ATKINS' PHYSICAL CHEMISTRY-ACCESS
ATKINS' PHYSICAL CHEMISTRY-ACCESS
11th Edition
ISBN: 9780198834700
Author: ATKINS
Publisher: OXF
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Chapter 1, Problem 1B.3P
Interpretation Introduction

Interpretation: The mean speed of the emerging beam relative to the initial value has to be calculated.

Concept introduction: The expression for the mean speed is derived from Maxwell Boltzmann distribution.  The mean speed for the gas particles is represented as,

  vmean=(8RTπM)1/2

Expert Solution & Answer
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Answer to Problem 1B.3P

The speed of the emerging beam relative to the initial value is vx=0.47vxinitial_.

Explanation of Solution

The expression for the velocity of the emerging beam is,

  vx=K0avxf(vx)dvx                                                                                       (1)

Where,

  • K is the constant of proportionality.
  • vx is the velocity.

The value of f(vx) is given by,

  f(vx)=(m2πkT)1/2emv2/2kT

Where,

  • m is the mass.
  • k is the Boltzmann constant.
  • T is the temperature.

The normalised function is written as,

  vx0avxdvx=K0af(vx)dvx1=K0af(vx)dvx

Substitute the value of f(vx) in the above equation.

    1=K(m2πkT)1/20aemv2/2kTdvx                                                                       (2)

The equation (2) is evaluated by defining an error function.

  mvx22kT=n2vx2=2kTmn2vx=(2kTm)1/2n

Or,

  dvx=(2kTm)1/2dn                                                                                          (3)

Substitute equation (3) in equation (2) to find n.

  1=K(m2πkT)1/2(2kTm)1/20ben2dn

Here, b=(m2kT)1/2×a

  1=K(2kTm)1/2(m2πkT)1/20ben2dn=(Kπ)1/20ben2dn

According to error function,

  erf(z)=2π1/20zen2dn

Therefore,

  1=(Kπ)1/20ben2dn=12Kerf(b)=2erf(b)

To calculate the mean velocity of the beam, substitute the values in equation (1).

    vx=K(m2πkT)1/20avxemvx2/2kTdvx=K(m2πkT)1/2(kTm)0addvx(emvx2/2kTdvx)

    vx=(kT2mπ)1/2(emvx2/2kT1) (4)

The initial velocity, a=vxinitial and the initial velocity is (2kT/mπ)1/2.  The expression for average magnitude of one dimensional velocity in the x direction is,

    vxinitial=20vxf(vx)dvx=20vx(m2πkT)1/2emv2/2kTdvx=(m2πkT)1/2(2kTm)=(2kTmπ)1/2

For a= the value of constant, b= and the value of error function, berf(b)=1. Substitute the value of a and berf(b)=erf(1/π1/2) in equation (4).

  vx=(2kTmπ)1/2×1e1/πerf(1/π1/2)                                                                           (5)

The value of erf(1/π1/2) is,

  erf(1/π1/2)=erf(0.56)=0.57

The value of e1/π is 0.73.

Substitute the values in equation (5) to calculate mean velocity.

    vx=(2kTmπ)1/2(10.730.57)=0.47vxinitial_

Thus, the value of mean velocity of emerging beam is vx=0.47vxinitial_.

