高等化工热力学习题与参考答案

1、1.1如果流体服从 van der Waals方程，请导出A和G的偏离函数以及逸度系数的表达式。解:Vm “RT(p vm带入van der Waal方程：p RT Vm bVm RT a RTn/ (2)dVmVm b V： Vm已知Am-AmAm-AmVm)dVm RTI nhVma 得.p '得.mRTI n 约V(RTln Vma) |VmVm b Vm 1V aRTln Vm b VmRTI唏RTlnRTInpVm p0Vm0需aRTaVm bVmVm bVV*2aRTVmDTVm bVmVm bRl则 GmGm0AmAm°Vmp0Vm0m .韵:傑-各 说)叭+(

2、ND 一加z=in _ Y竺 一 _Q + W吐山Vm-b BTV Va-b BfrV_ 丄hi (一 )1.2请根据下列丙酮(1)氯仿(2)溶液的实验数据 (35.17 C)计算(1) 以纯液态为标准态，丙酮的活度及活度系数，(2) 以无限稀释为参考态，氯仿的活度及活度系数。X20.00000.05880.12320.29100.42320.51430.66350.79970.91751.000p / kPa45.9343.0939.9030.7023.2818.0010.535.001.730.0p, / kPa0.0001.232.727.3911.8515.7122.6929.9235

3、.6139.08解:0 0Pi Pi ai Pi Xi x,iPiaai0 , x,iPixi(1)以纯液态为标准态，则p；45.93kPa1x2(2)以无限稀释为参考态，则p239.08kPaPl0，x,1Pla2P20 ,x,2P2a?*Jx,2x,2X20.00000.05880.12320.29100.42320.51430.66350.79970.91751.000p / kPa45.9343.0939.9030.7023.2818.0010.535.001.730.0p, / kPa0.0001.232.727.3911.8515.7122.6929.9235.6139.0810.

4、93820.86870.66840.50690.39190.22930.10890.03770a1x,110.99680.99080.94270.87870.80690.68130.54350.456600.03150.06960.18910.30320.40200.58060.76560.91121a20.53530.56490.64980.71650.78160.87510.95740.99311x,2x,20.50*1.00001.07051.12991.29971.43301.56331.75011.91471.98632.0000x,21.3已知对45 C时四氯化碳(1)-乙腈(2)

5、混合物的平衡气相组成，总蒸气压及混合热如下, 请分别计算：(1) 活度ai, a2与x的关系曲线，(2) 45C时 G及T S与x的关系曲线，exexex(3) H , G及TS与x的关系曲线。X100.1280.3170.4140.6310.8211y100.4000.5210.5600.6080.6701P /kPa27.5040.5348.0148.6549.0947.1034.50 H /J/mol04147458609307360解:(1)以纯液态为标准态，由表知0Pl34.50kPa,p227.50kPa，则a1P10P1PyiP2亍，a?0P1P2P(1 yJ0P2X10.000

6、.1280.3170.4140.6310.8211.000Gm J/mol0-541-594-646-591-4640095513391506152112000T SmX10.00000.1280.3170.4140.6310.8211.000X210.8720.6830.5860.3690.1790yi0.0000.4000.5210.5600.6080.6701y210.6000.4790.4400.392.03300P27.540.5348.0148.6549.0947.1034.5p / kPa016.2125.0127.2429.8531.5634.5p / kPa27.524.32

7、23.0020.7718.817.530ai00.450.720.790.870.911a210.880.840.760.680.640(2)由 GRTgln a1 n2ln a2)得GmRT(x-i ln a1 x21n a2)恒温下，T SmHmGm1HmRT(x In a1x2 In a2)J/mol2000-B-TM-1000 VV1.4已知NaCI水溶液中NaCI的浓度mB及平均离子活度系数m如下，请求出相应的 XB及(3)x,1a1,x,2a2X1X2exGmRT(*ln x,1X2ln x,2)eXH mHHidHTsmxHmxexGmX10.000.1280.3170.4140

8、.6310.8211.000GjJ/mol06241060115011537790H exH m04147458609307360exISm J/moI0-210-315-290-223-430解：Xb0.001mBM A10.001 mBM A18mB1000 36mBX,1 0.001vmMAX,1 0.036mX ,m 0.0010.01 0.10.51.02.04.06.0x ,并请讨论，这些数据可说明什么问题?m0.0010.010.10.51.02.04.06.0m0.96490.89600.77840.68110.65690.66760.79320.9862m0.96490.89

