BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY 第四章事组分亲统热力学美文司数及参考答素 1.At 310K,the partial vapour pressures ofa substance B dissolved ina liquidAare as follow XB 0.010 0.015 0.020 Po/kPa 82.0 122.0 166.1 Show that the solution obeys Henry's law in this range of mole fractions,and calculate Henry's law constant at 310 K. Check that pa/xp=a constant(KB) 0.010 0.015 0.020 (Pg/xB)/kPa 82×10 8.1×10 83×103 Ka=p/average value is82x10 2.The addition of 5.00g of a compound to 250g of naphthalene lowered the freezing point of the solvent by 0.780K.Calculate the molar mass of the compound K=6.94 for naphthalene Mn=maso时B n=mass of naphthalereb (mass of B)xK oM(mas of maphthalene)T M,=50gx694K 0.250kg)×(0.78 mol)=178gmol 3.Consider a container of volume 250 mL that is divided into two compartments of equal size.In the left compartment there is argon at 100 kPa and 0C:in the right compartment there is neon sure.Calculate the entropy and Gibbs energy of mixing wher the is removedAssume that the gases are perfect AmaG=nRT(xInx+xgInxg) AG=pr兮n+n2=-prih2 =-010m020h2=-173hw㎡-17 TEL:010-64434903 PDF文件使用"pdfFactory Pro”试用版本创建w,fineprint.,com.h
BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY TEL:010-64434903 1 第四章 多组分系统热力学英文习题及参考答案 1.At 310K，the partial vapour pressures of a substance B dissolved in a liquid A are as follows: xB 0.010 0.015 0.020 PB/kPa 82.0 122.0 166.1 Show that the solution obeys Henry's law in this range of mole fractions，and calculate Henry's law constant at 310 K. Check that pB/xB=a constant(KB) xB 0.010 0.015 0.020 (PB/xB)/kPa 8.2×103 8.1×103 8.3×103 KB=p/x，average value is 8.2 ×103 2.The addition of 5.00g of a compound to 250g of naphthalene lowered the freezing point of the solvent by 0.780 K. Calculate the molar mass of the compound. Kf = 6.94 for naphthalene B B n mass of B M = B bB n = mass of naphthalene × f B K T b D = so mass of naphthalene T mass of B K M f B ´D ´ = ( ) ( ) 1 1 178 (0.250 ) (0.780 ) (5.00 ) (6.94 ) - - = × ´ ´ × × = g mol kg K g K kg mol M B 3.Consider a container of volume 250 mL that is divided into two compartments of equal size. In the left compartment there is argon at 100 kPa and 0℃；in the right compartment there is neon at the same temperature and pressure. Calculate the entropy and Gibbs energy of mixing when the partition is removed. Assume that the gases are perfect. ( ln ln ) mix A A B B D G = nRT x x + x x Ar Ne n = n ， = = 0.5 Ar Ne x x ， RT pV n n n = Ar + Ne = ) ln 2 2 1 ln 2 1 2 1 ln 2 1 DmixG = pV ( + = - pV ln 2 10 1 (100 10 ) (0.250 ) 3 3 3 ÷ ÷ ø ö ç ç è æ = - ´ ´ L m Pa L 17.3Pa m 17.3J 3 = - × = - PDF 文件使用 "pdfFactory Pro" 试用版本创建 www.fineprint.com.cn
BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY 5=46-7-=64×102K -273K 4.Given that p*(H)=0.02308 atm and p*(H)=0.02239 atm in a solution in which 0.122kg of a non-volatile solute (M=241g-mol)is dissolved in 0.920kg water at 293K calculate the activity and activity coefficient of water in the solution Let A=water and B=solute 号2 0.920kg 0.122kg n,-001802 g:mol=5105m0l 0.50mol 5105 3105+006=090 5.Benzene and toluene form nearly ideal solutions.The boiling point of pure benzene is 80.1C Calculate the chemical potential of benzene relative to that of pure benzene when0.30 B=Benzene Ha()=Ha()+RTInxs RT1nxB=(8.314Kmor')x(3533K)xn0.30)=-3536Jmor Thus,its