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Chem 402 Lecture 19: L19 3:3:17
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Washington University in St. Louis

University College - Chemistry

University College - Chemistry Chem 402

Barnes Alexander

Spring

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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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