(a) Interpretation: The diffusion constant for He is to be calculated. Concept introduction: The diffusion of any gas particle among its own gas particles is known as self-diffusion. The diffusion constant is directly proportional to the product of mean free path and average velocity. The self-diffusion of a particle is given as, D = 3 8 d 2 ρ ⋅ R T π M Where, • D is the self-diffusion constant. • M is the molar mass. • d is the diameter of given particles. • ρ is the particle density. • R is the gas constant. • T is the temperature.

Question

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Answer 1

Question

Chapter 19, Problem 19.62E

Interpretation Introduction

(a)

Interpretation:

The diffusion constant for $He$ is to be calculated.

Concept introduction:

The diffusion of any gas particle among its own gas particles is known as self-diffusion. The diffusion constant is directly proportional to the product of mean free path and average velocity. The self-diffusion of a particle is given as,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant.

• $T$ is the temperature.

Expert Solution

Answer to Problem 19.62E

The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Explanation of Solution

It is given that the helium is at $1.00 atm$ , that is, standard pressure and $273 K$ . The molar mass of helium is $4.00 g/mol$ and the diameter of helium molecules is $2.65 A∘$ .

The diffusion constant can be calculated by the formula,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant and its value is $8.314 J⋅mol−1⋅K−1$ or $8.314 kg⋅m2⋅s−2⋅mol−1⋅K−1$ .

• $T$ is the temperature.

The diffusion constant can be represented in the terms of pressure as,

$D=3kT8d2p⋅RTπM$

Substitute the values of molar mass, diameter, pressure, gas constant and temperature.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(2.65 A∘)2×1.00 atm⋅8.314 m2⋅kg⋅s−2⋅K−1⋅mol−1×273 Kπ×4.00×10−3 kg⋅mol−1$

The unit of pressure is converted from $atm$ to $Pa$ . The conversion factor is $1 atm=101325 Pa$ .

The conversion factor of $Pa$ is,

$1 Pa=1 N/m21 N=kg⋅m⋅s−2$

Substitute the units of pressure and solve the square root as shown below.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(2.65 ×10−10 m)2×101325 kg⋅m−1⋅s−2×424.993 m⋅s−1D=8.444×10−5 m2⋅s−1$

The units of diffusion constant are usually expressed in $cm$ . The conversion factor is $1 cm=10−2m$ . The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Conclusion

The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Interpretation Introduction

(b)

Interpretation:

The diffusion constant for $Xe$ is to be calculated.

Concept introduction:

The diffusion any gas particle among its own gas particles is known as self-diffusion. The diffusion constant is directly proportional to the product of mean free path and average velocity. The self-diffusion of a particle is given as,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant.

• $T$ is the temperature.

Expert Solution

Answer to Problem 19.62E

The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

Explanation of Solution

It is given that the xenon is at $1.00 atm$ , that is, standard pressure and $273 K$ . The molar mass of xenon is $131.29 g/mol$ and the diameter of xenon molecules is $4.00 A∘$ .

The diffusion constant can be calculated by the formula,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant and its value is $8.314 J⋅mol−1⋅K−1$ or $8.314 kg⋅m2⋅s−2⋅mol−1⋅K−1$ .

• $T$ is the temperature.

The diffusion constant can be represented in the terms of pressure as,

$D=3kT8d2p⋅RTπM$

Substitute the values of molar mass, diameter, pressure, gas constant and temperature.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(4.00 A∘)2×1.00 atm⋅8.314 m2⋅kg⋅s−2⋅K−1⋅mol−1×273 Kπ×131.29×10−3 kg⋅mol−1$

The units of pressure are converted from $atm$ to $Pa$ . The conversion factor is $1 atm=101325 Pa$ .

The conversion factor of $Pa$ is,

$1 Pa=1 N/m21 N=kg⋅m⋅s−2$

Substitute the units of pressure and solve the square root as shown below.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(4.00 ×10−10 m)2×101325 kg⋅m−1⋅s−2×74.181 m⋅s−1D=6.47×10−6 m2⋅s−1$

The units of diffusion constant are usually expressed in $cm$ . The conversion factor is $1 cm=10−2m$ . The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

Conclusion

The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

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Answer 2

Question

Chapter 19, Problem 19.62E

Interpretation Introduction

(a)

Interpretation:

The diffusion constant for $He$ is to be calculated.

