Solving mixing tank problem with the Laplace transform The question I am trying to solve in my exercises reads as follows:

A mixing tank with a constant volume $V_{0}$ and flow rate $q$ is initially filled with pure water. If the inflow concentration is a constant $\alpha$ for $0 \leq t \leq T$ and zero afterwards, calculate the mass of the chemical in the tank at time $2T$. 

The answer of the problem is given, but there is no derivation, and needless to say I am very stuck in my own derivation. 

I know that the differential equation for the mixing tank is 
\begin{align*}
\frac{dm}{dt} &= r_{\text{in}} - r_{\text{out}} \\
&= f_{\text{in}}c_{\text{in}} - f_{\text{out}} c_{\text{out}}.
\end{align*}
Then here, to account for the inflow concentration at time $t$, I used the Heaviside function so that $c_{\text{in}} = \alpha(1-H(t-T))$. Now subbing in our known constant flow rate and constant volume in the tank, 
\begin{align*}
\require{cancel}
\frac{dm}{dt} &= q\cdot \alpha\left(1 - H(t-T)\right) - q \cdot \frac{m}{V_{0}}\\
\mathcal{L}\left[\frac{dm}{dt}\right] &= q\cdot \alpha\left(\frac{1}{s} -\frac{ e^{-Ts}}{s}\right) - \frac{q}{V_{0}}\mathcal{L}\left[m\right] \\
\mathcal{L}\left[m\right] \left(s + \frac{q}{V_{0}}\right) - \cancelto{0}{\mathcal{L}[0]} &= q\cdot \alpha \left(\frac{1}{s} - \frac{e^{-Ts}}{s}\right)
 \\
\mathcal{L}[m] &= \frac{q\cdot \alpha \left(\frac{1}{s} - \frac{e^{-Ts}}{s}\right)}{s + \frac{q}{V_{0}}}.
\end{align*}
But this to me looks entirely non-trivial to find an inverse Laplace for, especially because of the second term - I'm beginning to think that I've set up the entire equation incorrectly on which to apply the Laplace transform... can anyone spot what I've done wrong or what I am supposed to do from here?
 A: I realized that the easier approach than trying to evaluate what I had reached in my derivation is to introduce some dimensionless variables to simplify things. 

We have just like above that 
$$
\frac{dm}{dt} = q\cdot \alpha\left(1 - H(t-T)\right) - q \cdot \frac{m}{V_{0}}.
$$
Then by dimensional analysis, we have that $\left[\frac{V_{0}}{q}\right] = T$, and so we introduce a characteristic time by $t_{c} = \frac{V_{0}}{q}$. Then $\tau = \frac{t}{t_{c}}$ is dimensionless and by the Chain rule $\frac{dm}{dt} = \frac{dm}{d\tau}\frac{d\tau}{dt} = \frac{q}{V_{0}}\frac{dm}{d\tau}$. We also introduce a dimensionless mass $\mathcal{M}$ via $m_{c} = V_{0}\alpha$, and upon subbing dimensionless quantities into the DE have 
$$
\frac{d\mathcal{M}}{d\tau} = 1 - H\left(\tau - \frac{qT}{V_{0}}\right).
$$
From here, the Laplace transform is easily performed to acquire 
$$
\mathcal{L}[\mathcal{M}] = \frac{1}{s} - \frac{1}{s+1} - e^{\frac{-qTs}{V_{0}}}\left(\frac{1}{s} - \frac{1}{s+1}\right).
$$
This is much easier to apply the inverse Laplace operator to than what I had in my original question, and using the shift theorems 
$$
\mathcal{L}^{-1}\left[\mathcal{L}\left[\mathcal{M}\right]\right] = \mathcal{M} = 1 - e^{-\tau} - H\left(\tau - \frac{qT}{V_{0}}\right)\left[1 - e^{-\left(\tau - \frac{qT}{V_{0}}\right)}\right].
$$
From here, substituting dimensional values back in and evaluating at $t = 2T$ isn't too hard. 
