0.460×0.42)cm3/mol=124.76 cm3/mol

Calculating V¯2

  V2¯=V-x1dVdx1

  =105.92 cm3/mol - x1×d(x1V1+x2V2+x1x2×(45x1+25x2))dx1cm3/mol=105.92 cm3/mol - 0.4×d(110x1+90(1x1)+x(1x1)×(45x1+25(1x1)))/dx1 cm3/mol=105.92 cm3/mol - 0.4×(4510x160x12)cm3/mol=105.92 cm3/mol - 0.4×(4510×0.460×0.42)cm3/mol=93.36 cm3/mol

Conclusion

Thus, partial molar volume V¯1 =124.76 cm3/mol

Partial molar volume V¯2 =93.36 cm3/mol

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