In Example 5.1, liquid 4 evaporated and diffused through stagnant gas B. If for each mole of liquid 4 diffusing into gas B an equal amount of gas B dissolves and diffuses into the liquid A, the molar flux will change. Calculate the molar flux and convective mass transfer coefficient for the case of equimolar counter diffusion for A under the same conditions as stated in Example 5.1. PLEASE

Consider the case in which liquid A evaporates and diffuses through stagnant gas B as shown in Figure 5-2. Assuming that the process is carried out isothermally at 25C and the total pressure is 1atm, calculate the molar flux and convective mass transfer coefficient when the level of liquid A is 0.05m below the top of the container. Assume that DAB=1.0105m2/s. The mole fraction of A at the gas-liquid interface is 0.2 . At the top of the container, the mole fraction is 0.001. Solution: Starting with Eq. (5-20), we have NA=RT(PB)MPDAB(PA1PA)PA=yAPPB=yBPPA1=0.2atmPB1=0.8atmPA=0.001atmPB=0.999atm(PB)M=ln(PB/PB1)PBPB1(PB)M=ln(0.999/0.8)0.9990.8=0.896atmNA=(82.06103)(298)(0.05)(0.896)(1)(1.0103)(0.20.001)=1.82106m2skgmol The mass transfer coefficient is calculated by using the definitions given in Table 5-1. kG=RT(PB)MPDAB=(82.05103)(298)(0.05)(0.896)1.0105=9.13106m2satmkgmol