lecture 6: intro to entropy reading: zumdahl 10.1, 10.3 outline: –why enthalpy...

Download Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s

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  • Slide 1
  • Slide 2
  • Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: Why enthalpy isnt enough. Statistical interpretation of entropy Boltzmanns Formula
  • Slide 3
  • Enthalpy and Spontaneous Rxns Early on in the development of thermodynamics, it was believed that if a reaction was exothermic, it was spontaneous. But this cant be the whole story---consider an ice cube on a warm day. Consider the following reaction: H 2 O(s) H 2 O(l) H rxn = +6.02 kJ Endothermic (ice feels cold) ..yet spontaneous (and it melts) !
  • Slide 4
  • Enthalpy and Spontaneous Rxns Consider the following problem: Mixing of a gas inside a bulb adiabatically (q = 0). q = 0, w = 0, E = 0, and H = 0 This process is NOT exothermic, but it still happens.
  • Slide 5
  • Statistical Interpretation of Entropy Imagine that we have a collection of 3 distinguishable particles who have a total energy of 3 . Lets ask the question, How will this fixed amount of energy distribute itself over the particles?
  • Slide 6
  • Statistics and Entropy (cont.) Our system consists of three distinguishable particles. There are three quanta of energy ( ) available for a total of energy of 3
  • Slide 7
  • First Arrangement: All on one The first possible arrangement we consider is one in which all energy resides on one particle There are three ways to do this
  • Slide 8
  • Second Arrangement: 2, 1, 0 Next arrangement: 2 on 1, 1 on another, and the third has 0 Six ways to do this
  • Slide 9
  • Third Arrangement The final possible arrangement is 1 on each particle. Only one way to do this.
  • Slide 10
  • Which Arrangement? Which arrangement is most probable? Ans: The arrangement which the greatest number of possibilities In this case: 2, 1, 0
  • Slide 11
  • The Dominant Configuration Configuration: a type of energy distribution. Microstate: a specific arrangement of energy corresponding to a configuration. Which configuration will you see? The one with the largest # of microstates. This is called the dominant configuration (Why does a rope tangle?)
  • Slide 12
  • Determining Weight Weight ( ): the number of microstates associated with a given configuration. We need to determine , without having to write down all the microstates. A = the number of particles in your system. a i is the number of particles with the same amount of energy. ! = factorial, and means take the product.
  • Slide 13
  • Determining Weight (cont.) Consider 300 students where 3 students have 1 of energy, and the other 297 have none. A = 300 a 1 = 3 a 0 = 297 = 4.5 x 10 6
  • Slide 14
  • Weight and Entropy The connection between weight ( ) and entropy (S) is given by Boltzmanns Formula: S = k(ln k = Boltzmanns constant = R/N a = 1.38 x 10 -23 J/K The dominant configuration will have the largest ; therefore, S is greatest for this configuration
  • Slide 15
  • Young Ludvig Boltzmann
  • Slide 16
  • A devoted father And husband. His wife called him My sweet, fat darling
  • Slide 17
  • Boltzmann at 58 Troubled by severe bouts of depression, and criticism of his scientific ideas
  • Slide 18
  • Boltzmann took his own life while on a family vacation in Switzerland.
  • Slide 19
  • Example: Crystal of CO Consider the depiction of crystalline CO. There are two possible arrangements for each CO molecule. Each arrangement of CO is possible. For a mole of CO: = N a !/(N a /2!) 2 = 2 Na
  • Slide 20
  • Example: Crystal of CO For a mole of CO: = N a !/(N a /2!) 2 = 2 Na Then, S = k ln( ) = k ln (2 Na ) = N a k ln(2) = R ln(2) = 5.64 J/mol.K
  • Slide 21
  • Another Example: Expansion What is S for the expansion of an ideal gas from V 1 to 2V 1 ? Focus on an individual particle. After expansion, each particle will have twice the number of positions available.
  • Slide 22
  • Expansion (cont.) Original Weight = Final Weight = Then S = S 2 -S 1 = k ln(2 ) - kln( ) = k ln(2 / ) = k ln(2)
  • Slide 23
  • Expansion (cont.) Therefore, the S per particle = k ln (2) For a mole of particles: S = k ln (2 Na ) = N a k ln(2) = R ln(2) = 5.64 J/mol.K
  • Slide 24
  • Expansion (cont.) Note in the previous example that weight was directly proportional to volume. Generalizing: S = k ln ( final ) - kln( initial ) = k ln( final / initial ) = Nk ln( final / initial ) for N molec. = Nkln(V final /V initial )

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