Concept introduction:

The diffusion of any gas particle among its own gas particles is known as self-diffusion. The diffusion constant is directly proportional to the product of mean free path and average velocity. The self-diffusion of a particle is given as,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant.

• $T$ is the temperature.

Expert Solution

Answer to Problem 19.62E

The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Explanation of Solution

It is given that the helium is at $1.00 atm$ , that is, standard pressure and $273 K$ . The molar mass of helium is $4.00 g/mol$ and the diameter of helium molecules is $2.65 A∘$ .

The diffusion constant can be calculated by the formula,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant and its value is $8.314 J⋅mol−1⋅K−1$ or $8.314 kg⋅m2⋅s−2⋅mol−1⋅K−1$ .

• $T$ is the temperature.

The diffusion constant can be represented in the terms of pressure as,

$D=3kT8d2p⋅RTπM$

Substitute the values of molar mass, diameter, pressure, gas constant and temperature.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(2.65 A∘)2×1.00 atm⋅8.314 m2⋅kg⋅s−2⋅K−1⋅mol−1×273 Kπ×4.00×10−3 kg⋅mol−1$

The unit of pressure is converted from $atm$ to $Pa$ . The conversion factor is $1 atm=101325 Pa$ .

The conversion factor of $Pa$ is,

$1 Pa=1 N/m21 N=kg⋅m⋅s−2$

Substitute the units of pressure and solve the square root as shown below.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(2.65 ×10−10 m)2×101325 kg⋅m−1⋅s−2×424.993 m⋅s−1D=8.444×10−5 m2⋅s−1$

The units of diffusion constant are usually expressed in $cm$ . The conversion factor is $1 cm=10−2m$ . The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Conclusion

The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Interpretation Introduction

(b)

Interpretation:

The diffusion constant for $Xe$ is to be calculated.

Concept introduction:

The diffusion any gas particle among its own gas particles is known as self-diffusion. The diffusion constant is directly proportional to the product of mean free path and average velocity. The self-diffusion of a particle is given as,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant.

• $T$ is the temperature.

Expert Solution

Answer to Problem 19.62E

The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

Explanation of Solution

It is given that the xenon is at $1.00 atm$ , that is, standard pressure and $273 K$ . The molar mass of xenon is $131.29 g/mol$ and the diameter of xenon molecules is $4.00 A∘$ .

The diffusion constant can be calculated by the formula,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant and its value is $8.314 J⋅mol−1⋅K−1$ or $8.314 kg⋅m2⋅s−2⋅mol−1⋅K−1$ .

• $T$ is the temperature.

The diffusion constant can be represented in the terms of pressure as,

$D=3kT8d2p⋅RTπM$

Substitute the values of molar mass, diameter, pressure, gas constant and temperature.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(4.00 A∘)2×1.00 atm⋅8.314 m2⋅kg⋅s−2⋅K−1⋅mol−1×273 Kπ×131.29×10−3 kg⋅mol−1$

The units of pressure are converted from $atm$ to $Pa$ . The conversion factor is $1 atm=101325 Pa$ .

The conversion factor of $Pa$ is,

$1 Pa=1 N/m21 N=kg⋅m⋅s−2$

Substitute the units of pressure and solve the square root as shown below.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(4.00 ×10−10 m)2×101325 kg⋅m−1⋅s−2×74.181 m⋅s−1D=6.47×10−6 m2⋅s−1$

The units of diffusion constant are usually expressed in $cm$ . The conversion factor is $1 cm=10−2m$ . The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

Conclusion

The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

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Answer 3

Question

Chapter 19, Problem 19.62E

Interpretation Introduction

(a)

Interpretation:

The diffusion constant for $He$ is to be calculated.

Concept introduction:

The diffusion of any gas particle among its own gas particles is known as self-diffusion. The diffusion constant is directly proportional to the product of mean free path and average velocity. The self-diffusion of a particle is given as,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant.

• $T$ is the temperature.

Expert Solution

Answer to Problem 19.62E

The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Explanation of Solution

It is given that the helium is at $1.00 atm$ , that is, standard pressure and $273 K$ . The molar mass of helium is $4.00 g/mol$ and the diameter of helium molecules is $2.65 A∘$ .

The diffusion constant can be calculated by the formula,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant and its value is $8.314 J⋅mol−1⋅K−1$ or $8.314 kg⋅m2⋅s−2⋅mol−1⋅K−1$ .

• $T$ is the temperature.

