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 V J .

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Answer 1

Question

Chapter 5, Problem 5B.3P

(a)

Interpretation Introduction

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)

Expert Solution

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,

$VJ,2=Vm,2+aox12+a1(3x1−x2)x12$

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,

$VJ=Videal+VtotalE (2)$

Where,

$VJ$ is the partial molar volume.

$Videal$ is the molar volume of the ideal mixture.

$VtotalE$ is the total excess volume.

The expression for $VtotalE$ is given as,

$VtotalE=(x1+x2)VE (3)$

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).

$VtotalE=n1n2(n1+n2)(ao+a1(n1-n2)n1+n2)$

The value of $Videal$ is,

$Videal=(x1Vm,1+x2Vm,2)$

Substitute the value of $Videal$ and $VtotalE$ 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,

$VJ,2=(∂Vtotal∂x2)p,T,x1=Vm,2+aox12+a1(3x1−x2)x12$

Hence, the partial volume of propionic acid is,

$VJ,1=Vm,1+aox22+a1(3x1-x2)x22$

Hence, the partial volume of oxane is,

$VJ,2=Vm,2+aox12+a1(3x1-x2)x12$

(b)

Interpretation Introduction

Interpretation:

The partial molar volume for each component in equimolar mixture has to be calculated.

(b)

Expert Solution

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,

$VJ,2=Vm,2+aox12+a1(3x1-x2)x12 (2)$

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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Answer 2

Question

Chapter 5, Problem 5B.3P

(a)

Interpretation Introduction

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)

Expert Solution

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,

$VJ,2=Vm,2+aox12+a1(3x1−x2)x12$

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,

$VJ=Videal+VtotalE (2)$

Where,

$VJ$ is the partial molar volume.

$Videal$ is the molar volume of the ideal mixture.

$VtotalE$ is the total excess volume.

The expression for $VtotalE$ is given as,

$VtotalE=(x1+x2)VE (3)$

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).

$VtotalE=n1n2(n1+n2)(ao+a1(n1-n2)n1+n2)$

The value of $Videal$ is,

$Videal=(x1Vm,1+x2Vm,2)$

Substitute the value of $Videal$ and $VtotalE$ 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,

$VJ,2=(∂Vtotal∂x2)p,T,x1=Vm,2+aox12+a1(3x1−x2)x12$

Hence, the partial volume of propionic acid is,

$VJ,1=Vm,1+aox22+a1(3x1-x2)x22$

Hence, the partial volume of oxane is,

$VJ,2=Vm,2+aox12+a1(3x1-x2)x12$

(b)

Interpretation Introduction

Interpretation:

The partial molar volume for each component in equimolar mixture has to be calculated.

(b)

Expert Solution

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,

$VJ,2=Vm,2+aox12+a1(3x1-x2)x12 (2)$

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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Answer 3

Question

Chapter 5, Problem 5B.3P

(a)

Interpretation Introduction

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)

Expert Solution

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,

$VJ,2=Vm,2+aox12+a1(3x1−x2)x12$

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,

$VJ=Videal+VtotalE (2)$

Where,

$VJ$ is the partial molar volume.

$Videal$ is the molar volume of the ideal mixture.

$VtotalE$ is the total excess volume.

The expression for $VtotalE$ is given as,

$VtotalE=(x1+x2)VE (3)$

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).

$VtotalE=n1n2(n1+n2)(ao+a1(n1-n2)n1+n2)$

The value of $Videal$ is,

$Videal=(x1Vm,1+x2Vm,2)$

Substitute the value of $Videal$ and $VtotalE$ 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,

$VJ,2=(∂Vtotal∂x2)p,T,x1=Vm,2+aox12+a1(3x1−x2)x12$

Hence, the partial volume of propionic acid is,

$VJ,1=Vm,1+aox22+a1(3x1-x2)x22$

Hence, the partial volume of oxane is,

$VJ,2=Vm,2+aox12+a1(3x1-x2)x12$

(b)

Interpretation Introduction

Interpretation:

The partial molar volume for each component in equimolar mixture has to be calculated.

(b)

Expert Solution

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,

$VJ,2=Vm,2+aox12+a1(3x1-x2)x12 (2)$

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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Answer 4

Question

Chapter 5, Problem 5B.3P

(a)

Interpretation Introduction

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)

Expert Solution

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,

$VJ,2=Vm,2+aox12+a1(3x1−x2)x12$

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,

$VJ=Videal+VtotalE (2)$

Where,

$VJ$ is the partial molar volume.

$Videal$ is the molar volume of the ideal mixture.

$VtotalE$ is the total excess volume.

The expression for $VtotalE$ is given as,

$VtotalE=(x1+x2)VE (3)$

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).

$VtotalE=n1n2(n1+n2)(ao+a1(n1-n2)n1+n2)$

The value of $Videal$ is,

$Videal=(x1Vm,1+x2Vm,2)$

Substitute the value of $Videal$ and $VtotalE$ 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,

$VJ,2=(∂Vtotal∂x2)p,T,x1=Vm,2+aox12+a1(3x1−x2)x12$

Hence, the partial volume of propionic acid is,

$VJ,1=Vm,1+aox22+a1(3x1-x2)x22$

Hence, the partial volume of oxane is,

$VJ,2=Vm,2+aox12+a1(3x1-x2)x12$

(b)

Interpretation Introduction

Interpretation:

The partial molar volume for each component in equimolar mixture has to be calculated.

