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Lecture 19

Chem 402 Lecture 19: L19 3:3:17 Premium

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School
Washington University in St. Louis
Department
University College - Chemistry
Course
University College - Chemistry Chem 402
Professor
Barnes Alexander
Semester
Spring

Description
3 March 2017 L19: Entropy and Fundamental Equations I. Entropy and Fundamental Equations A. Entropy Calculation 1. In General a. ∆S uni ∆S sys ∆S surr b. For a spontaneous process  ∆S uni 0, ∆Ssys+ ∆Ssurr 0, , ∆sys> -∆Ssurr c. You can only calculate ∆S for a reversible process! d. **IN ALL CASES, WE MUST FIND A REVERSIBLE PATH TO ð𝑞𝑟𝑒𝑣 CALCULATE ∫ 𝑇 ** 2. Entropy of Mixing: QM/Classical Method 𝑁 𝑁 𝑞 𝑞 a. ∆S mix= 𝑘𝑙𝑛Ω𝑓− 𝑘𝑙𝑛Ω =𝑖𝑘𝑙𝑛( ) −𝑁!𝑙𝑓( ) 𝑁! 𝑖 𝑉 b. For a monoatomic ideal gas, 𝑞 = 𝑣(volume of gas / voxels) or just use the 3/2 translational partition function 𝑞 = 𝑉 ( 2𝜋𝑚𝑘𝑇 ) 𝑡𝑟𝑎𝑛𝑠 ℎ2 3. Entropy of Mixing: ideal gases at constant T and P a. ∆S demix -∆S mixsince entropy (S) is a function of state b. For demixing process, ∆U = 0, qrev= -wrev= pAdV A p dB (wBrk of compression for each gas) c. ∆S demix ∫𝑑𝑞𝑟𝑒𝑣= ∫𝑉𝐴 𝑝𝐴𝑑𝑉𝐴 + ∫𝑉𝐵 𝑝𝐵𝑑𝑉𝐵 = ∫ 𝑉𝐴𝑛𝐴𝑅𝑇𝑑𝑉 𝐴+ ∫ 𝑉𝐵𝑛𝐵𝑅𝑇𝑑𝑉 𝐵 = 𝑇 𝑉 𝑇 𝑉 𝑇 𝑉 𝑇𝑉 𝑉 𝑉𝑇 𝑉𝐴 𝑉𝐵 𝑛𝐴𝑅𝑙𝑛( )𝑉+ 𝑛 𝑅𝑙𝐵( ) 𝑉 • Since this is an ideal gas, sub in pV = nRT • Then we want to get this in terms of mole fractions 𝑉𝐴 𝑛𝐴 𝑉𝐵 𝑛𝐵 • For an ideal gas, 𝐴 = = ,𝑋𝐵= = , so we multiply through by n/n 𝑉 𝑛 𝑉 𝑛 d. ∆𝑆 𝑚𝑖𝑥= −𝑛𝑅[𝑋 𝑙𝑛𝐴 + 𝐴 𝑙𝑛𝑋 𝐵 𝐵 • Notice that ∆S = -∆S , also that this is the same result from the QM mix demix derivation • The negative in front ensures that the entropy of mixing is always positive • Since X And X aBe always < 1, ∆S mix> 0, mixing is always spontaneous 4. Heating (or cooling) at constant V a. ∆S = ð𝑞𝑟𝑒𝑣= 𝑇2𝐶𝑣𝑑𝑇(if C is T independent, 𝐶 ln⁡( )) ∫ 𝑇 ∫𝑇1 𝑇 v 𝑣 𝑇1 5. Reversible Phase Change at Constant T and P a. ∆H vap= qp 𝑞𝑣𝑎𝑝 ∆𝐻 𝑣𝑎𝑝 b. ∆S vap(100°C) = 𝑝 = (where Tbis boiling point at 1 bar) 𝑇 𝑏 𝑇𝑏 6. Irreversible Phase Change at constant T and p a. Spontaneous and irreversible, so we must find a reversible path between the two states to calculate ∆S • Use a thermocycler to bring to a fusion or vaporization temp, then use 𝑟𝑒𝑣 ⁡𝑞𝑝 = -∆H fusnd then bring the temp back to the original one • Works because ∆S is a state functions nd B. More on the 2 Law 1. dS is greater than the differential heat for an irreversible process divided by T 𝑖𝑟𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒 a. For the following closed path: State 1 → ⁡State 2 → State 1 𝑑𝑞𝑟𝑒𝑣 b. For a closed path, ∮ 𝑇 ≤ 0 • Where dq is an inexact differential, and the inequality holds for this irreversible closed path:∮,𝑑𝑞𝑖𝑟𝑟𝑒< 0 𝑇 c. Although we have a reversible step in the closed path, the ONE irreversible path makes the entire closed path irreversible d. By itself, 𝑑𝑞𝑟𝑒𝑣 is inexact, but𝑞𝑟𝑒𝑣is exact because it is a state function 𝑇 • Since there are many irreversible paths that exist (unspecified!) and all 𝑑𝑞𝑟𝑒𝑣 reversible paths return the same value (entropy change), 𝑇 is exact 2𝑑𝑞𝑖𝑟𝑟𝑒𝑣 1𝑑𝑞𝑟𝑒𝑣 e. For this closed path:∫1 𝑇 + ∫2 𝑇 < 0 2 𝑑𝑞𝑖𝑟𝑟𝑒𝑣 2𝑑𝑞𝑟𝑒𝑣 • Change integration limits and sign of reversible ste∫1 − ∫1 < 0 2 𝑑𝑞𝑖𝑟𝑟𝑒𝑣 2 𝑑𝑞𝑟𝑒𝑣 𝑑𝑞𝑟𝑒𝑣 𝑑𝑞𝑖𝑟𝑟𝑒𝑣 𝑇 𝑇 f. ∫1 < ∫1 and = 𝑑𝑆, thus 𝑑𝑆 >⁡ 𝑇
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