A plug$-$flow reactor can be designed by reading the Levenspiel plot $(-\frac{1}{r_A}vs.x_A)$ of the reaction. $r_A$ is the rate of disappearance of A and $x_A$ is the percent conversion of $A.$ The area from a Levenspiel plot is related to the reactor volume via the following relation: $V=F_{A0}*A_{PFR}.F_{AO}$ is the molar flowrate and APFR $($in $\frac{m^6}{\text{k}\text{m}\text{o}\text{l}})$ is the area under the Levenspiel plot's curve. If it is desired to design a reactor that would convert $80\%$ of a fresh reactant to $x_A=0\mathrm{.}80).$ If $F_{AO}=0\mathrm{.}125$kmo$\frac{l}{m^3},$ find the volume of the

reactor in $m^3.$

Answer: $A=148\mathrm{.}42345$ and $V=18\mathrm{.}5529375$