chemical potential is lowered by this amount. Pe=asPa=YaX8Ps =(0.93)x(0.30)x(760Torr)=212Torr Question:What is the lowering of the chemical potential in the nonideal solution with y=0.932 6.By measuring the equilibrium between liquid and vapour phases of a solution at 30C at 1.00 atm,it was found that xA=0.220 when y=0.314.Calculate the activities and activity coefficients of both components in this solution on the Raoult's law basis.The vapour pressures of the pure components at this temperature are:p=73.0kPa and p=92.kPa (xA is the mole fraction in the liquid andy the mole fraction in the vapour.) TEL:010-64434903 2 PDF文件使用"pdfFactory Pro”试用版本创建，fineprint,com,cn
BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY TEL:010-64434903 2 = = -D D = K J T G S mix mix 273 17.3 6.34 ×10-2 J×K -1 4.Given that *( ) 0.02308 p H2O = atm and *( ) 0.02239 p H2O = atm in a solution in which 0.122kg of a non-volatile solute (M=241g×mol-1) is dissolved in 0.920kg water at 293K calculate the activity and activity coefficient of water in the solution. Let A=water and B=solute. 0.9701 0.02308 0.02239 * = = = atm atm p p a A A A A A A x a g = and A B A A n n n x + = mol kg mol kg nA 51.05 0.01802 0.920 1 = × = - mol kg mol kg nB 0.506 0.241 0.122 1 = × = - 0.990 51.05 0.506 51.05 = + xA = 0.980 0.990 0.9701 g A = = 5.Benzene and toluene form nearly ideal solutions. The boiling point of pure benzene is 80.1℃. Calculate the chemical potential of benzene relative to that of pure benzene when xbenzene=0.30 at its boiling point. If the activity coefficient of benzene in this solution were actually 0.93 rather than 1.00, what would be its vapour pressure? B=Benzene B B B (l) (l) RT ln x * m = m + 1 1 1 ln (8.314 ) (353.3 ) (ln0.30) 3536 - - - RT x = J ×K ×mol ´ K ´ = - J ×mol B Thus，its chemical potential is lowered by this amount. p a p x p Torr Torr B B B B B B (0.93) (0.30) (760 ) 212 * * = = g = ´ ´ = Question: What is the lowering of the chemical potential in the nonideal solution with g = 0.93 ? 6.By measuring the equilibrium between liquid and vapour phases of a solution at 30℃ at 1.00 atm，it was found that xA=0.220 when yA=0.314. Calculate the activities and activity coefficients of both components in this solution on the Raoult's law basis. The vapour pressures of the pure components at this temperature are： p kPa A 73.0 * = and p kPa B 92.1 * = . （xA is the mole fraction in the liquid and yA the mole fraction in the vapour.） PDF 文件使用 "pdfFactory Pro" 试用版本创建 www.fineprint.com.cn
BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY D D 4p十pg700n0314 P=(760Tor)×(0.314)=238.64Tom Pg=760Tor-238.64Tor=521.36Tom 238.647or (730x10 Pa)T015Pa)x(607206 521.36Tom ae=-921x10P阳)x101325Pax(60Mo℃0755 latm atm -号-8-0%8 TEL:010-64434903 PDF文件使用"pdfFactory Pro'”试用版本创建wm,fineprint.com.cn
BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY TEL:010-64434903 3 0.314 760 = = + = Torr p p p p y A A B A A p Torr Torr A = (760 )´(0.314) = 238.64 p Torr Torr Torr B = 760 - 238.64 = 521.36 0.436 ) 760 ) ( 101325 1 (73.0 10 ) ( 238.64 3 * = ´ ´ ´ = = atm Torr Pa atm Pa Torr p p a A A A 0.755 ) 760 ) ( 101325 1 (92.1 10 ) ( 521.36 3 * = ´ ´ ´ = = atm Torr Pa atm Pa Torr p p a B B B 1.98 0.220 0.436 = = = A A A x a g 0.968 0.780 0.755 = = = B B B x a g PDF 文件使用 "pdfFactory Pro" 试用版本创建 www.fineprint.com.cn