The diffusion constant can be represented in the terms of pressure as,

$D=3kT8d2p⋅RTπM$

Substitute the values of molar mass, diameter, pressure, gas constant and temperature.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(2.65 A∘)2×1.00 atm⋅8.314 m2⋅kg⋅s−2⋅K−1⋅mol−1×273 Kπ×4.00×10−3 kg⋅mol−1$

The unit of pressure is converted from $atm$ to $Pa$ . The conversion factor is $1 atm=101325 Pa$ .

The conversion factor of $Pa$ is,

$1 Pa=1 N/m21 N=kg⋅m⋅s−2$

Substitute the units of pressure and solve the square root as shown below.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(2.65 ×10−10 m)2×101325 kg⋅m−1⋅s−2×424.993 m⋅s−1D=8.444×10−5 m2⋅s−1$

The units of diffusion constant are usually expressed in $cm$ . The conversion factor is $1 cm=10−2m$ . The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Conclusion

The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Interpretation Introduction

(b)

Interpretation:

The diffusion constant for $Xe$ is to be calculated.

Concept introduction:

The diffusion any gas particle among its own gas particles is known as self-diffusion. The diffusion constant is directly proportional to the product of mean free path and average velocity. The self-diffusion of a particle is given as,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant.

• $T$ is the temperature.

Expert Solution

Answer to Problem 19.62E

The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

Explanation of Solution

It is given that the xenon is at $1.00 atm$ , that is, standard pressure and $273 K$ . The molar mass of xenon is $131.29 g/mol$ and the diameter of xenon molecules is $4.00 A∘$ .

The diffusion constant can be calculated by the formula,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant and its value is $8.314 J⋅mol−1⋅K−1$ or $8.314 kg⋅m2⋅s−2⋅mol−1⋅K−1$ .

• $T$ is the temperature.

The diffusion constant can be represented in the terms of pressure as,

$D=3kT8d2p⋅RTπM$

Substitute the values of molar mass, diameter, pressure, gas constant and temperature.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(4.00 A∘)2×1.00 atm⋅8.314 m2⋅kg⋅s−2⋅K−1⋅mol−1×273 Kπ×131.29×10−3 kg⋅mol−1$

The units of pressure are converted from $atm$ to $Pa$ . The conversion factor is $1 atm=101325 Pa$ .

The conversion factor of $Pa$ is,

$1 Pa=1 N/m21 N=kg⋅m⋅s−2$

Substitute the units of pressure and solve the square root as shown below.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(4.00 ×10−10 m)2×101325 kg⋅m−1⋅s−2×74.181 m⋅s−1D=6.47×10−6 m2⋅s−1$

The units of diffusion constant are usually expressed in $cm$ . The conversion factor is $1 cm=10−2m$ . The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

Conclusion

The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

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Answer 4

Question

Chapter 19, Problem 19.62E

Interpretation Introduction

(a)

Interpretation:

The diffusion constant for $He$ is to be calculated.

Concept introduction:

The diffusion of any gas particle among its own gas particles is known as self-diffusion. The diffusion constant is directly proportional to the product of mean free path and average velocity. The self-diffusion of a particle is given as,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant.

• $T$ is the temperature.

Expert Solution

Answer to Problem 19.62E

The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Explanation of Solution

It is given that the helium is at $1.00 atm$ , that is, standard pressure and $273 K$ . The molar mass of helium is $4.00 g/mol$ and the diameter of helium molecules is $2.65 A∘$ .

The diffusion constant can be calculated by the formula,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant and its value is $8.314 J⋅mol−1⋅K−1$ or $8.314 kg⋅m2⋅s−2⋅mol−1⋅K−1$ .

• $T$ is the temperature.

The diffusion constant can be represented in the terms of pressure as,

$D=3kT8d2p⋅RTπM$

Substitute the values of molar mass, diameter, pressure, gas constant and temperature.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(2.65 A∘)2×1.00 atm⋅8.314 m2⋅kg⋅s−2⋅K−1⋅mol−1×273 Kπ×4.00×10−3 kg⋅mol−1$

The unit of pressure is converted from $atm$ to $Pa$ . The conversion factor is $1 atm=101325 Pa$ .

The conversion factor of $Pa$ is,

$1 Pa=1 N/m21 N=kg⋅m⋅s−2$

Substitute the units of pressure and solve the square root as shown below.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(2.65 ×10−10 m)2×101325 kg⋅m−1⋅s−2×424.993 m⋅s−1D=8.444×10−5 m2⋅s−1$

The units of diffusion constant are usually expressed in $cm$ . The conversion factor is $1 cm=10−2m$ . The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Conclusion

The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Interpretation Introduction

(b)

Interpretation:

The diffusion constant for $Xe$ is to be calculated.

