solutions

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Solutions1. The activity coefficient of zinc in liquid brass is given (in joules ) by the following equationfor temperature 1000-1500K: 238300ln Cu Zn x RT -=γ, where x Cu is the mole fraction ofcopper. Calculate the partial pressure of zinc P Zn over a solution of 60 mol % copper and 40 mol % zinc at 1200K. The vapor pressure of pure zinc is 1.17 atm at 1200K. (7.1, p196) Solution:)(117.017.11.01.04.025.025.038.11200314.86.03830038300ln 22atm a P P x a RTx Zn pZn Zn Zn Zn Zn Zn Cu CuZn =⨯===⨯===-=⨯⨯-=-=γγγ2. Using the equation give in Problem 7.1, for the activity coefficient of zinc in liquid brass, derive an equation for the activity coefficient of copper using the Gibbs-Duhem equation. (7.2, 196)Solution: According to Gibbs –Duhem Equation:2111ln 0Cu Cu 38300ln 238300238300238300)(ln )(ln )(ln 0)(ln )(ln 0)1()(ln )(ln 0)(ln )(ln )(ln )(ln 0)(ln )(ln ZnCu Znx Zn Cu x Zn Cunx Cu CuZn Cu Zn CuZn Cu Cu ZnZn Zn Zn Cu Zn Cu Zn Zn Cu Cu Cu Cu Zn Zn Zn Zn Cu Cu Zn Zn x RTdx x RTdx x RTdx x RTx x d d x x d d x d x x d dx dx dx x d x x d x x d x a d x x d x d x a d x a d x nZ CuCu-=⋅-=⋅=⋅-⋅-=-==∴=-+=+==+++=+⎰⎰⎰⎰γγγγγγγγ++3.(a) At 900K, is Fe3C a stable compound relative to pure Fe and graphite?( 7.3, 196)(b) At 900K, what is the thermodynamic activity of carbon in equilibrium with Fe and Fe3C ?Carbon as graphite is taken as the standard state.(c) In the Fe-C phase diagram, the carbon content of α -iron in equilibrium with Fe3C is0.0113 wt. %. What is the solubility of graphite in α-iron at 900K? DA TA A T 900K, J GCFe C Fe ographite346333)(+=∆=+4. From vapor pressure measurements, the following values have been determined for the activity of mercury in liquid mercury-bismuth alloys at 593K. Calculate the activity of bismuth in a 40 atom % alloy at this temperatureN Hg 0.949 0.893 0.851 0.753 0.653 0.537 0.437 0.330 0.207 0.063 a Hg 0.961 0.9290.9080.8400.7650.6500.5420.4320.2780.092(7.4,196) Solution: N Hg 0.949 0.893 0.851 0.753 0.653 0.537 0.437 0.330 0.207 0.063 a Hg 0.961 0.929 0.908 0.840 0.765 0.650 0.542 0.432 0.278 0.092 γHg1.0131.041.061.121.171.211.241.311.341.46Plot ln γHg ~x Hg0.00.20.40.60.8 1.00.000.050.100.150.200.250.300.350.40l n γH gx H g)(ln 391.0ln )11(391.0)(ln )391.0()(ln )(ln 1ln 0Hg Bi BiBi x BiBiHgBiHg HgBiHg Bix x dx x d dx x x d x x d Bi+-=--=-⨯-=-=⎰⎰γγγγγwhen x Bi = 0.4, x Hg =0.6107.11.0)6.04.0(ln 391.0ln ==+-=Bi Biγγ7.5 For a given binary system at constant T and P , the liquid molar volume of the solution (cm3/mol) is given by : B A B A x x x x V 5.280100++=(a) Compute the partial molar volumes of A and B and plot them, together with the molar volumeof the solution, as a function of the composition of the solution;(b) Compute the volume of mixing as a function of composition. (7.5, 196)Solution: the calculated partial variables are as follows: A A B A x x x x V 5.280100++=B x P T A A x xV V B5.2100,,+=⎪⎪⎭⎫ ⎝⎛∂∂= B A x P T BB x x xVV B5.25.825.280,,-=+=⎪⎪⎭⎫ ⎝⎛∂∂= 0.00.20.40.60.8 1.080859095100105c m 3/m o lx B(b) BA B B B MBA x x x V x x V VVV5.2)100(80)1(100(80,100=--=+--=== 4. For an ideal binary solution of A and B atoms, plot schematically the chemical potential of both species as a function of the composition of the solution. Indicate on the plot the molar Gibbs free energy of pure A and B. (7.6,196)5. At 473︒C, the system Pb-Sn exhibits regular solution behavior, and the activity coefficient of Pb is given by: ()()2132.0log PbPb x --=γ. Write the corresponding equation of the variation ofSn γ with composition at 473︒C. (7.7, p196)6. MgCl 2 and MgF 2 are two salts that can form solutions. The Gibbs free energy of fusion(J/mol) for both compounds is given by:For MgCl 2 : ∆G = 43905-43.644T, Melting point =987KFor MgF 2: ∆G = 58702-38.217T, Melting point = 1536K The free energy of mixing (J/mol) for liquid mixture MgCl 2 and MgF 2 is given by:))(252556()ln ln (222222222MgCl MgF MgF MgCl MgF MgF MgClMgCl Mix x x x x x x x x RT G -+-++=∆.Compute the maximum solubility of MgF 2 in liquid MgCl 2 at 900︒C. MgCl 2 does not dissolve insolid MgF 2. (7.8, 197)7. The thermodynamic properties of Al-Mg solution at 1000K are given in accompanying table. (a) If one mole of pure liquid aluminum and one mole of pure liquid magnesium, each at 1000K, are mixed adiabatically, what will be the final temperature of the solution that is formed ? (b) What is the total change in entropy for the process ?DA TA Quantities of Mixing Liquid Alloys at 1000KMg x)/(mol cal G M ∆)/(mol cal H M ∆ )/(mol cal S M ∆)]./([K mol cal C p0.1 -800 -300 0.5 7.1 0.2 -1250 -600 0.65 7.18 0.3 -1550 -750 0.8 7.26 0.4 -1700 -850 0.85 7.34 0.5 -1800 -900 0.9 7.42 0.6 -1700 -850 0.85 7.5 0.7 -1550 -750 0.8 7.58 0.8 -1250 -600 0.65 7.66 0.9 -800gf-3000.57.74。