9、600.77840.68110.65690.66760.79320.9862XB0.0000180.00020.00180.00880.01740.03360.06290.0888*x,0.96490.89630.78120.69340.68050.71570.90741.19922.1由pV kTIn 和d(pV) SdT pdV Nd ,导出巨正则系综热力学函数关系InI nklnkT(人，N kT ()v,tkT(-) t, G N , E kT2(-)VT v,(_)+ pdfpdVVdpSdTpdV NdNdVdpSdTdGSdTVdpkidnii 1SdTVdpdNdGNddNd(

10、N )GN又由于：d(”)= Nd” + Sd 丁 + pdf(kTIn )VkT( ln,TV),T(kTIn )TklnkT(InT)(kT In )kT (亘)v_V,TNkEiP =i NjNkEiNjEi "kT e或者:InkT2In讦,VTSpVV,kTkT2(由麦克斯韦变换关系式:InInInTV,k?V,V,TT -kTInInV,TT v,E kT2InV'kT32.3 在 1atm, 300K, V= 1cm条件下,求Ne的分子配分函数q,此值有无量纲？并请求出正则 27 配分函数（以In Q表示）。（已知m=1.660 x 10- kg,MN=20.1

11、79 )解：Ne为单原子分子：只有平动加电子配分函数q2723220.179 1.660 101.38 10300trans32 1068.84310256.626 10 34 2qelecge1qqtrans qelec无量纲8.843 1025N qN!In QNq ln( ) N InN!q In N! Nlnq N In N NnNAPVNaRT01 105 10 6 6.02 10238.314 3002.44 1019In QNlnq N In NN 3.9 10202.4请证明Cv毎_2(2)请证明N1/2 EE（式中22），并由此说明，尽管E可以忽略,E是一个不小的量。解:(1

12、) (EE)E22EE2E2PjE2pjQjE2eEQkT2皿EjEjejEjkTkTN,VkT22kT2CvN,VkT2N,VkT2(2)ENl/2ENN1/2kT2Cv33若 Cv -Nk , E -NkT2 2i321E N(N 2)E 22大于o34.2将Ikmol氮气压缩于0.04636m3的钢瓶中，温度为273.15K ,请分别使用理想气体公式、RK方程、SRK方程和PR方程计算此时的压力，及计算值与实验值比较的相对偏差 （实验值：101.33MPa）。（氮气的临界参数为Tc 126.2K , pc3.394 MPa ,0.040)解：理想气体状态方程:pV nRTp 48.99M

13、PaRK方程：Vmmr2T 2.50.427481.55766RT_ _ b T0X VPc0.08664空 2.6784 10Pc88.2MPaSRK方 程：RT a(T) pVm b Vm(Vm b)Tr T 2.1644r Tcm 0.480 1.5740.176 20.5432(T)1 m 1052Tr=0.5536a(T) 0.427488.314 126.22 (T)0.1387 0.55360.07683.394 106b 2.6784 105p 93.34MPa3PR方程:RTa(T)Vm b Vm(Vm b) b(Vm b)77.16MPa仃)1m(1 Tr0.5)20.63

14、14r2t2a(T) ac (T)0.45724(T)0.09364Pcb 0.07780RTc 2.4051 10 5Pc2m 0.3746 1.542260.269920.435954.3有压力为10 Pa和温度为25C的氮气，现将其由100 L压缩到1 L,其温度降为100C, 试求此时压力。(提示：可将初始状态的氮气视为理想气体，终了状态的氮气则以RK方程计算)解：始态:p1V1 nRT1 n 4.0342molVm V10.02479n终态：1 10 °4.03420.2479 10RTaVm b T 0.5Vm(Vm b)4.77MPa4.4用下列公式计算 N2(1)-

15、n-GHw(2)混合体系(xi=0.30 )在188C, 6.89 X 10 kPa时的摩尔体积(1)理想气体状态方程B(2)维里方程pV RT 1 -VCV2混合规则：Byi yi jjBij，CiyiyjykCjkj k3B1114 cm/mol,B22265cm3/mol,B12B219.5cm3/mol32Cm 1300(cm /mol),C2223230250(cm /mol),32C1124950(cm /mol),C122327270(cm3/mol)2(3) Pitzer对比态关联式 Z Z(0)z(1)已知n- C4H0的参数为Tc425.2K, Pc3.800 MPa,0.