Concept introduction:

The diffusion any gas particle among its own gas particles is known as self-diffusion. The diffusion constant is directly proportional to the product of mean free path and average velocity. The self-diffusion of a particle is given as,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant.

• $T$ is the temperature.

Expert Solution

Answer to Problem 19.62E

The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

Explanation of Solution

It is given that the xenon is at $1.00 atm$ , that is, standard pressure and $273 K$ . The molar mass of xenon is $131.29 g/mol$ and the diameter of xenon molecules is $4.00 A∘$ .

The diffusion constant can be calculated by the formula,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant and its value is $8.314 J⋅mol−1⋅K−1$ or $8.314 kg⋅m2⋅s−2⋅mol−1⋅K−1$ .

• $T$ is the temperature.

The diffusion constant can be represented in the terms of pressure as,

$D=3kT8d2p⋅RTπM$

Substitute the values of molar mass, diameter, pressure, gas constant and temperature.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(4.00 A∘)2×1.00 atm⋅8.314 m2⋅kg⋅s−2⋅K−1⋅mol−1×273 Kπ×131.29×10−3 kg⋅mol−1$

The units of pressure are converted from $atm$ to $Pa$ . The conversion factor is $1 atm=101325 Pa$ .

The conversion factor of $Pa$ is,

$1 Pa=1 N/m21 N=kg⋅m⋅s−2$

Substitute the units of pressure and solve the square root as shown below.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(4.00 ×10−10 m)2×101325 kg⋅m−1⋅s−2×74.181 m⋅s−1D=6.47×10−6 m2⋅s−1$

The units of diffusion constant are usually expressed in $cm$ . The conversion factor is $1 cm=10−2m$ . The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

Conclusion

The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

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Answer 5

Chapter 19, Problem 19.62E

Interpretation Introduction

(a)

Interpretation:

The diffusion constant for $He$ is to be calculated.

Concept introduction:

The diffusion of any gas particle among its own gas particles is known as self-diffusion. The diffusion constant is directly proportional to the product of mean free path and average velocity. The self-diffusion of a particle is given as,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant.

• $T$ is the temperature.

Expert Solution

Answer to Problem 19.62E

The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Explanation of Solution

It is given that the helium is at $1.00 atm$ , that is, standard pressure and $273 K$ . The molar mass of helium is $4.00 g/mol$ and the diameter of helium molecules is $2.65 A∘$ .

The diffusion constant can be calculated by the formula,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant and its value is $8.314 J⋅mol−1⋅K−1$ or $8.314 kg⋅m2⋅s−2⋅mol−1⋅K−1$ .

• $T$ is the temperature.

The diffusion constant can be represented in the terms of pressure as,

$D=3kT8d2p⋅RTπM$

Substitute the values of molar mass, diameter, pressure, gas constant and temperature.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(2.65 A∘)2×1.00 atm⋅8.314 m2⋅kg⋅s−2⋅K−1⋅mol−1×273 Kπ×4.00×10−3 kg⋅mol−1$

The unit of pressure is converted from $atm$ to $Pa$ . The conversion factor is $1 atm=101325 Pa$ .

The conversion factor of $Pa$ is,

$1 Pa=1 N/m21 N=kg⋅m⋅s−2$

Substitute the units of pressure and solve the square root as shown below.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(2.65 ×10−10 m)2×101325 kg⋅m−1⋅s−2×424.993 m⋅s−1D=8.444×10−5 m2⋅s−1$

The units of diffusion constant are usually expressed in $cm$ . The conversion factor is $1 cm=10−2m$ . The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Conclusion

The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Interpretation Introduction

(b)

Interpretation:

The diffusion constant for $Xe$ is to be calculated.

Concept introduction:

The diffusion any gas particle among its own gas particles is known as self-diffusion. The diffusion constant is directly proportional to the product of mean free path and average velocity. The self-diffusion of a particle is given as,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant.

• $T$ is the temperature.

Expert Solution

Answer to Problem 19.62E

The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

Explanation of Solution

It is given that the xenon is at $1.00 atm$ , that is, standard pressure and $273 K$ . The molar mass of xenon is $131.29 g/mol$ and the diameter of xenon molecules is $4.00 A∘$ .