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Chapter 1 Solutions

ATKINS' PHYSICAL CHEMISTRY-ACCESS

Ch. 1 - Prob. 1A.3BECh. 1 - Prob. 1A.4AECh. 1 - Prob. 1A.4BECh. 1 - Prob. 1A.5AECh. 1 - Prob. 1A.5BECh. 1 - Prob. 1A.6AECh. 1 - Prob. 1A.6BECh. 1 - Prob. 1A.7AECh. 1 - Prob. 1A.7BECh. 1 - Prob. 1A.8AECh. 1 - Prob. 1A.8BECh. 1 - Prob. 1A.9AECh. 1 - Prob. 1A.9BECh. 1 - Prob. 1A.10AECh. 1 - Prob. 1A.10BECh. 1 - Prob. 1A.11AECh. 1 - Prob. 1A.11BECh. 1 - Prob. 1A.1PCh. 1 - Prob. 1A.2PCh. 1 - Prob. 1A.3PCh. 1 - Prob. 1A.4PCh. 1 - Prob. 1A.5PCh. 1 - Prob. 1A.6PCh. 1 - Prob. 1A.7PCh. 1 - Prob. 1A.8PCh. 1 - Prob. 1A.9PCh. 1 - Prob. 1A.10PCh. 1 - Prob. 1A.11PCh. 1 - Prob. 1A.12PCh. 1 - Prob. 1A.13PCh. 1 - Prob. 1A.14PCh. 1 - Prob. 1B.1DQCh. 1 - Prob. 1B.2DQCh. 1 - Prob. 1B.3DQCh. 1 - Prob. 1B.1AECh. 1 - Prob. 1B.1BECh. 1 - Prob. 1B.2AECh. 1 - Prob. 1B.2BECh. 1 - Prob. 1B.3AECh. 1 - Prob. 1B.3BECh. 1 - Prob. 1B.4AECh. 1 - Prob. 1B.4BECh. 1 - Prob. 1B.5AECh. 1 - Prob. 1B.5BECh. 1 - Prob. 1B.6AECh. 1 - Prob. 1B.6BECh. 1 - Prob. 1B.7AECh. 1 - Prob. 1B.7BECh. 1 - Prob. 1B.8AECh. 1 - Prob. 1B.8BECh. 1 - Prob. 1B.9AECh. 1 - Prob. 1B.9BECh. 1 - Prob. 1B.1PCh. 1 - Prob. 1B.2PCh. 1 - Prob. 1B.3PCh. 1 - Prob. 1B.4PCh. 1 - Prob. 1B.5PCh. 1 - Prob. 1B.6PCh. 1 - Prob. 1B.7PCh. 1 - Prob. 1B.8PCh. 1 - Prob. 1B.9PCh. 1 - Prob. 1B.10PCh. 1 - Prob. 1B.11PCh. 1 - Prob. 1C.1DQCh. 1 - Prob. 1C.2DQCh. 1 - Prob. 1C.3DQCh. 1 - Prob. 1C.4DQCh. 1 - Prob. 1C.1AECh. 1 - Prob. 1C.1BECh. 1 - Prob. 1C.2AECh. 1 - Prob. 1C.2BECh. 1 - Prob. 1C.3AECh. 1 - Prob. 1C.3BECh. 1 - Prob. 1C.4AECh. 1 - Prob. 1C.4BECh. 1 - Prob. 1C.5AECh. 1 - Prob. 1C.5BECh. 1 - Prob. 1C.6AECh. 1 - Prob. 1C.6BECh. 1 - Prob. 1C.7AECh. 1 - Prob. 1C.7BECh. 1 - Prob. 1C.8AECh. 1 - Prob. 1C.8BECh. 1 - Prob. 1C.9AECh. 1 - Prob. 1C.9BECh. 1 - Prob. 1C.1PCh. 1 - Prob. 1C.2PCh. 1 - Prob. 1C.3PCh. 1 - Prob. 1C.4PCh. 1 - Prob. 1C.5PCh. 1 - Prob. 1C.6PCh. 1 - Prob. 1C.7PCh. 1 - Prob. 1C.8PCh. 1 - Prob. 1C.9PCh. 1 - Prob. 1C.10PCh. 1 - Prob. 1C.11PCh. 1 - Prob. 1C.12PCh. 1 - Prob. 1C.13PCh. 1 - Prob. 1C.14PCh. 1 - Prob. 1C.15PCh. 1 - Prob. 1C.16PCh. 1 - Prob. 1C.17PCh. 1 - Prob. 1C.18PCh. 1 - Prob. 1C.19PCh. 1 - Prob. 1C.20PCh. 1 - Prob. 1C.22PCh. 1 - Prob. 1C.23PCh. 1 - Prob. 1C.24PCh. 1 - Prob. 1.1IACh. 1 - Prob. 1.2IACh. 1 - Prob. 1.3IA
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