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南昌大学2014上半年度材料热力学重点题(附氧势图).doc

南昌大学2014上半年度材料热力学重点题(附氧势图).doc

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南昌大学2014上半年度材料热力学重点题(附氧势图)

solution: Solution:,exothermic reaction For the reaction : Solution: (a) When the equilibrium is reached, T = 500?C = 773K at T=300?C=573K, Although the equilibrium PO2 is very low, kinetically the reaction is not favoured and reaction speed is very slow. So 300?C is not suitable at At T=800?C=1073K, lnPO2 =-22.2, PO2 =2.28?10-10 atm. At 800?C, if the equilibrium is reached, nitrogen can be of high purity level. However, at this high temperature , particles of Cu will weld together to reduce effective work surface. So it is not suitable to use this high temperature in purification either. (c ) The equilibrium oxygen pressure remains the same when the total pressure increases, which means a higher purity level of N2 . Solution: N2 =2N, H2 = 2H , For N2 dissolving : For H2 dissolving: (a)For dissolving N2, PN2 = 1 atm, [N]=35cm3/100g melt, similarly: [H]’ =24.75cm3/100g melt total gas : [H]?+[N]? = 49.5 cm3/100g melt (b) [H]? =24.75 cm3/100g melt (c ) [H]?+[N]? = [N](0.33)1/2 /1+[H](0.33)1/2 /1=20.10+20.10 = 40.2cm3/100g melt Solution: (1)0.2 (2) (3) Solution:(a) (b) (c) Solution:(b) (c) 答案更正为: Temperature Phase Composition Fraction 1300 Liquid 60 61.5 α 8 38.5 β 99 0 1000+ Liquid 70 50.8 α 9 49.2 β 98 0 1000- Liquid _ 0 α 7 63.7 β 98 36.3 Solution: Solution Solution: Solution Solution: solution: Solution: Solution: (a) At 0?C, ?G =0, ? Tm?S = ?H (b) At 0?C, ?G =0 ? a reversible process can be designed as follows to do the calculation: (d) Solution: (a) (b) No, diamond is not thermodynamically stable relative to graphite at 298K. (c ) (c ) Assuming N atm , ?G = 0, reversible processes as following can be designed to realize this,

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