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 system) du is exact differential Uisastatefunction 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 3 1
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 4 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 and at =constant and dh H dq q as advertised!! dw 0 H H H q 0 0 at const no w endmic (heat gained by system) exmic (heat lost by system) 5
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 6 manipulating thermodynamic functions: fun and games for example: HW# 1. Derive the following for any closed system, with only - work: C U U 7 3
total differential for U(,,n) and H(,,n) Un (,,, n,..., n ) 1 N N U U U du d d dni n Hn (,,, n,..., n ) 1, n, n i1 i,, n n N N H H H dh d d dni n n, n, i1 i,, n n j j i i for now closed system all dn i =0 8 H(,): some manipulations and relationships (closed system) dh dq d closed system, dw 0 'divide by d, holding constant' H dq d H dq _ nc d math handout #6 'divide by d, holding constant' H dq d H dq d H H dh d d H dh ncd d eqns. 3.30-3.3 E&R (p. 56 [5] ) nd 9 4
U(,): some manipulations and relationships (closed system) du dq d closed system, dw 0 'divide by d, holding constant' U dq d U dq _ nc d 'divide by d, holding constant' U dq d U dq d U U du d d U du nc d d eqn. 3.1-3.15 E&R (p 50 [46] ) nd 10 save for later when we have tools from nd Law of hermodynamics U du nc d d need 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 HW #13* 11 5
some important relationships between C and C 1 _ 1 _ H U 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 1 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 13 6
let s finish C vs C (very general relationship) ~E&R 3.37 U nc 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 14 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 15 7
experimental C and C for selected gasses R=8.31 J mol -1 K -1 ideal gas C C R 3 monatomic C R 5 diatomic C R 1 1 J mol K 3 R 1.47 5 R 0.78 7 R 9.10 able from: http://www.scribd.com/doc/33638936/ncer-book-hysics-class-xi- 16 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!! 17 8