Chem. 412 – Phys. Chem. I T e m p era tu re P h a se T ran sitio n s H e a t E n g in es Z e ro E n tro p y fo r P e rfe c t C ry sta ls D o w n to Z e ro K 3 rd L a w E ntro p ies 3 rd L a w o f T herm odynam ics E n tro p y o f th e U n ive rse C a rn o t C ycles H e a t P um ps S u p e rc o o le d C alcu latio n s R e ve rsib le vs . Irre ve rsib le V olum e P re s su re E n tro p y Changes 2 n d L a w o f T herm odynam ics E n tro p y/D iso rd er D isb u rs e m e n t o f E nergy E n tro p y - 2 n d & 3 rd L a w s o f T herm odynam ics
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Chem. 412 – Phys. Chem. I. Spontaneous Processes Mixing of gases Heat flow from hot to cold (Most) macroscopic events are irreversible Key Sign of Spontaneity:
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Spontaneous Mixing of GasesSpontaneous Mixing of Gases
• Driving force due to Entropy• Compare Interdiffusion of gases to playing
“Bridge” (Cards): Chance of getting 13 cards of same suit (after proper shuffles) is <<<<<<<< Chance of getting some mixture of cards.
• S = -k pi ln pi (via Statistical TD)
cardsforx
ob 521035.6
1.Pr
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3rd Law of Thermodynamics3rd Law of Thermodynamics So = 0 for any physical or chemical change involving perfect crystals at absolute zero.
• 3rd Law Entropies:
• For Irreversible Processes:
)(lnTdCT
dTCS
PP
0 surro
syso
univo SSS
Supercool ExampleSupercool ExampleAt constant pressure of 1 atm, calculate ΔSsys , ΔSsurr and ΔSuniv upon the sudden freezing of 1 mole of H2O at -10.00°C. (sudden freezing of supercooled water)
Given: CP(ℓ) = 75.3 J K-1 mol-1
CP(s) = 36.9 J K-1 mol-1
ΔHf° = 5950. J mol-1 (s → ℓ)
Sketch a Hess’ Law diagram replacing the irreversible process with three reversible steps.
Calculate ΔSsys [ ΔS(H2O) ]
Calculate ΔSsurr [ Need to find ΔHsurr at 263.15 K ]