16、193(可采用混合规则 XmiXi，式中X为参数)解：(1)理想气体状态方程Xi0.30,x20.70T461.15K,p6.89MPaVRT5.5646 10 4m3/molP(2)维里方程pVB CRT 1-V V2B132.58cm3/mol,C14552.47cm3/molVm4.284 10 4m3/mol(3)Pitzer对比态关联式 Z ZZT1c 126.2K,T2c 425.2Kp1c 3.394MPa,p2c 3.800MPa10.040, 20.193T；0.3T；c 0.7T2c 335.5Kpc 3.678MPa0.1471Tr461.15335.51.37456.8

17、9 106Pr3.67821.87320 1査得Z0.7679,Z0.1785Z z0Z10.7942、，ZRTVm4.510 4m3/molP4.5请用下列状态方程计算乙烷（1）-丙烷混合气体（x10.75）在5.066 MPa和 100C时的摩尔体积。 （1 ）理想气体方程；（2） vdWTc (K)p c (MPa)方程；（3） RK方程。C2H6305.44.884（乙烷和丙烷的临界参数见右表）GH8369.84.246（1）理想气体方程VmRT6.123910 4m3/molP（2）vdW方 程PRTaVm bVm2由：a27(RT；)2b1 RT：64 Pc8 Pc得b10.556

18、9,a26.4985100.93925,b29.05121050.5i0.6432, bVm4.43 104m3/molxb 7.1367 10 5(3) RK方程RT由:Vm b T0.5Vm(Vm b)r2t2*5a 0.42748Pc0.08664RTcPc得:ai9.8613,a218.3009Vm5.14.5042105,b26.27361050.54.45 1011.7285, bm3/mol4.9466 10 5G ex已知池RT2 2A2X1X2A21X1X2，式中A12和A21为方程参数，请导出ln x,1的表达式。解：Gex mRTA12X1 (1 X1)A1X2(1 xj

19、Gex mRTA2nt(1亠)2 An1n2n1 £ (122n1n2A2(n1 H)3An1 n221 3(n1 H)Gex mRT2八门小2A|23(n1 乜)3niRT3A12n22mn22(n1n2 )4A21(n1 匕)42A212x1x22X1 x2RT32 Ai2(x2 2x1x2 )nRTIn x,1GexP取小於财严魏12x1 x225.2已知苯（1）环己烷（2）体系在 77.6c和液相组成x10.525时形成共沸物，总压力为 p 101.33kPa，且 p1sat99.3kPa 和 p；at 98.0kPa。请用 van Laar 方程计算此温度下 Xi分别为0.

20、0, 0.2, 0.4, 0.6, 0.8和1.0时的各组分活度系数、平衡压力和汽相组成。az、 y1 )（提示，汽相可视为理想气体，平衡条件为pyixi x,ipisat,且共沸点时X；Z解:77.6 C和液相组成x10.525p 101.33kPa，且 p1sat99.3kPa 和 98.0kPa汽相可视为理想气体，平衡条件为pyixi x,i pT ,且共沸点时 xazazy1汽液平衡时:satpy1人 x,1P1X,1satP1satpy2X2 x,2 P2Px,2satP2Inx,1A21x2A,2x!A21x2Inx,2A2X1A12x1A21x2In101.3399.3A12A2

21、10.475In101.3398.0A2 0.525 A21 0.4752A12 0.525A2 0.525 A21 0.475则：A2°1258A210.093xx2x,1x,2*y2P1P2011.1341100.96710980.20.81.07281.0060.210.77821.2878.830.40.61.03541.02110.4060.59441.1460.090.60.41.01381.0426.05960.40460.3940.840.80.21.00311.0685.07860.20779.6520.981.0011.09750.98.099.305.3请用UNIFAC方程计算丙酮(1)正戊烷(2)体系在307K,X1 = 0.047时丙酮的活度系数。 有关的基团参数为解:基团编号RQCH10.90110.848CH20.67440.540CHCO221.67241.48881,22822,1476.4K26.76K = 1x0.9011+ 1X1,6721 a 2.5735 -1x0.848+1x1.488 = 2.336% = 2 x0 901.1 +8x0.6744 = 38264 彳仝= 2x0.848+3 x 0-540 士 3.316= 0.03210,03358