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Chem 2 - The Second Law of Termodynamics: Entropy Microstates and the Boltzmann Equation II

Chem 2 - The Second Law of Termodynamics: Entropy Microstates and the Boltzmann Equation II

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Chem 2 - The Second Law of Termodynamics: Entropy Microstates and the Boltzmann Equation II

  1. 1. The Second Law of Thermodynamics: Entropy, Microstates, and the Boltzmann Equation (Pt. 2) By Shawn P. Shields, Ph.D. This work is licensed by Shawn P. Shields-Maxwell under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
  2. 2. The Second Law of Thermodynamics The total entropy change for any spontaneous process is positive (+S) This increase in entropy involves an evolution from less probable configurations to more probable.
  3. 3. Entropy and Microstates Entropy is related to the number of possible arrangements or “microstates.” Microstates are instantaneous properties of the system. Examples include instantaneous velocities and positions of molecules (atoms, etc.)
  4. 4. Entropy and Microstates The number of microstates can be increased by Increasing the volume Increasing the temperature How?
  5. 5. Example 1: Counting Microstates Suppose we have four distinguishable particles in one side of a two-compartment container. How many possible ways are there to distribute them between the two sides?
  6. 6. The Possible Microstates for a Sample of Four Gas Molecules in Two Bulbs of Equal Volume" from http://2012books.lardbucket.org/books/principles-of- general-chemistry-v1.0m/s22-03-the-second-law-of-thermodynami.html Microstates Highest probability to observe state III (Largest number of microstates)
  7. 7. Boltzmann's Equation Relates the Microscopic to the Macroscopic The relationship between the number of microstates and the entropy is S = 𝑘 𝐵 ln Ω Where  is the number of microstates and kB (Boltzmann constant) = 1.38  1023 J/K
  8. 8. Boltzmann's Equation and S The change in entropy for a process can be calculated as ΔS = 𝑘 𝐵 ln Ω2 Ω1 Where 1 is the initial number of microstates and 2 is the final number of microstates (after the process)
  9. 9. Example 2: Calculating the Number of Microstates Four distinguishable particles are initially sealed in the right side of a two- compartment container. Suppose the compartment is opened, and the particles are allowed to distribute throughout both compartments. How many microstates are there initially, and finally? Calculate S for the process.
  10. 10. Example 2: Calculating the Number of Microstates and S How many microstates are there initially? There is only one possible arrangement for all 4 particles in the right compartment. Edited “The Possible Microstates for a Sample of Four Gas Molecules in Two Bulbs of Equal Volume" from http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0m/s18-06-reaction-rates-a-microscopic-v.html BE SURE TO CHANGE LINK!!!!!!!!
  11. 11. Example 2 (continued): Calculating the Final Number of Microstates How many microstates are there in the final state? The volume was doubled: 2# 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 = Ω 𝑓 Ω 𝑓 = 24 = 16 This is the same answer… 16 microstates (or possible arrangements).
  12. 12. Example 2 (continued): Calculating S Calculate S for this process Δ𝑆 = 𝑘 𝐵 ln Ω2 Ω1 ∆𝑆 = ln 1.38 × 10−23 ln 16 1 = 3.83 × 10−23 J The entropy increased 3.8310-23 J when we doubled the volume.
  13. 13. Next up… Predicting Entropy Changes Qualitatively, plus Examples (Pt 3)

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Chem 2 - The Second Law of Termodynamics: Entropy Microstates and the Boltzmann Equation II

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