The diffusion constant can be calculated by the formula,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant and its value is $8.314 J⋅mol−1⋅K−1$ or $8.314 kg⋅m2⋅s−2⋅mol−1⋅K−1$ .

• $T$ is the temperature.

The diffusion constant can be represented in the terms of pressure as,

$D=3kT8d2p⋅RTπM$

Substitute the values of molar mass, diameter, pressure, gas constant and temperature.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(4.00 A∘)2×1.00 atm⋅8.314 m2⋅kg⋅s−2⋅K−1⋅mol−1×273 Kπ×131.29×10−3 kg⋅mol−1$

The units of pressure are converted from $atm$ to $Pa$ . The conversion factor is $1 atm=101325 Pa$ .

The conversion factor of $Pa$ is,

$1 Pa=1 N/m21 N=kg⋅m⋅s−2$

Substitute the units of pressure and solve the square root as shown below.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(4.00 ×10−10 m)2×101325 kg⋅m−1⋅s−2×74.181 m⋅s−1D=6.47×10−6 m2⋅s−1$

The units of diffusion constant are usually expressed in $cm$ . The conversion factor is $1 cm=10−2m$ . The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

Conclusion

The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

Answer 6

Chapter 19, Problem 19.62E

Interpretation Introduction

(a)

Interpretation:

The diffusion constant for $He$ is to be calculated.

Concept introduction:

The diffusion of any gas particle among its own gas particles is known as self-diffusion. The diffusion constant is directly proportional to the product of mean free path and average velocity. The self-diffusion of a particle is given as,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant.

• $T$ is the temperature.

Expert Solution

Answer to Problem 19.62E

The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Explanation of Solution

It is given that the helium is at $1.00 atm$ , that is, standard pressure and $273 K$ . The molar mass of helium is $4.00 g/mol$ and the diameter of helium molecules is $2.65 A∘$ .

The diffusion constant can be calculated by the formula,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant and its value is $8.314 J⋅mol−1⋅K−1$ or $8.314 kg⋅m2⋅s−2⋅mol−1⋅K−1$ .

• $T$ is the temperature.

The diffusion constant can be represented in the terms of pressure as,

$D=3kT8d2p⋅RTπM$

Substitute the values of molar mass, diameter, pressure, gas constant and temperature.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(2.65 A∘)2×1.00 atm⋅8.314 m2⋅kg⋅s−2⋅K−1⋅mol−1×273 Kπ×4.00×10−3 kg⋅mol−1$

The unit of pressure is converted from $atm$ to $Pa$ . The conversion factor is $1 atm=101325 Pa$ .

The conversion factor of $Pa$ is,

$1 Pa=1 N/m21 N=kg⋅m⋅s−2$

Substitute the units of pressure and solve the square root as shown below.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(2.65 ×10−10 m)2×101325 kg⋅m−1⋅s−2×424.993 m⋅s−1D=8.444×10−5 m2⋅s−1$

The units of diffusion constant are usually expressed in $cm$ . The conversion factor is $1 cm=10−2m$ . The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Conclusion

The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Answer 7

Chapter 19, Problem 19.62E

Interpretation Introduction

(a)

Interpretation:

The diffusion constant for $He$ is to be calculated.

Concept introduction:

The diffusion of any gas particle among its own gas particles is known as self-diffusion. The diffusion constant is directly proportional to the product of mean free path and average velocity. The self-diffusion of a particle is given as,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant.

• $T$ is the temperature.

Answer 8

Expert Solution

Answer to Problem 19.62E

The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

It is given that the helium is at $1.00 atm$ , that is, standard pressure and $273 K$ . The molar mass of helium is $4.00 g/mol$ and the diameter of helium molecules is $2.65 A∘$ .

The diffusion constant can be calculated by the formula,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant and its value is $8.314 J⋅mol−1⋅K−1$ or $8.314 kg⋅m2⋅s−2⋅mol−1⋅K−1$ .

• $T$ is the temperature.

The diffusion constant can be represented in the terms of pressure as,

$D=3kT8d2p⋅RTπM$

Substitute the values of molar mass, diameter, pressure, gas constant and temperature.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(2.65 A∘)2×1.00 atm⋅8.314 m2⋅kg⋅s−2⋅K−1⋅mol−1×273 Kπ×4.00×10−3 kg⋅mol−1$

The unit of pressure is converted from $atm$ to $Pa$ . The conversion factor is $1 atm=101325 Pa$ .

