Explanation According to the Boltzmann distribution, the velocity components, v x and v y is directly proportional to the exponential function to their kinetic energy, f ( v ) = K e − E / k T (1) Where, Ø K is the constant of proportionality. Ø E is the kinetic energy. Ø k is the Boltzmann constant. Ø T is the temperature. The kinetic energy of the system of molecules that move in a two-dimensional plane is given by, Kinetic Energy = 1 2 m v x 2 + 1 2 m v y 2 Substitute the value of kinetic energy in equation (1). f ( v ) = K e − ( m v x 2 + m v y 2 ) / 2 k T = K e − m v x 2 / 2 k T e − m v y 2 / 2 k T The distribution factorizes into two terms, f ( v ) = f ( v x ) f ( v y ) and K = K x K y Therefore, f ( v x ) = K x e − m v x 2 / 2 k T To calculate the constant K x , the molecule should have the velocity over the range, − ∞ < v x < ∞ . In order to integrate over the full range, the probability should be 1 . f ( v x ) = 1 Substitute the expression for f ( v x ) . 1 = K x ∫ − ∞ ∞ e − m v x 2 / 2 k T d v x = K x ( 2 π k T m ) 1 / 2 Therefore, K x = ( m 2 π k T ) 1 / 2 and f ( v x ) = ( m 2 π k T ) 1 / 2 e − m v x 2 / 2 k T Similarly the expression for f ( v x ) can be derived

11th Edition

ISBN: 9780198814740

Author: ATKINS, P. W. (peter William), 1940- (author.)

Publisher: Oxford University Press,

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Kinetic-Theory of Gases

Kinetic theory of gases is the model or method which was developed in the 19th century by Maxwell and Boltzmann. This is possible as the interatomic force which is of short-range forces that are important for solids and liquids can be neglected for gases…

Question

Chapter 1, Problem 1B.2P

Interpretation Introduction

Interpretation: The distribution of speeds and the mean speed of the molecules at a temperature T moving in a two-dimensional plane has to be calculated.

Concept introduction: The speeds of individual molecules spread across a wide range in a gas and the gas collisions ensure that their speeds change constantly. The distribution of speeds is identified by recognizing that the molecules possess kinetic energy and the distribution of energy over the molecules is determined with the help of Maxwell Boltzmann distribution.

Expert Solution & Answer

Chapter 1 Solutions

Atkins' Physical chemistry

Ch. 1 - Prob. 1A.1ST Ch. 1 - Prob. 1A.2ST Ch. 1 - Prob. 1B.1ST Ch. 1 - Prob. 1C.1ST Ch. 1 - Prob. 1A.1DQ Ch. 1 - Prob. 1A.1AE Ch. 1 - Prob. 1A.1BE Ch. 1 - Prob. 1A.2AE Ch. 1 - Prob. 1A.2BE Ch. 1 - Prob. 1A.3AE

