Chapter 66.1. Carbon dioxide is diffusing through nitrogen in one direction at atmospheric pressure and 0C. The mole fraction of CO2 at point A is 0.2; at point B, 3 m away, in the direction of diffusion, it is 0.02. Diffusivity D is 0.144 cm2/s. The gas phase as a whole is stationary; that is, nitrogen is diffusing at the same rate as the carbon dioxide, but in the opposite direction. (a) What is the molal flux of CO2, in kilogram moles per square meter per hour? (b) What is the net mass flux, in kilograms per square meter per hour? (c) At what speed, in meters per second, would an observer have to move from one point to the other so that the net mass flux, relative to him or her, would be zero? (d) At what speed would the observer have to move so that, relative to him or her, the nitrogen is stationary? (e) What would be the molal flux of carbon dioxide relative to the observer under condition (d)?solution: (a)from(6.1-8)from equation(6.1-19)(b) net mass fluxfor carbon dioxide (molecular weight=44)mass flux of CO2= 441.38810-4 kg/m2hfor nitrogen (molecular weight=28)mass flux of N2= 281.38810-4 kg/m2hso the net mass flux in the direction of CO2 diffusionm=(44-28) 1.38810-4 =2.22110-3 kg/m2h(c) Here JA=NA=NB, since the diffusion is equimolal. The concentration at any point depends on position due to the concentration profile of the equimolal diffusion, so does velocity based on equations (6.1-3a) and (6.1-3b). To select two points, yA=0.2 and 0.02, respectively, to calculate the positions of observerfor yA=0.2, CA=Cm yA=from equation (6.1-3a)the diffusing velocity of A: for B: CB=Cm (1-yA)=the diffusing velocity of B: let uo is the velocity of the observer moving in the direction of CO2 diffusion, then net velocity (uA-uo) gives a mass transfer rate mA equal to that in opposite direction mB corresponding to (uB+uo) from equations (6.1-4) and (6.1-5)for mA = mB, and rearranging two equations above gives at yA=0.2,It is similar to calculating uo at the point of yA=0.02(d) When the velocity of observer moving is equal to that of nitrogen diffusing, the nitrogen is stationaryuo=(e) When the velocity of observer moving is equal to that of nitrogen diffusing, the molal flux of carbon dioxide diffusing is indicated byChapter77.1 solution:The data from third column in the table are used as calculating Mole fraction x of ammonia in water:molar ratio of ammonia to water X in liquidmolar ratio of ammonia to inert gas the results of calculation are list in the tablep/kPa0 0.4 0.8 1.2 1.6 2.0 2.43 3.32 4.23 6.67 9.28x0 0.00527 0.01048 0.01563 0.02074 0.0258 0.03079 0.04063 0.05028 0.07357 0.09574X0 0.00530 0.01059 0.01588 0.02118 0.0265 0.03177 0.04235 0.05294 0.07940 0.1059Y0 0.00396 0.00796 0.01198 0.01600 0.0201 0.02453 0.03387 0.04353 0.07040 0.1008The data in the table are used to plot the mole fraction x versus partial pressure p diagram and molar ratio X-Y diagram虚线范围表示符合Herrys law虚线范围表示符合Herrys law7.3 Vapor-pressure data for a mixture of pentane (C5H12) and hexane (C6H14) are given by the table. Calculate the vapor and liquid composition in equilibrium for the pentane-hexane at 13.3kpa pressure on assuming that the vapor of mixture approaches ideal behavior and liquid follows Raoults low. t,K260.6265270275280285289PA,kPa13.317.321.926.534.542.548.9PB, kPa2.833.54.265.08.5311.213.3Solution: From Raoults low （equations (7.1-4) and (7.1-5) ） And the total pressure of system is equal to sum of the partial pressures of two substancesrearranging equation above 1and rearranging equation above gives 2substituting the data for the table into the equations 1 and 2 gives the results in the following tablet,K260.6265270275280285289PA,kPa13.317.321.926.534.542.548.9PB,kPa, 2.833.54.265.08.5311.213.3x1.00.7100.5130.3860.1840.0670y1.00.9240.8450.7690.4770.21407.4 Using the conditions in the problem 7.3 for the pentane-hexane mixture, do as follows: (a) Calculate the relative volatility.(b) Use the relative volatility to calculate the vapor and liquid composition in equilibrium and compare with the result given by the problem 7.3.Solution: (a) calculate the value of by The results are given in the table :t,K260.6265270275280285289PA,kPa13.317.321.926.534.542.548.9PB,kPa, 2.833.54.265.08.5311.213.34.704.945.145.34.043.793.68Average relative volatilityand equilibrium relation with relative volatility for a mixture of pentane (C5H12) and hexane (C6H14) is expressed by(b) Using the equat