The conversion factor of $Pa$ is,

$1 Pa=1 N/m21 N=kg⋅m⋅s−2$

Substitute the units of pressure and solve the square root as shown below.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(2.65 ×10−10 m)2×101325 kg⋅m−1⋅s−2×424.993 m⋅s−1D=8.444×10−5 m2⋅s−1$

The units of diffusion constant are usually expressed in $cm$ . The conversion factor is $1 cm=10−2m$ . The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Answer 13

Conclusion

The diffusion constant of helium is $0.844 cm2⋅s−1$ .

Answer 14

Interpretation Introduction

(b)

Interpretation:

The diffusion constant for $Xe$ is to be calculated.

Concept introduction:

The diffusion any gas particle among its own gas particles is known as self-diffusion. The diffusion constant is directly proportional to the product of mean free path and average velocity. The self-diffusion of a particle is given as,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant.

• $T$ is the temperature.

Expert Solution

Answer to Problem 19.62E

The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

Explanation of Solution

It is given that the xenon is at $1.00 atm$ , that is, standard pressure and $273 K$ . The molar mass of xenon is $131.29 g/mol$ and the diameter of xenon molecules is $4.00 A∘$ .

The diffusion constant can be calculated by the formula,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant and its value is $8.314 J⋅mol−1⋅K−1$ or $8.314 kg⋅m2⋅s−2⋅mol−1⋅K−1$ .

• $T$ is the temperature.

The diffusion constant can be represented in the terms of pressure as,

$D=3kT8d2p⋅RTπM$

Substitute the values of molar mass, diameter, pressure, gas constant and temperature.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(4.00 A∘)2×1.00 atm⋅8.314 m2⋅kg⋅s−2⋅K−1⋅mol−1×273 Kπ×131.29×10−3 kg⋅mol−1$

The units of pressure are converted from $atm$ to $Pa$ . The conversion factor is $1 atm=101325 Pa$ .

The conversion factor of $Pa$ is,

$1 Pa=1 N/m21 N=kg⋅m⋅s−2$

Substitute the units of pressure and solve the square root as shown below.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(4.00 ×10−10 m)2×101325 kg⋅m−1⋅s−2×74.181 m⋅s−1D=6.47×10−6 m2⋅s−1$

The units of diffusion constant are usually expressed in $cm$ . The conversion factor is $1 cm=10−2m$ . The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

Conclusion

The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

Answer 15

Interpretation Introduction

(b)

Interpretation:

The diffusion constant for $Xe$ is to be calculated.

Concept introduction:

The diffusion any gas particle among its own gas particles is known as self-diffusion. The diffusion constant is directly proportional to the product of mean free path and average velocity. The self-diffusion of a particle is given as,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant.

• $T$ is the temperature.

Answer 16

Expert Solution

Answer to Problem 19.62E

The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

It is given that the xenon is at $1.00 atm$ , that is, standard pressure and $273 K$ . The molar mass of xenon is $131.29 g/mol$ and the diameter of xenon molecules is $4.00 A∘$ .

The diffusion constant can be calculated by the formula,

$D=38d2ρ⋅RTπM$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $ρ$ is the particle density.

• $R$ is the gas constant and its value is $8.314 J⋅mol−1⋅K−1$ or $8.314 kg⋅m2⋅s−2⋅mol−1⋅K−1$ .

• $T$ is the temperature.

The diffusion constant can be represented in the terms of pressure as,

$D=3kT8d2p⋅RTπM$

Substitute the values of molar mass, diameter, pressure, gas constant and temperature.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(4.00 A∘)2×1.00 atm⋅8.314 m2⋅kg⋅s−2⋅K−1⋅mol−1×273 Kπ×131.29×10−3 kg⋅mol−1$

The units of pressure are converted from $atm$ to $Pa$ . The conversion factor is $1 atm=101325 Pa$ .

The conversion factor of $Pa$ is,

$1 Pa=1 N/m21 N=kg⋅m⋅s−2$

Substitute the units of pressure and solve the square root as shown below.

$D=3×1.381×10−23 m2⋅kg⋅s−2⋅K−1×273 K8(4.00 ×10−10 m)2×101325 kg⋅m−1⋅s−2×74.181 m⋅s−1D=6.47×10−6 m2⋅s−1$

The units of diffusion constant are usually expressed in $cm$ . The conversion factor is $1 cm=10−2m$ . The diffusion constant of xenon is $0.0647 cm2⋅s−1$ .