Chemstry 391 Fall 2007 Exam I KEY (Monday September 17) 1. (25 Ponts) ***Do 5 out of 6***(If 6 are done only the frst 5 wll be graded)*** a). Defne the terms: open system, closed system and solated system n terms of flows between a system and ts surroundngs. Open: matter can transfer between system and surroundngs. Closed: matter cannot transfer between system and surroundngs. Isolated: no transfer of anythng between system and surroundngs. b). Classfy the energy and the square of the pressure as ntensve, extensve or nether: Explan your reasonng. Energy s ntensve snce t s proportonal to the number of atoms n the materal. The pressure s ndependent of the system sze so t s ntensve. Takng ts square does not change ths feature. c). What s the condton for thermal equlbrum between ten thousand systems? The temperatures of all ten thousand systems must have the same value. 1
d). Defne mathematcally the relatonshp between an exact dfferental and a state functon wth two ndependent varables. In order for a functon f(x,y) to be a state functon, t must be possble to wrte the total f f dfferental df n the form df = dx + dy x y y. If the form df as wrtten exsts, t s x an exact dfferental. e). Two deal gas systems undergo reversble expansons startng from the same P and. At the end of the expansons, the two systems have the same volume. The pressure n the system that has undergone an adabatc expanson s lower than that n the system that has undergone an sothermal expanson. Explan ths result wthout usng equatons. In the system undergong adabatc expanson, all the work done must come through the lowerng of ΔU, and therefore of the temperature. By contrast, some of the work done n the sothermal expanson can come at the expense of the heat that has flowed across the boundary between the system and surroundngs. f). What s wrong wth the followng statement (that often appears n newspapers)?: Because the well nsulated house stored a lot of heat, the temperature ddn't fall much when the furnace faled. Rewrte the sentence to convey the same nformaton n a correct way. Heat cannot be stored t s not a state functon. Somethng lke the followng s safe. Because the house was well nsulated, the walls were nearly adabatc. Therefore, the temperature of the house dd not fall as rapdly when n contact wth the surroundngs at a lower temperature as would have been the case f the walls were dathermal. 2
2. (26 ponts) The nternal energy U s a functon of the ndependent varables T and, U = f ( T, ). As a consequence, du = dt + d. The enthalpy H = U + P. T A. Develop (all detals shown) an expresson for. P T Consderng P fxed and dvdng by dt for fxed P yelds, snce = 1, T P U U U = + T P T P B. Usng A and the defnton of the heat capactes C = develop an expresson for CP C. H and C P = P C P ( U P) H + C = = = + P T P P P P Substtute n from A to get CP C = + + P T T T P P = + P = + P T T T T P P T P 3
3. (25 ponts) For ammona measurement gves = T 3 840J/m mole at 300 K and = 37.3J/K mole. Estmate the change n molar energy of ammona when t s heated through 2 K and compressed through 100 cm 3. (Hnt. Use the general expresson for du n queston 2 and replace dfferentals by small but fnte changes). From du = dt + d and assumng over small T and ranges that the T coeffcents are constants, then Δ U = Δ T + Δ. T Pluggng n the data, and notng that n a compresson Δ < 0 and for heatng Δ T > 0 : 3 ( ) 3 2 3 ( 37.3J/Kmole)( 2 K) ( 840J/m mole) 10 cm ( m/100cm) Δ U = + = 74.52 J/mole 4
4. (24 ponts) For the followng processes explan whether each of the quanttes q, w, postve, zero, negatve or ambguous. Δ U and Δ H s a) Reversble adabatc expanson of an deal gas. q=0 snce adabatc. w negatve for an expanson snce w= pd and d s postve for expanson. Snce Δ U = q+ w Δ U s negatve. H U ( P) U nr T Δ =Δ +Δ =Δ + Δ, for deal gas. Also, du = C dt. Snce du < 0 and C > 0 then dt < 0 and Δ T s negatve. Therefore, wth Δ U negatve, ΔH s negatve. b) Reversble heatng of a gas at constant pressure, p. q s postve snce the gas s heated. From the heat capacty defnton, dqp = CPdT, and Cp>0; so dt >0. Then Δ T >0. From du = C dt and C > 0 ΔU s postve. The deal gas equaton of state p Then, snce p s fxed, w = pδ s negatve. = nrt and the fact that Δ T >0 then gves Δ > 0. From Δ H =Δ U +Δ ( P) =Δ U + nrδ T and ΔU and Δ T are postve; Δ H s postve. Or, Δ H =Δ U +Δ ( P) =Δ U + pδ and Δ U and pδ are postve, so Δ H s postve. 5
Mathematcal expressons: n ( n+ 1) xdx x n n = /( + 1); = 0,1,2,... 1 dx = ln x x ( n 1) ( ) n x dx= x / n 1 ; n= 2,3,... x y z Euler: = 1 y zx xy z x y Inverter: = 1 y x z z 6
n dμ + SdT dp = 0 (Gbbs-Duhem) νμ = 0 (reacton equlbrum). μ = μ + RTln a ; a = γ x; 0 7