A gas that contains $C{O}_2$ is contacted with liquid water in an agitated batch absorber$.$ Henry's law gives the equilibrium solubility of $C{O}_2$ in water, $C_A=\frac{p_A}{h_A},$ where $C_A(mo\frac{l}{L})$ is the $C{O}_2$ concentration in solution, $p_A(atm)$ is the partial pressure of $C{O}_2$ in the gas phase and $h_A$ is Henry's law constant. The rate of transfer of $C{O}_2$ from the gas to the liquid per unit area of gas$-$liquid interface is given by $r{a}_a(mo\frac{l}{c}{m}^2(s))=k(C^{**}-C_A)$ where $C^{**}A$ is the concentration of $C{O}_2$ that would be in equilibrium with the $C{O}_2$ in the gas phase $($:$p\frac{A}{h_A}\}.$ Suppose the gas$-$phase total pressure is $P(atm)$ and contains $y_A$ mol fraction of $C{O}_2,$ and the liquid phase initially has $V(c{m}^3)$ of pure water with the agitation of the liquid phase sufficient to neglect spatial composition dependency, and if the amount of absorbed $C{O}_2$ is low enough for $P,V,$ and $y_A$ to be assumed constant. Write a differential balance on $C{O}_2$ in the liquid phase and solve the differential equation to show that $C_A(t)=C^{**}A[1-$exp$(kS\frac{t}{V})].$