
(a)
Interpretation:
The expression for the partial volume of each component at the given temperature has to be derived.
Concept Introduction:
The partial volume of the substance in a mixture is defined as the change in volume per mole of the substance added to the large volume of the mixture. It is denoted by VJ.
(a)

Answer to Problem 5B.3P
The partial volume of propionic acid is,
VJ,1=Vm,1+aox22+a1(3x1−x2)x22
The partial volume of oxane is,
VJ,2=Vm,2+aox21+a1(3x1−x2)x21
Explanation of Solution
The Propionic acid is assumed as component 1 and THP is assumed as component 2.
The excess volume is given as,
VE=x1x2{ao+a1(x1−x2)} (1)
Where,
x1 and x2 are the mole fractions of components 1 and component 2, respectively.
VE is the excess volume.
ao and a1 are the constants.
The volume of the ideal mixture is,
Vtotal=x1Vm,1+x2Vm,2
Where,
Vtotal is the total volume of the ideal mixture.
x1 and x2 are the mole fractions of components 1 and component 2, respectively.
Vm,1 and Vm,2 are the molar volumes of the component 1 and component 2, respectively.
Thus, the partial molar volume of the real mixture is given by the formula,
VJ=Videal+VEtotal (2)
Where,
VJ is the partial molar volume.
Videal is the molar volume of the ideal mixture.
VEtotal is the total excess volume.
The expression for VEtotal is given as,
VEtotal=(x1+x2)VE (3)
The mole fraction of component 1 is,
x1= n1n1+n2 (4)
Where,
- n1 is the number of moles of component 1.
- n2 is the number of moles of component 2.
The mole fraction of component 2 is,
x2= n2n1+n2 (5)
Substitute the equation (1), (4) and (5) in equation (3).
VEtotal=n1n2(n1+n2)(ao+a1(n1-n2)n1+n2)
The value of Videal is,
Videal=(x1Vm,1+x2Vm,2)
Substitute the value of Videal and VEtotal in equation (2).
VJ=n1Vm,1+n2Vm,2+n1n2(n1+n2)(ao+a1(n1-n2)n1+n2)
Thus, the partial volume of component 1 is,
VJ,1=(∂Vtotal∂n1)p,T,n2=Vm,1+aon22(n1+n2)2+a1(3n1−n2)n22(n1+n2)3=Vm,1+aox22+a1(3x1−x2)x22
Thus, the partial volume of component 2 is,
VJ,2=(∂Vtotal∂x2)p,T,x1=Vm,2+aox21+a1(3x1−x2)x21
Hence, the partial volume of propionic acid is,
VJ,1=Vm,1+aox22+a1(3x1-x2)x22
Hence, the partial volume of oxane is,
VJ,2=Vm,2+aox21+a1(3x1-x2)x21
(b)
Interpretation:
The partial molar volume for each component in equimolar mixture has to be calculated.
(b)

Answer to Problem 5B.3P
The molar volume of the component 1 is 75.73 cm3 mol−1_. The molar volume of the component 2 is 99.07 cm3 mol−1_.
Explanation of Solution
The Propionic acid is assumed to be component 1 and THP is assumed to be component 2.
The molar volume of component 1 is given by the formula,
Vm,1 = Molar massDensity
The value of molar mass of component 1 is 74.08 g mol−1.
The density of the component 1 is 0.97174 g cm−3.
Substitute the value of molar mass and density in the above expression.
Vm,1 = 74.08 g mol−10.97174 g cm−3=76.23 cm3 mol−1
The formula for the partial molar volume for component 1 is,
VJ,1=Vm,1+aox22+a1(3x1-x2)x22 (1)
Where,
VJ,1 is the molar volume of component 1.
x2 is the molar fraction of component 2.
x1 is the molar fraction of component 1.
Vm,1 is the molar volume of the component 1.
ao and a1 are the constants.
The value of ao is −2.467 cm3 mol−1.
The value of a1 is 0.0608 cm3 mol−1.
The value of Vm,1 is 76.23 cm3 mol−1.
Since, the mixture is equimolar, assume the value of x1=x2. Hence, the value of x1 is 0.5 and the value of x2 is 0.5.
Substitute the values of x1, x2, Vm,1, ao and a1 in the equation (1).
VJ,1=76.23 cm3 mol−1+(−2.467 cm3 mol−1)×(0.5)2+0.5(3×0.5−0.5)(0.5)2=76.23 cm3 mol−1−0.616+0.125=75.73 cm3 mol−1_
Hence, the molar volume of the component 1 is 75.73 cm3 mol−1_.
The molar volume of component 2 is given by the formula,
Vm,2 = Molar massDensity
The value of molar mass of component 1 is 86.13 g mol−1.
The density of the component 1 is 0.86398 g cm−3.
Substitute the value of molar mass and density in the above expression.
Vm,1 = 86.13 g mol−10.86398 g cm−3=99.68 cm3 mol−1
The formula molar volume for component 1 is,
VJ,2=Vm,2+aox21+a1(3x1-x2)x21 (2)
Where,
VJ,2 is the molar volume of component 2.
x2 is the molar fraction of component 2.
x1 is the molar fraction of component 1.
Vm,2 is the molar volume of component 2.
ao and a1 are the constants.
The value of ao is −2.467 cm3 mol−1.
The value of a1 is 0.0608 cm3 mol−1.
The value of Vm,2 is 99.68 cm3 mol−1.
Since, the mixture is equimolar, assume the value of x1=x2. Hence, the value of x1 is 0.5 and the value of x2 is 0.5.
Substitute the values of x1, x2, Vm,2, ao and a1 in the equation (2).
VJ,2=99.68 cm3 mol−1+(−2.467 cm3 mol−1)(0.5)2+(0.0608 cm3 mol−1)(3×0.5−0.5)(0.5)2=99.68 cm3 mol−1−0.61675+0.0152=99.07 cm3 mol−1_
Hence, the molar volume of the component 2 is 99.07 cm3 mol−1_.
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Chapter 5 Solutions
Atkins' Physical Chemistry
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