Chemistry 163B. Lecture 5 Winter Challenged Penmanship. Notes

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1 Chemistry 163B Lecture 5 Winter 2014 Challenged enmanship Notes 1

2 total differential (math handout #4; E&R ch. 3) infinitesimal change in value of state function (well behaved function) total change in f f x, y a well behaved function f f df dx dy x y y x change in f per unit change in x (along x direction) amount of change in x + change in f per unit change in y (along y direction) amount of change in y 2

3 differential of product (product rule) d( xy) ydx xdy 3

4 example of implication of total differentials First Law du = dq + dw + dn sys sys sys sys (n=number of moles; dn=0 for closed system) U is state function du is exact differential sys dn 0 (closed system) math first law U U du (, ) d d dq sys + dw OR math first law U U du (, ) d d dq sys +dw sys sys 4

5 divide through by?? math handout #6 U U du (, ) d d divide through by d holding x (something else) constant U U U X X X later special simplification if x= or 5

6 two relationships for ideal gasses: a (msec) look ahead (will prove rigorously in next lecture, but gthis is the next lecture) for any substance du dq nc d and U = nc d for a constant volume process but for an ideal gas du nc d and U nc for ANY path (not only constant process) [other parts of path, changes of and with constant, give zero contribution to U] for ideal gas C C R monatomic ideal gas 3 5 C R C R 2 2 [simple proof coming soon] 6

7 ideal gas U nc for ANY path (not only constant process) du dq nc d and U = nc d for a constant volume process du nc d and U nc for ANY path (not only constant process) (general, w other =0, dn=0) U U du d d du dq - d U du d dq nc d U nc U du nc d d ideal gas U but du nc d d U 0 even if not constant (i.e. any path) du nc d U nc U along general path where both (const ) and (const ) vary U nc nada 7

8 C = C +nr for ideal gas for only - work and closed system ( dw du dq d du dq ext _ =n C d dq U d nc du dq d other p p dq du d p p 0, dn=0) dq U nc d for monatomic ideal gas 3 nr and U nr 2 U 3 U nr 2 3 nr 5 nc nr nr 2 2 8

9 1 st Law recapitulation U internal energy du = dq + dw + dn sys sys sys sys sys surr du du energy conserved (n=number of moles; dn=0 for closed system) du is exact differential U is a state function completely general for only - work and closed system (dn=0) du dq d ext dq Constant volume process du = ΔU = dw Adiabatic process du = ΔU = q w 9

10 enthalpy: q for process at constant ressure HªU+ int (definition of enthalpy, H) since U is state function and, are state variables, H is also a SAE FUNCION why a new state function you might ask?? du dq ; U q heat at constant volume but most reactions and many physical processes are carried out at constant completely general desire state function for q, heat at constant pressure 10

11 enthaply: H, a state function for heat transfer at constant pressure H U int dh du d d dh dq d dw d d dh dq d dw other other and at =constant and dh H dq q as advertised!! dw other 0 H H H q 0 0 at const no w other endothermic (heat gained by system) exothermic (heat lost by system) 11

12 H ideal gas H q p ncd nc (general, w other =0, dn=0) ideal gas H U U nr dh du nrd (general for ideal gas) dh nc d nrd v dh n( C R) d dh nc d v (general for idel gas, even not const) IDEAL GAS ANYIME, EEN IF NO CONSAN H= ideal gas general (w =0, dn=0) nc other 12

13 manipulating thermodynamic functions: fun and games for example: HW#3 12. Derive the following for any closed system, with only - work: C U U 13

14 total differential for U(,,n) and H(,,n) Un (,,, n,..., n ) 1 2 N N U U U du d d dn n Hn (,,, n,..., n ) 1 2, n, n i1 i,, n n N N H H H dh d d dn n n, n, i1 i,, n n j j i i i i for now closed system all dn i =0 14

15 H(,): some manipulations and relationships (closed system) dh dq d closed system, dw other 0 'divide by d, holding constant' H dq d H dq d _ nc math handout #6 'divide by d, holding constant' H dq d H dq d H H dh d d H dh ncd d eqns E&R (p. 56 [52] ) 2nd 15

16 U(,): some manipulations and relationships (closed system) du dq d closed system, dw 0 'divide by d, holding constant' U dq d U dq d _ nc 'divide by d, holding constant' U dq d U dq d other U U du d d U du nc d d eqn E&R (p 50 [46] ) 2nd 16

17 save for later when we have tools from 2 nd Law of hermodynamics U du nc d d need 2 nd Law to evaluate this in terms of,, U E& R eqn.3.19 many of the results in E&R ch 3 use this [yet] unproven result; we will derive later class should use result in HW3 #13* 17

18 some important relationships between C and C _ 1H 1U C and C n n _ to get relationship between C and C one needs to have relationship involving both H and U; soooo H U dh du d d 18

19 continuing with relating C and C H U dh du d d divide by d, constant H U nc U now to get U U du nc d d U U nc 19

20 let s finish C vs C (very general relationship) ~E&R 3.37 nc U U nc nc U nc nc volume change per change of 1º energy to raise 1º const (vol changes) energy to raise 1º const potential energy as molecules separate per unit volume change energy lost as - work per unit volume change 20

21 C vs C for ideal gas U nc nc for ideal gas nr Energy, U is function of ONLY, U() U nr 0 nr nc nc 0 nc nc nr C C R for ideal gas 21

22 experimental C and C for selected gasses R=8.31 J mol -1 K -1 ideal gas C C R 3 monatomic C R 2 5 diatomic C R J mol K 3 R R R able from: 22

23 in section derive equation following equation H nc nc start with du dh d d divide by d with constant and then boogie along as we just did!! 23

24 First Law: ideal gas calculations relationships that apply to ideal gasses for all conditions with w other =0 and constant composition (some also apply more generally): U qw w extd nr q? n C d nc q? n C d nc C C R H U monatomic ideal gas U nc H nc any conditions C 3 2 R any conditions C 5 2 R 24

Chemistry 163B Winter 2013 notes for lecture 4

Chemistry 163B Winter 2013 notes for lecture 4 Chemistry 163B Lecture 4 Winter 013 Challenged enmanship Notes 1 st Law recapitulation U internal energy du = dq + dw + dn sys sys sys sys sys surr du du energy conserved (n=number of moles; dn=0 for closed