Thermodynamics of Materials

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Thermdynamcs f Materals 14th Lecture 007. 4. 8 (Mnday) FUGACITY dg = Vd SdT dg = Vd at cnstant T Fr an deal gas dg = (RT/)d = RT dln Ths s true fr deal gases nly, but t wuld be nce t have a smlar frm fr real fluds. dg = RT dln ƒ where ƒ s the fugacty φ = ƒ/ 1 as 0 φ: the fugacty ceffcent, ƒ = φ Fugacty may be thught f as a thermdynamc pressure; t has unts f pressure.

NON-IDEAL MIXTURES OF NON- IDEAL GASES Fr a perfect gas mxture: μ = μ + RT ln = μ + RT ln + RT ln X T Fr an deal mxture f real gases: μ = μ + RT lnf = μ + RT lnf + RT ln X f = Xf = XφT Fr a real mxture f real gases: μ = μ + RT lnf f = X χ f = X χφ T Lews Fugacty Rule Crrectn fr nn-deal mxng Crrectn fr nn-deal gas equl. vapr pressure f pure A = p equl. vapr pressure f pure B = p mxture X, X A B What wuld be the equl. vapr pressure f A and B fr the mxture? A B p = X p A A A p = X p B B B Rault' s law applcable when the magntudes f the A-A, B-B, and A-B bnd energes n the slutn are dentcal. nrmally applcable when the cncentratn s hgh

RAOULT S LAW Rault (1887) studed vapr-lqud systems n whch tw r mre lqud cmpnents were mxed n knwn prprtns and the lqud was equlbrated wth ts wn vapr. The cmpstn f the vapr was then determned. The ttal vapr pressure f the system was lw, s the vapr behaved deally and cnfrmed t Daltn s law. In such systems, the partal pressures f the gaseus cmpnents were fund t be a lnear functn f the ther mle fractn n the lqud. Fr a bnary system A-B, A = X A A º and B = X B B º A 10 8 ttal 6 B 4 0 0.0 0. 0.4 0.6 0.8 1.0 A X B B

Ideal slutn Hw abut nn-deal slutn? case 1) E > E and E > E A-B A-A A-B B-B case ) E < E and E < E A-B A-A A-B B-B What wuld be the equl. vapr pressure f A and B? case 1) k A < 1and kb < 1 negatve devatn case ) k > 1and k > 1 pstve devatn A Henry' s law B p = X p, p = X p A A A B B B p A= kax A, p B= kbxb applcable t dlute slutn HENRY S LAW Henry (1803) was studyng the slublty f gases n lquds. He fund that the amunt f gas dsslved n a lqud n cntact wth t was drectly prprtnal t the pressure n the gas,.e., = K X K s a cnstant called the Henry s Law cnstant. Ths law hld s nly at relatvely lw values f.

Actvty The actvty f a cmpnent n any state at T s frmally defned as beng the rat f the fugacty f the substance n that state t ts fugacty n ts standard states,.e. fr the speces r substance, f a = f In a cndensed slutn, f s the fugacty at T and f s the fugacty f pure (the standard state) at T. If the vapr n equlbrum wth the cndensed slutn s deal, then f = p p = a p

Actvty If the vapr n equlbrum wth the cndensed slutn s deal, then f = p p = a p If the cmpnent exhbts Raultan (deal) behavr, p = Xp a = X If the cmpnent exhbts Henryan behavr, a= kx Actvty f Real Systems Gaskell

Derve Δμ fr the fllwng reactn n terms f standard Gbbs free energy and actvtes. L+ mm = rr+ ss ( r s S ) ( m ) Δμ = μ + μ μ + μ R L M μ = μ + RTln a = G + RTln a k k k k k ( rg ( R RTlnaR) sg ( S RTlnaS) ) ( GL RTlnaL) m( GM RTlnaM) Δμ = + + + ( ) + + + = ( rgr + sgs ( GL + mgm) ) + ( rrtlnar + srtlnas ( RTlnaL + mrtlnam) ) r s aa R S = ( rgr + sgs ( GL + mgm) ) + RTln m aa L M r s aa R S =Δ G + RTln m aa L M ( R S ( L M) ) ΔG rg + sg G + mg = Δμ standard Gbbs free energy Gaskell, Table A-1 used fr rreversblty f chemcal reactns

r s aa R S Δμ = Δ G + RTln m aa L M Δμ=Δ G =Δ G + RTlnQ Q aa r s R S aa m L M : prper qutent f actvtes fr the reactn 0 ΔG Q < exp rreversble reactn RT Δμ = 0 at equlbrum Δμ = 0 = Δ G + RTlnQ equl Q equl r s aa R S = K Equlbrum Cnstant m aa L M equl 0 =Δ G + RTlnK Δ G = RTlnK Q Δμ = Δ G = RTln K

