= Solution
For $0<s<1$, differentiation gives
$$
V'(t)=\frac{Q}{1-s}\left(e^{-t/\tau}-s e^{-st/\tau}\right).
$$
Its unique zero solves $e^{-(1-s)t/\tau}=s$. Therefore the <mobile-volume peak of an exponentially forced porous current> occurs at
$$
\boxed{t_{\max}=\frac{\tau\log(1/s)}{1-s}.}
$$
The derivative is initially positive and is negative after this time. More explicitly, at the stationary point $V''=-Qs e^{-st/\tau}/\tau<0$. Thus it is the global positive maximum, with
$$
\boxed{V_{\max}=Q\tau s^{s/(1-s)}.}
$$
The claim of a finite maximum requires nonzero <capillary residual trapping>. At $s=0$ the mobile volume rises monotonically towards $Q\tau$ and the peak time tends to infinity. As $s\to1$, the limiting peak time is $\tau$ and the limiting maximum is $Q\tau/e$.
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