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Unformatted text preview: Department of Chemical Engineering ChE 170 University of California, Santa Barbara Fall 2010 Handout 2: Molecular Thermodynamics Boltzmann Law Consider all possible molecular configurations of a system. Let a particular configuration be denoted by the index g . At some temperature G , Boltzmann’s law states that the fraction of the time the system will spend in that configuration is given by ¡ ¢ £ ¤¥ ¦§ ¨ © ª « where ¬ ¢ is the energy of the configuration and ¤ is a constant. These probabilities must sum to unity when one considers all possible configurations of a system. Thus, we must have ¡ ¢ ¢ £ 1 Plugging in the expression above, ¤¥ ¦§ ¨ © ª « ¢ £ 1 ® ¤ ¥ ¦§ ¨ © ª « ¢ £ 1 ® ¤ £ ¯¥ ¦§ ¨ © ª « ¢ ° ¦± Therefore, a general expression for the probability, without the constant, is ¡ ¢ £ ¥ ¦§ ¨ © ª « ∑ ¥ ¦§ ¨ ² © ª « ¢ ² Configurations versus states Oftentimes we are interested not in specific configurations, but rather, more general states that could involve a collection of different configurations. Some examples of different states: bound versus unbound folded versus unfolded phosphorylated versus unphosphorylated To find the fraction of the time that a system spends in a specific state, we need to sum up the fraction of the time that it spends in all configurations that belong to that state: ¡³state´ £ ¡ ¢ ¢µ¶state· £ ¤ ¥ ¦§ ¨ © ª « ¢µ¶state· Now we consider a simplification. We will assume that all of the configurations in a state have the same energy ¬ ¢ £ ¬ state . That means that each term in the sum above will have the same value. We will denote the number of configurations in the state by ΩgstateG . Then, we can write, ¡gstateG ¢ £ΩgstateG¤ ¥¦ state § ¨ © But we recognize that ªgstateG ¢ « ¬ lnΩgstateG and so we can write, ¡gstateG ¢ £¤ © § ¨ © ¥ ¦ state § ¨ © ¢ £¤ ¥ ® state § ¨ © where we have identified ¯ state ¢ ° state ± ²ª state . If the system is at constant pressure, this expression takes on a slightly different form (which we present here without derivation): ¡gstateG ¢ £¤ ¥ ³ state § ¨ © where ´ state gives the Gibbs free energy of that state. Now, to find the constant of normalization, we take a similar approach as before. We consider that the sum of probabilities over all possible states must equal unity, µ ¡g¶G states · ¢ 1 ¸ £ µ ¤ ¥ ³ ¹ § ¨ © states · ¢ 1 ¸ £ ¢ º µ ¤ ¥ ³ ¹ § ¨ © states · » ¥¼ Therefore we find that the probability of different states is given by, ¡gstateG ¢ ¤ ¥ ³ state § ¨ © ∑ ¤ ¥ ³ ¹ § ¨ © states · Department of Chemical Engineering ChE 170 University of California, Santa Barbara Fall 2010 Handout 2: Molecular Thermodynamics Boltzmann Law Consider all possible molecular configurations of a system. Let a particular configuration be denoted by the index g . At some temperature G , Boltzmann’s law states that the fraction of...
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 Fall '10
 Ceweb
 Chemical Engineering, Thermodynamics, Mole, Santa Barbara, Department of Chemical Engineering, molecular thermodynamics, Boltzmann Law

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