Physical Chemistry
Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
bartleby

Concept explainers

Question
Chapter 19, Problem 19.48E
Interpretation Introduction

(a)

Interpretation:

The mean free paths for nitrogen and oxygen atoms are to be calculated.

Concept introduction:

The mean free path of collisions between gaseous atoms is given by the formula given below.

λ=kT2πd2p

The average collision frequency of one atom is given by the formula,

z=4πρd2kTπm

The total number of collisions is given by the formula,

Z=2πρ1ρ2[d1+d22]2kTπm

Expert Solution
Check Mark

Answer to Problem 19.48E

The mean free paths for nitrogen and oxygen atoms are 8.44×108m and 9.43×108m respectively.

Explanation of Solution

The mean free path of collisions between gaseous atoms is given by the formula given below.

λ=kT2πd2p …(1)

Where,

p is the pressure.

T is the temperature.

d is the collision diameter.

k is the Boltzmann constant.

Substitute the values in the equation (1) for nitrogen atom as given below.

λ=kT2πd2p=(1.381×1023J/K)(273K)2×3.14×(3.15A×1010m1A)2×1atm×1Latm101.33J×(0.1m)31L=8.44×108m

Substitute the values in the equation (1) for oxygen atom as given below.

λ=kT2πd2p=(1.381×1023J/K)(273K)2×3.14×(2.98A×1010m1A)2×1atm×1Latm101.33J×(0.1m)31L=9.43×108m

Conclusion

The mean free paths for nitrogen and oxygen atoms are 8.44×108m and 9.43×108m respectively.

Interpretation Introduction

(b)

Interpretation:

The average collision frequencies for nitrogen and oxygen atoms are to be calculated.

Concept introduction:

The mean free path of collisions between gaseous atoms is given by the formula given below.

λ=kT2πd2p

The average collision frequency of one atom is given by the formula,

z=4πρd2kTπm

The total number of collisions is given by the formula,

Z=2πρ1ρ2[d1+d22]2kTπm.

Expert Solution
Check Mark

Answer to Problem 19.48E

The average collision frequencies for nitrogen and oxygen atoms are 96.18×108s1 and 317.55×108s1.

Explanation of Solution

The average collision frequency of one atom is given by the formula,

z=4πρd2kTπm …(2)

Where,

ρ is the density.

T is the temperature.

m is the mass of an atom.

d is the collision diameter.

k is the Boltzmann constant.

The ratio of nitrogen to oxygen in air is 4:1. Thus, the number of moles of nitrogen and oxygen in the mixture is 0.8 and 0.2 respectively. This gives the density of argon and xenon at STP as follows.

ρnitrogen=6.022×10230.8×0.001m3×22.4=0.34×1026m3

ρoxygen=6.022×10230.2×0.001m3×22.4=1.34×1026m3

The mass of nitrogen and oxygen is 2.33×1026kg and 2.66×1026kg respectively.

Substitute the values in the equation (2) for nitrogen atom as given below.

z=4πρd2kTπm=4×3.14×0.34×1026m3×(3.15A×1010m1A)2(1.381×1023J/K)(273K)(2.33×1026kg)(3.14)=260.176×1052.705×1013s1=96.18×108s1

Substitute the values in the equation (2) for oxygen atom as given below.

z=4πρd2kTπm=4×3.14×1.34×1026m3×(2.98A×1010m1A)2(1.381×1023J/K)(273K)(2.66×1026kg)(3.14)=917.709×1052.890×1013=317.55×108s1

Conclusion

The average collision frequencies for nitrogen and oxygen atoms are 96.18×108s1 and 317.55×108s1.

Interpretation Introduction

(c)

Interpretation:

The total number of collisions between nitrogen and oxygen atoms is to be calculated.

Concept introduction:

The mean free path of collisions between gaseous atoms is given by the formula given below.

λ=kT2πd2p

The average collision frequency of one atom is given by the formula,

z=4πρd2kTπm

The total number of collisions is given by the formula,

Z=2πρ1ρ2[d1+d22]2kTπm.

Expert Solution
Check Mark

Answer to Problem 19.48E

The total number of collisions between nitrogen and oxygen atoms is 32.93×1034m3s1.

Explanation of Solution

The total number of collisions is given by the formula,

Z=2πρ1ρ2[d1+d22]2kTπμ …(3)

Where,

ρ is the density.

T is the temperature.

μ is the reduced mass.

d is the collision diameter.

k is the Boltzmann constant.

The ratio of nitrogen to oxygen in air is 4:1. Thus, the number of moles of nitrogen and oxygen in the mixture is 0.8 and 0.2 respectively. This gives the density of argon and xenon at STP as follows.

ρnitrogen=6.022×10230.8×0.001m3×22.4=0.34×1026m3

ρoxygen=6.022×10230.2×0.001m3×22.4=1.34×1026m3

The mass of nitrogen and oxygen is 2.33×1026kg and 2.66×1026kg respectively.

The reduced mass is calculated as follows:

μ=mN2mO2mN2+mO2

Substitute the mass of nitrogen and oxygen in above formula.

μ=(2.33×1026kg)(2.66×1026kg)2.33×1026kg+2.66×1026kgμ=6.1978×1052kg24.99×1026kgμ=1.242×1026kg

Thus, the reduced mass of nitrogen and oxygen is 1.242×1026kg.

Substitute the values in the equation (3) as given below.

Z=(2×3.14×1.34×1026m3×0.134×1026m3×((3.15A× 10 10m1A)+(2.98A× 10 10m1A)2)2×(1.381×1023J/K)(273K))(1.242×1026kg)(3.14)=65.042×10211.975×1013m3s1=32.93×1034m3s1

Conclusion

The total number of collisions between nitrogen and oxygen atoms is 32.93×1034m3s1.

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Chapter 19 Solutions

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