Chem 402 Lecture 5: L5 1:27:17

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Washington University in St. Louis
University College - Chemistry
University College - Chemistry Chem 402
Barnes Alexander

25 January 2017 L5: When Terms Drop Out, Boltzman Statistics, and QM vs. Classical I. Thermodynamic Variable Derivations from Q A. Derivation of the Ideal Gas Law + Insights! 1. 𝑑𝐴 = βˆ’π‘†π‘‘π‘‡ βˆ’ 𝑝𝑑𝑉 + 𝑑𝑁, ( ) 𝑑𝐴 = βˆ’π‘, A = -kTlnQ, Q = q /N! 𝑑𝑉 𝑇,𝑁 πœ•π‘™π‘›(π‘ž /𝑁!) πœ•π‘™π‘›(π‘ž )βˆ’π‘™π‘›π‘!) 2. 𝑃 = π‘˜π‘‡( πœ•π‘‰ ) = π‘˜π‘‡( πœ•π‘‰ ) 𝑇,𝑁 𝑇,𝑁 a. After separating the ln term, the N power goes in front of the ln πœ• 2πœ‹π‘šπ‘˜π‘‡ 3/2 3. 𝑃 = π‘π‘˜π‘‡ πœ•π‘‰ (ln π‘ž βˆ’ ln⁑(𝑁!))𝑇,𝑁, then sub in π‘‘π‘Ÿπ‘Žπ‘›π‘ π‘‰,𝑇) = 𝑉 βˆ— ( β„Ž2 ) a. kT is important because it represents the energy of molecules hitting walls b. The N can drop out in front because pressure scales with the number of molecules hitting the wall 3 πœ• 2πœ‹π‘šπ‘˜π‘‡ 2 4. 𝑃 = π‘π‘˜π‘‡ πœ•π‘‰ (ln(𝑉 βˆ— ( β„Ž2 ) ) βˆ’ ln 𝑁! ) 𝑇,𝑁 πœ• 2πœ‹π‘šπ‘˜π‘‡ 3 πœ•ln⁑(𝑉) 1 = π‘π‘˜π‘‡ (ln((𝑉) + ln⁑( ) ) βˆ’ ln⁑(𝑁!)) = π‘π‘˜π‘‡( ) = π‘π‘˜π‘‡( ) πœ•π‘‰ β„Ž2 𝑇,𝑁 πœ•π‘‰ 𝑇,𝑁 𝑉 a. After you sub in transV, T), separate the ln term b. (lnN!) drops out because it doesn’t matter how many distinguishable particles you have with respect to volume β€’ It doesn’t matter if they are distinguishable for pressure β€’ Pressure is just molecules hitting walls, so this term goes to zero 2πœ‹π‘šπ‘˜π‘‡ c. The 𝑙𝑛( )3/2 term will drop out because it goes to zero with respect to V β„Ž2 β€’ Also, kT and N were already accounted for once, so this cancels d. Then we are left with P = kT (d/dV of lnV), which is just = 1/V β€’ Makes sense since as volume decreases pressure must increase (physical insight) 1 𝑅 1 5. 𝑃 = π‘π‘˜π‘‡( ) 𝑉 𝑁( )𝑇( )𝑁 PV𝑉= nRNT 𝑛 a. For n moles, N =nN And k = R/N = A = R/(N/n), gives us the ideal gas law! B. Calculating U = from the Translational Degree of Freedom 1. 𝑒 =⁑ 𝐸 ⁑= π‘˜π‘‡ 2 πœ• (𝑙𝑛𝑄) = π‘˜π‘‡ 2 πœ• (𝑙𝑛(q /N!) ) = π‘˜π‘‡ 2 πœ• (𝑙𝑛(q ) βˆ’ πœ•π‘‡ 𝑉,𝑁 πœ•π‘‡ 𝑉,𝑁 πœ•π‘‡ ln⁑(N!) = π‘˜π‘‡ 2 πœ• (𝑁𝑙𝑛(q) βˆ’ ln⁑(N!)) 𝑉,𝑁 πœ•π‘‡ 𝑉,𝑁 a. Sub in U and then Q = q /N! and separate out the ln 3 2 πœ• 2πœ‹π‘šπ‘˜π‘‡ 2 2 πœ• 3/2 2. = π‘˜π‘‡ πœ•π‘‡ (𝑁𝑙𝑛(𝑉 βˆ— ( β„Ž2 ) ) βˆ’ ln⁑(N!)) = π‘˜π‘‡ πœ•π‘‡(𝑁𝑙𝑛(𝑇 ) + 𝑉,𝑁 3 (𝑁𝑙𝑛( 2πœ‹π‘šπ‘˜ ) βˆ— V) βˆ’ ln⁑(N!)) = π‘˜π‘‡ 2 πœ• ( 𝑁𝑙𝑛𝑇) = π‘π‘˜π‘‡ 2πœ• (𝑙𝑛𝑇) β„Ž 2 πœ•π‘‡ 2 𝑉,𝑁 2 πœ•π‘‡ 𝑉,𝑁 𝑉,𝑁 a. Sub in qtransthen separate out T 3 2πœ‹π‘šπ‘˜ 2 b. The (𝑁𝑙𝑛( β„Ž2 ) βˆ— V) term cancels to zero because this is a constant and will not change with respect to T c. lnN! Also goes to zero since it doesn’t matter if the particles are distinguishable when trying to calculate energy d. The (3/2) on T then comes down (ln rules) and can be pulled out since it is a constant, N is also constant here and can be pulled out 3. 3 π‘π‘˜π‘‡ 2 πœ• (𝑙𝑛𝑇 ) = π‘π‘˜π‘‡ 2 1 𝐸 ⁑= π‘π‘˜π‘‡ = 𝑛𝑅𝑇 3 2 πœ•π‘‡ 𝑉,𝑁 2 𝑇 2 2 a. First derivative of lnT is (1/T) b. For n moles (N = nN ) and k = R/N , gives us the for a monoatomic ideal A A gas Μ…Μ…Μ…Μ… 3 c. In molar quantity: 𝐸 ⁑= 𝑅𝑇 2 C. Specific Heat Capacity at Constant Volume (C ) v 1. Since C pf water = 4.184 J/Β°Cβ€’g, essentially energy over temp 2. Molar C (vor a monoatomic gas in the electronic ground state) πœ•( 𝑅𝑇) a. 𝐢 = ( ) πœ•π‘ˆ = ( 2 ) = 𝑅3
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