Ex. A gas mxture at ne atmsphere ttal pressure has the fllwng mle fractn: X H = 0.01, X O = 0.03, X HO = 0.96 At 700 C, fr the reactn H + O = H O ΔG = - 393 KJ (a) Determne the drectn f spntaneus change fr ths system (b) Determne the equlbrum fractn (cmpstn). (a) Δ ( G RT) ( ) K = e = e = 1.5 10 / 393,000 / 8.314 973 1 ah O XH O (0.96) Q = = = = 3.1 10 a a X X (0.01) (0.03) H O H O Q Δμ = Δ G = RTln K Frward reactn 5 (b) H + O = H O ΔG = - 393 KJ 0.01-x 0.03-x 0.96+x K (0.96 + x) = = 1.5 10 (0.01 x) (0.03 x) 1 x 0.005 X H = 0, X O = 0.051, X HO = 0.9749

Fr a system f a cndensed phase and a gas phase ( ) μ (T, ) = μ (T, ) + RTln M(s) M(g) M(g) M(g) M(g) =μ (T, ) + RTln = 1 M(g) M(g) M(g) M(g) dμ = Vd μ (T,) = μ (T, ) T,n + Vd 0 at T <<1 μ(t,) μ(t, ) + RT lna = 0, a = 1 μ M(g) M( g) T, =1atm : STD ( ) μ M(s) M( s) At equlbrum μ (T,) =μ M(s) M(g) (T,) μ μ = μ + RTln M(s) M(s) M(g) M(g) μ = μ + RTln a M(s) M(s) M(s) μ = μ + RTln M(s) M(g) M(g) μ actvty = 1 μ M(s) M(g) 를 standard state 로할때는 를 standard state 로할때는 actvty = equlbrum vapr pressure

Derve the expressn fr rreversblty n the cas e f M 1 + O(g = MO (s) ) ( s) G (n the gas phase) = G (n the sld metal phase) M MO M G (n the gas phase) = G (n the sld xde phase) MO 1 M + O = MO (s) (g) 1 1 Δ G = G + RTln a G RTln = MO(s) MO(s) O (g) O G RTln a 0 M (s) M(s) 1 a MO Δ G = G G G = RTln MO(s) O (g) M(s) 1 a (s) O (s) M(s) Δ G 1 = RTln 1 O sld 를 standard state 로선택할때는 a = 1, a = 1 MO(s) M( s) 1 Δ G = RTln = RTlnO = RTlnK O

Ex) The standard Gbbs free energy change fr the reactn f N s + O = NO s ΔG = - 471,00 + 17 T (J) 98 < T < 176. Fnd the maxmum xygen partal pressure t avd the xdatn f nckel at 1500 K. a 1 N O O Ns + O = NO NO = = Δ G = RTlnK = RTln O K a a Δ G = 1300 (J) = RTln = 8.314 1500ln 1500K O O 8 O = 3.76 10 (atm) Hw can we make O < 10-10 atm r 10-0 atm? Hw can we cntrl O? Equlbrum n H O-H and CO -CO Mxtures Determne the rat f HO t H t make O = 10-10 atm at 000 K. 1 H + O = HO, ΔG = 47, 500 + 55. 85T J 47, 500 55. 85 ln K = = 8. 167 at 8. 3144T 8. 3144 000 K

K HO 3 = 3.51 10 = O H H HO 1 O wth = 3.51 10 10 = 3.51 10 = 3 5 10 10 atm, Hw can yu prepare ths cndtn n the labratry? If H = 1 atm, HO must be 0.035 atm. 0.035 atm s the saturated vapr pressure f water at 7.0 C. bubblng hydrgen gas at 1 atm pressure thrugh pure lqud water at 7 C. Hw can yu make O = 10-0 atm at 1000 K? 1 CO + O = CO, Δ G = 8, 400 + 86.81T J Determne the rat f CO t CO t make O = 10-0 atm at 1000 K. ln K = 8,400 86.81 3.5 at 1000 K 8.3144T 8.3144 =

wth O = 0 10 atm, K 10 CO CO = 1.646 10 = = 1 10 COO CO 10 CO 1.646 = CO = 1 atm = CO + CO CO = 0.6 atm, 0.378 atm CO = Thermdynamcs f Chemcal Vapr Depstn Hw d we defne the rreversblty crtern fr depstn n CVD? Hw abut n VD?