1 problem found
A liquid of fixed volume \(V\) is made up of two chemicals \(A\) and \(B\,\). A reaction takes place in which \(A\) converts to \(B\,\). The volume of \(A\) at time \(t\) is \(xV\) and the volume of \(B\) at time \(t\) is \(yV\) where \(x\) and \(y\) depend on \(t\) and \(x+y=1\,\). The rate at which \(A\) converts into \(B\) is given by \(kVxy\,\), where \(k\) is a positive constant. Show that if both \(x\) and \(y\) are strictly positive at the start, then at time \(t\) \[ y= \frac {D\e^{kt}}{1+D \e^{kt}} \;, \] where \(D\) is a constant. Does \(A\) ever completely convert to \(B\,\)? Justify your answer.
Solution: We have \(\dot{A} = -kVxy\) or \(\dot{x}V = -kVxy\), ie \(\dot{x} = -kxy\) and similarly \(\dot{y} = kxy = k(1-y)y\). \begin{align*} && \frac{\d y}{\d t} &= ky(1-y) \\ \Rightarrow && \int k \d t &= \int \frac{1}{y(1-y)} \d y \\ \Rightarrow && kt &= \int \left ( \frac{1}{y} + \frac{1}{1-y} \right) \d y \\ &&&= \ln y - \ln (1-y) + C\\ \Rightarrow && kt &= \ln \frac{y}{D(1-y)} \\ \Rightarrow && De^{kt} &= \frac{y}{1-y} \\ \Rightarrow && y(1+De^{kt}) &= De^{kt} \\ \Rightarrow && y &= \frac{De^{kt}}{1+De^{kt}} \end{align*} As \(t \to \infty\) \(y \to \frac{D}{D} = 1\) so depending on how fine grained we want to go we might say that 'yes it completely converts' when there is an immeasurably small amount of \(A\) left, or we might say it doesn't since it only tends to \(1\) and never actually reaches it.