30_562ln08

30_562ln08 - MIT OpenCourseWare http:/ocw.mit.edu 5.62...

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MIT OpenCourseWare http://ocw.mit.edu 5.62 Physical Chemistry II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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5.62 Spring 2008 Lecture #30 Page 1 Kinetic Theory of Gases: Collision Dynamics and Scattering The goal of kinetic theory is to understand the collision process between a pair of molecules. In this lecture we give a simple description of the classical scattering process. Although we cannot completely develop this subject, it is important to know that scattering experiments are an invaluable source of information about molecular interactions and that the results of scattering experiments can be used to predict transport coefficients for g incoming flux scattering angle ases. The typical scattering experiment consists in sending a flux of particles into a stationary target gas and observing the deflection of particles at a distance from the target region at various scattering angles. We describe this process at the molecular level by considering a particular collision pair. We assume the following for the colliding pair: In center of mass coordinates the energy of the pair is given by: E = μ 2 ! x(t) 2 + ! y(t) 2 ! " # $ + V[r(t)] μ = m 1 m 2 m 1 + m 2 and r(t) is the distance between the colliding particles. Only two components (x and y) are needed to characterize the collision because it is possible to show that two particle scattering is confined to a plane in the center of mass coordinate system. The energy is the sum of the x and y kinetic energy of the fictitious particle with reduced mass µ . There is a potential energy, V[r(t)], that depends only on the distance between the two centers. The collision is assumed to be “elastic,” which means that no energy is transferred into the internal modes of the two collision partners. In this case the total E is constant at every point along the trajectory. revised 4/22/08 5:33 PM
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5.62 Spring 2008 Lecture #30 Page 2 In polar coordinates, x(t) = r(t)cos ! (t) [ ] , y(t) = r(t)sin ! (t) [ ] , and the energy is: E = μ 2 ! r(t) 2 + r(t) 2 ! ! (t) 2 " # $ % + V[r(t)] . In addition, because the force is central there is conservation of angular momentum ! L
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30_562ln08 - MIT OpenCourseWare http:/ocw.mit.edu 5.62...

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