I'm having a hard time understanding how to determine V for this problem.

Here are two chegg solutions that don't explain it good enough for me: https://www.chegg.com/homework-help/questions-and-answers/total-50-g-wet-microorganisms-1-100-kg-m3-diameter-0667-m-added-0100-l-aqueous-solution-37-q88515781 https://www.chegg.com/homework-help/questions-and-answers/total-50-g-wet-microorganisms-density-1100-kg-m3-diameter-0667-m-added-0100-l-aqueous-solu-q31424686

I've determined the mass transfer coefficient to be 9.909x10^-3 m/s but I don't understand how to derive the fomula to figure out N_AA. The answer for N_AA is 9.07*10^-5.

Please don't use the solutions already posted on chegg. I tried understanding them but how they are written is not helpful for me. Geankoplis (Transport Processes and Separation Process Principles, 5th edition, chapter 21) mentions things about void fractions, A, J_D, and a, but I don't have any void fractions here. What do I do? What is being accomplished by combining the volume and surface area of the sphere? Why did we bother to calculate the volume of the microorganisms? Why are we using the diameter of the microbe in the volume of the sphere equation?

21.4-2. Mass Transfer of O2 in Fermentation Process. A total of 5.0 g of wet microorganisms having a density of 1100 kg/m and a diameter of 0.667 um are added to 0.100 L of aqueous solution at 37C in a shaker flask for a fermentation. Air can enter through a porous stopper. Use physical property data from Example 21.4-1. a. Calculate the maximum rate possible for mass transfer of oxygen in kg mol O2/s to the surface of the microorganism, assuming that the solution is saturated with air at 101.32 kPa abs pressure. b. By material balances on other nutrients, the actual utilization of O2 by the microorganism is 6.30 x 10-6 kg mol O2/s. What would be the actual concentration of O2 in the solution as percent saturation during the fermentation