(b)

Expert Solution

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,

$VJ,2=Vm,2+aox12+a1(3x1-x2)x12 (2)$

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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Answer 5

Chapter 5, Problem 5B.3P

(a)

Interpretation Introduction

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)

Expert Solution

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,

$VJ,2=Vm,2+aox12+a1(3x1−x2)x12$

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,

$VJ=Videal+VtotalE (2)$

Where,

$VJ$ is the partial molar volume.

$Videal$ is the molar volume of the ideal mixture.

$VtotalE$ is the total excess volume.

The expression for $VtotalE$ is given as,

$VtotalE=(x1+x2)VE (3)$

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).

$VtotalE=n1n2(n1+n2)(ao+a1(n1-n2)n1+n2)$

The value of $Videal$ is,

$Videal=(x1Vm,1+x2Vm,2)$

Substitute the value of $Videal$ and $VtotalE$ 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,

$VJ,2=(∂Vtotal∂x2)p,T,x1=Vm,2+aox12+a1(3x1−x2)x12$

Hence, the partial volume of propionic acid is,

$VJ,1=Vm,1+aox22+a1(3x1-x2)x22$

Hence, the partial volume of oxane is,

$VJ,2=Vm,2+aox12+a1(3x1-x2)x12$

(b)

Interpretation Introduction

Interpretation:

The partial molar volume for each component in equimolar mixture has to be calculated.

(b)

Expert Solution

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,

$VJ,2=Vm,2+aox12+a1(3x1-x2)x12 (2)$

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 $99.07 cm3 mol−1_$ .

Answer 6

Chapter 5, Problem 5B.3P

(a)

Interpretation Introduction

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)

Expert Solution

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,

$VJ,2=Vm,2+aox12+a1(3x1−x2)x12$

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,

$VJ=Videal+VtotalE (2)$

Where,

$VJ$ is the partial molar volume.

$Videal$ is the molar volume of the ideal mixture.

$VtotalE$ is the total excess volume.

The expression for $VtotalE$ is given as,

$VtotalE=(x1+x2)VE (3)$

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).

$VtotalE=n1n2(n1+n2)(ao+a1(n1-n2)n1+n2)$

The value of $Videal$ is,

$Videal=(x1Vm,1+x2Vm,2)$

Substitute the value of $Videal$ and $VtotalE$ 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,

$VJ,2=(∂Vtotal∂x2)p,T,x1=Vm,2+aox12+a1(3x1−x2)x12$

Hence, the partial volume of propionic acid is,

$VJ,1=Vm,1+aox22+a1(3x1-x2)x22$

Hence, the partial volume of oxane is,

$VJ,2=Vm,2+aox12+a1(3x1-x2)x12$

Answer 7

Chapter 5, Problem 5B.3P

(a)

Interpretation Introduction

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 $VJ$ .

Answer 8

(a)

Expert Solution

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,

$VJ,2=Vm,2+aox12+a1(3x1−x2)x12$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

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,

$VJ=Videal+VtotalE (2)$

Where,

$VJ$ is the partial molar volume.

$Videal$ is the molar volume of the ideal mixture.

$VtotalE$ is the total excess volume.

The expression for $VtotalE$ is given as,

$VtotalE=(x1+x2)VE (3)$

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).

$VtotalE=n1n2(n1+n2)(ao+a1(n1-n2)n1+n2)$

The value of $Videal$ is,

$Videal=(x1Vm,1+x2Vm,2)$

Substitute the value of $Videal$ and $VtotalE$ 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,

$VJ,2=(∂Vtotal∂x2)p,T,x1=Vm,2+aox12+a1(3x1−x2)x12$

Hence, the partial volume of propionic acid is,

$VJ,1=Vm,1+aox22+a1(3x1-x2)x22$

Hence, the partial volume of oxane is,

$VJ,2=Vm,2+aox12+a1(3x1-x2)x12$

Answer 13

(b)

Interpretation Introduction

Interpretation:

The partial molar volume for each component in equimolar mixture has to be calculated.

(b)

Expert Solution

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,

$VJ,2=Vm,2+aox12+a1(3x1-x2)x12 (2)$

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 $99.07 cm3 mol−1_$ .

Answer 14

(b)

Interpretation Introduction

Interpretation:

The partial molar volume for each component in equimolar mixture has to be calculated.

Answer 15

(b)

Expert Solution

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 $99.07 cm3 mol−1_$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

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,

$VJ,2=Vm,2+aox12+a1(3x1-x2)x12 (2)$

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 $99.07 cm3 mol−1_$ .

Answer 20

Ch. 5 - Prob. 5A.1ST Ch. 5 - Prob. 5A.2ST Ch. 5 - Prob. 5A.3ST Ch. 5 - Prob. 5A.4ST Ch. 5 - Prob. 5B.2ST Ch. 5 - Prob. 5C.2ST Ch. 5 - Prob. 5F.1ST Ch. 5 - Prob. 5A.1DQ Ch. 5 - Prob. 5A.2DQ Ch. 5 - Prob. 5A.3DQ

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 V J .

Answer to Problem 5B.3P

Explanation of Solution

Interpretation: The partial molar volume for each component in equimolar mixture has to be calculated.

Answer to Problem 5B.3P

Explanation of Solution

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Chapter 5 Solutions

Interpretation:

The partial molar volume for each component in equimolar mixture has to be calculated.