Ch. 1 - Prob. 1A.3BE Ch. 1 - Prob. 1A.4AE Ch. 1 - Prob. 1A.4BE Ch. 1 - Prob. 1A.5AE Ch. 1 - Prob. 1A.5BE Ch. 1 - Prob. 1A.6AE Ch. 1 - Prob. 1A.6BE Ch. 1 - Prob. 1A.7AE Ch. 1 - Prob. 1A.7BE Ch. 1 - Prob. 1A.8AE Ch. 1 - Prob. 1A.8BE Ch. 1 - Prob. 1A.9AE Ch. 1 - Prob. 1A.9BE Ch. 1 - Prob. 1A.10AE Ch. 1 - Prob. 1A.10BE Ch. 1 - Prob. 1A.11AE Ch. 1 - Prob. 1A.11BE Ch. 1 - Prob. 1A.1P Ch. 1 - Prob. 1A.2P Ch. 1 - Prob. 1A.3P Ch. 1 - Prob. 1A.4P Ch. 1 - Prob. 1A.5P Ch. 1 - Prob. 1A.6P Ch. 1 - Prob. 1A.7P Ch. 1 - Prob. 1A.8P Ch. 1 - Prob. 1A.9P Ch. 1 - Prob. 1A.10P Ch. 1 - Prob. 1A.11P Ch. 1 - Prob. 1A.12P Ch. 1 - Prob. 1A.13P Ch. 1 - Prob. 1A.14P Ch. 1 - Prob. 1B.1DQ Ch. 1 - Prob. 1B.2DQ Ch. 1 - Prob. 1B.3DQ Ch. 1 - Prob. 1B.1AE Ch. 1 - Prob. 1B.1BE Ch. 1 - Prob. 1B.2AE Ch. 1 - Prob. 1B.2BE Ch. 1 - Prob. 1B.3AE Ch. 1 - Prob. 1B.3BE Ch. 1 - Prob. 1B.4AE Ch. 1 - Prob. 1B.4BE Ch. 1 - Prob. 1B.5AE Ch. 1 - Prob. 1B.5BE Ch. 1 - Prob. 1B.6AE Ch. 1 - Prob. 1B.6BE Ch. 1 - Prob. 1B.7AE Ch. 1 - Prob. 1B.7BE Ch. 1 - Prob. 1B.8AE Ch. 1 - Prob. 1B.8BE Ch. 1 - Prob. 1B.9AE Ch. 1 - Prob. 1B.9BE Ch. 1 - Prob. 1B.1P Ch. 1 - Prob. 1B.2P Ch. 1 - Prob. 1B.3P Ch. 1 - Prob. 1B.4P Ch. 1 - Prob. 1B.5P Ch. 1 - Prob. 1B.6P Ch. 1 - Prob. 1B.7P Ch. 1 - Prob. 1B.8P Ch. 1 - Prob. 1B.9P Ch. 1 - Prob. 1B.10P Ch. 1 - Prob. 1B.11P Ch. 1 - Prob. 1C.1DQ Ch. 1 - Prob. 1C.2DQ Ch. 1 - Prob. 1C.3DQ Ch. 1 - Prob. 1C.4DQ Ch. 1 - Prob. 1C.1AE Ch. 1 - Prob. 1C.1BE Ch. 1 - Prob. 1C.2AE Ch. 1 - Prob. 1C.2BE Ch. 1 - Prob. 1C.3AE Ch. 1 - Prob. 1C.3BE Ch. 1 - Prob. 1C.4AE Ch. 1 - Prob. 1C.4BE Ch. 1 - Prob. 1C.5AE Ch. 1 - Prob. 1C.5BE Ch. 1 - Prob. 1C.6AE Ch. 1 - Prob. 1C.6BE Ch. 1 - Prob. 1C.7AE Ch. 1 - Prob. 1C.7BE Ch. 1 - Prob. 1C.8AE Ch. 1 - Prob. 1C.8BE Ch. 1 - Prob. 1C.9AE Ch. 1 - Prob. 1C.9BE Ch. 1 - Prob. 1C.1P Ch. 1 - Prob. 1C.2P Ch. 1 - Prob. 1C.3P Ch. 1 - Prob. 1C.4P Ch. 1 - Prob. 1C.5P Ch. 1 - Prob. 1C.6P Ch. 1 - Prob. 1C.7P Ch. 1 - Prob. 1C.8P Ch. 1 - Prob. 1C.9P Ch. 1 - Prob. 1C.10P Ch. 1 - Prob. 1C.11P Ch. 1 - Prob. 1C.12P Ch. 1 - Prob. 1C.13P Ch. 1 - Prob. 1C.14P Ch. 1 - Prob. 1C.15P Ch. 1 - Prob. 1C.16P Ch. 1 - Prob. 1C.17P Ch. 1 - Prob. 1C.18P Ch. 1 - Prob. 1C.19P Ch. 1 - Prob. 1C.20P Ch. 1 - Prob. 1C.22P Ch. 1 - Prob. 1C.23P Ch. 1 - Prob. 1C.24P Ch. 1 - Prob. 1.1IA Ch. 1 - Prob. 1.2IA Ch. 1 - Prob. 1.3IA

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