Chemistry 163B q rev, Clausius Inequality and calculating ΔS for ideal gas P,V, changes (HW#6) Challenged Penmanship Notes 1
statements of the Second Law of hermodynamics 1. Macroscopic properties of an isolated system eventually assume constant values (e.g. pressure in two bulbs of gas becomes constant; two block of metal reach same ) [Andrews. p37] 2. It is impossible to construct a device that operates in cycles and that converts heat into work without producing some other change in the surroundings. Kelvin s Statement [Raff p 157]; Carnot Cycle 3. It is impossible to have a natural process which produces no other effect than absorption of heat from a colder body and discharge of heat to a warmer body. Clausius s Statement, refrigerator 4. In the neighborhood of any prescribed initial state there are states which cannot be reached by any adiabatic process ~ Caratheodory s statement [Andrews p. 58] 2
four steps to exactitude wtotal L ql I. CARNO [ ideal gas] 1 1 q q U U U II. ANY REVERSIBLE ' WO EMPERAURE ' MACHINE CARNO [ ideal gas] III. cycle dq rev 0 IV. cycle dq rev 0 3
Entropy ds dq rev 4
goals of lecture 1. Relate ΔS and q irrev 2. Calculate ΔS for P,V, changes of ideal gas (HW#6) a. using REVERSIBLE path (q rev ) [even for irreversible processes] b. using partial derivatives of S with respect to P, V, [ a look ahead] 5
entropy and heat for actual (irreversible processes): q irrev an irreversible (actual) cyclic engine ε irrev coupled with a Carnot heat pump of ε C will not violate 2 nd Law if ε irrev < ε C (viz section, HO#25 SL 29;HO #28) ΔU cyclic =0 -w total =q U +q L for both rev and irrev irrev ql irrev L qu irrev U q q L L BU what about q irrev with ε irrev < ε C?? w total qu q q L L irrev L 1 1 qu qu qu U irrev U irrev irrev irrev irrev 0 U S cyclic engine ( reversible or irreversible ) qu rev ql rev Scyclic ( ) engine 0 ( ) ( reversible or irreversible ) U L reversible dqrev dq dq irrev ds ds ds 6
2 nd Law of hermodynamics in terms of entropy S is a SAE FUNCION x dq rev S rev irrev dq irrev E&R eqn 5.33 Clausius inequality 7
Lecture 3: Pressure-Volume work reversible isothermal expansion; P ext =P int P=10 atm isothermal P=P int =1 10 atm 1atm 9atm expansion 9atm 1atm int V final 1mole 300K 10 atm V 1 2 1 1mole 300K 1 atm V 2 q rev =-w =+5.743 kj irrev 8
Lecture 3: Isothermal expansion: P ext = const ideal gas (irreversible) P=10 atm isothermal =0 P=1 atm 1atm expansion ΔV 1atm 9atm 9atm 1mole 300K 10 atm V 1 1atm final initial Pext dv 2 1 1mole 300K 1 atm V 2 q irrev =-w=+2.244 kj 9
EXAMPLE from early lectures: isothermal expansion initial final same initial and final dq q S P ext = P int ; q = 5743 J rev q 19. 14 19.14? J K -1 P ext = const 1 atm; q irrev = 2244 J q 7. 48 19.14? J K -1 S final initial some reversible path dq rv e 10
S universe 0 S S 0 system surroundings universe disorder increases 11
calculating entropy (see summary on review handout) 12
dqrev look ahead - ΔS for changes in,v; (always S ) also: S(, V ) : S S ds d dv V V final initial coming very soon S ncv S P V V V so: nc P V ds d dv V ideal gas always (no w other, closed system) nc nr nc nr V V V V ds d dv S d dv rev const V path rev const path final Vfinal S ncv ln nr ln E&R eqn 5.18 initial Vinitial q rev vary const V path q rev V vary const path 13
dqrev look ahead- ΔS for changes in,p; (always S ) also: S(, P) : S S ds d dp P P final initial coming very soon S ncp S V P P P so: nc V P ds d dp P ideal gas always (no w other, closed system) nc nr nc nr P P P P ds d dp S d dp rev const P path rev const path final Pfinal S ncp ln nr ln E&R eqn 5.19 initial Pinitial q rev vary const P path q rev vary P const path 14
End of Lecture 15
Return of Midterm #1 can t wait? pick up exam from A in section or during office hours ianyu Secs B and E FRON OF ROOM Gabe Secs A and D 16
midterm#1 return: the scene no grades on individual midterms (DON ASK!!) class continues to do well, near class average B- As and Amy have keys- speak with them about any questions on exam notes on exams from E.S. ( me) Let me know who you are - props for good scores Need to do better - I m happy to meet with you - can always do better, and we do care about you too!! 17
Midterm #1 Winter 2014 total pts for qtr (115 + 138 + 162 + 46) = 115+ 347 = 464 18
MIDERM #2 MIDERM #2 MIDERM #1 MIDERM #1 MIDERM #2 MIDERM #2 FINAL MIDERM #1 FINAL FINAL FINAL MIDERM #1 MIDERM #1 A CAUIONARY ALE? 78%? SUDY Winter 2014 Winter 2013 Winter 2010 19
Return of Midterm #1 can t wait? pick up exam from A in section or during office hours ianyu Secs B and E FRON OF ROOM Gabe Secs A and D 20