Physical Chemistry
Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
Question
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Chapter 19, Problem 19.22E
Interpretation Introduction

Interpretation:

The validation of the statement that the value of constant K, defined by 1vxgx(vx)vxgx(vx) and 1vΓ(v)vΓ(v) is the same for 1vygy(vy)vygy(vy) and 1vzgz(vz)vzgz(vz) is to be stated.

Concept introduction:

The probability distribution function of the velocities of the gas molecules in each dimension is given by gx(vx), gy(vy) and gz(vz). This is a vector quantity. This probability function in the range of to + is represented by,

+gx(vx)dvx=1

Expert Solution & Answer
Check Mark

Answer to Problem 19.22E

The given statement that the value of constant K, defined by 1vxgx(vx)vxgx(vx) and 1vΓ(v)vΓ(v) is the same for 1vygy(vy)vygy(vy) and 1vzgz(vz)vzgz(vz) has been validated.

Explanation of Solution

The distribution function gx is given by,

gx=(m2πkT)1/2emvx2/2kT

It is given that the value of constant K is defined by,

1vxgx(vx)vxgx(vx) …(1)

Substitute the value of gx in equation (1).

K=1vx(m2πkT)1/2emvx2/2kTvx(m2πkT)1/2emvx2/2kT

Keep all the constant terms out of the differentiation.

K=1vx(m2πkT)1/2emv2/2kTvx(m2πkT)1/2emv2/2kT

K=1vxemv2/2kTvxemv2/2kT …(2)

The differentiation of the exponential term with respect to vx is shown below.

emvx2/2kTvx=emvx2/2kT.(2mvx2kT)

=emvx2/2kT.(mvxkT) …(3)

Substitute equation (3) into equation (2).

K=1vxemvx2/2kT.(mvxkT)emv2/2kT

Cancel the common terms and thus, the value of K is,

K=(mkT)

The value of K obtained is a constant, since the terms that contain velocity vx cancels out.

Similarly for gy(vy) and gz(vz) the terms that contain velocity vy and vz respectively cancels out and the value of K obtained is constant.

It is given that the value of constant K is defined by,

1vΓ(v)vΓ(v) …(4)

The term Γ(v) is given by,

Γ(v)=gx(vx)gy(vy)gz(vz) …(5)

Since Γ(v) is the product of three uni-dimensional distribution function the value of 1vΓ(v)vΓ(v) is same for each uni-dimensional distribution function g. Thus, the value of K is same for each uni-dimensional distribution function g.

Conclusion

The given statement that the value of constant K, defined by 1vxgx(vx)vxgx(vx) and 1vΓ(v)vΓ(v) is the same for 1vygy(vy)vygy(vy) and 1vzgz(vz)vzgz(vz) has been validated.

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Chapter 19 Solutions

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