PHYSICAL CHEMISTRY-STUDENT SOLN.MAN.
PHYSICAL CHEMISTRY-STUDENT SOLN.MAN.
2nd Edition
ISBN: 9781285074788
Author: Ball
Publisher: CENGAGE L
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Chapter 19, Problem 19.49E
Interpretation Introduction

(a)

Interpretation:

The mean free paths for argon and xenon 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 19.49E

The mean free paths for argon and xenon atoms are 1.24×1016m and 5.23×1015m 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 argon atom as given below.

λ=kT2πd2p=(1.381×1023J/K)(273K)2×3.14×(2.60A×1010m1A)2×1atm×1Latm101.33J×(0.1m)31L=1.24×1016m

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

λ=kT2πd2p=(1.381×1023J/K)(273K)2×3.14×(4.00A×1010m1A)2×1atm×1Latm101.33J×(0.1m)31L=5.23×1015m

Conclusion

The mean free paths for argon and xenon atoms are 1.24×1016m and 5.23×1015m respectively.

Interpretation Introduction

(b)

Interpretation:

The average collision frequencies for argon and xenon 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 19.49E

The average collision frequencies for argon and xenon atoms are 15.311×108s-1 and 199.86×107s-1.

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 mixture given is a 50:50 mixture. Thus, the number of moles of argon and xenon in the mixture is 0.5 each. This gives the density of argon and xenon at STP as follows.

ρxenon=6.022×10230.5×0.001m3×22.4=0.134×1026m3

ρargon=6.022×10230.5×0.001m3×22.4=0.134×1026m3

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

z=4πρd2kTπm=4×3.14×0.134×1026m3×(2.60A×1010m1A)2(1.381×1023J/K)(273K)(6.63×1026kg)(3.14)=15.311×108s1

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

z=4πρd2kTπm=4×3.14×0.134×1026m3×(4.00A×1010m1A)2(1.381×1023J/K)(273K)(2.18×1025kg)(3.14)=199.86×107s-1

Conclusion

The average collision frequencies for argon and xenon atoms are 15.311×108s1 and 199.86×107s1.

Interpretation Introduction

(c)

Interpretation:

The total number of collisions between argon and xenon 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 19.49E

The total number of collisions between argon and xenon atoms is 37.79×1033m3s1.

Explanation of Solution

The total number of collisions is given by the formula,

Z=2πρ1ρ2[d1+d22]2kTπm …(3)

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 mixture given is a 50:50 mixture. Thus, the number of moles of argon and xenon in the mixture is 0.5 each. This gives the density of argon and xenon at STP as follows.

ρxenon=6.022×10230.5×0.001m3×22.4=0.134×1026m3

ρargon=6.022×10230.5×0.001m3×22.4=0.134×1026m3

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

Z=2πρ1ρ2[d1+d22]2kTπm=(4×3.14×0.134×1026m3×0.134×1026m3×((4.00A× 10 10m1A)+(2.60A× 10 10m1A)2)2×(1.381×1023J/K)(273K))(0.508×1025kg)(3.14)=37.79×1033m3s1

Conclusion

The total number of collisions between argon and xenon atoms is 37.79×1033m3s1.

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Chapter 19 Solutions

PHYSICAL CHEMISTRY-STUDENT SOLN.MAN.

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