What is the concentration as a function of time for an open system I know how to find the concentration as a function of time for a closed (or constant volume) system.
if x is the amount of solute;
X'=(rate inflow)(concentration in)-(rate outflow)(concentration outflow)
However I have trouble with systems that have different inflows and outflows (changing volume).
The (concentration outflow) term is the one I am haveing difficulty writing. I know it is (amount of solute out/volume out), but what is the amount of solute out?
 A: Suppose we have a system with a constant inflow of a solute and a constant volume outflow, although not necessarily constant concentration. The change in volume per unit of time of the system is then
$$\frac{dV_S}{dt}=RV_{in}-RV_{out} \; .$$
This one can be solved easily giving
$$V_S(t)=V_S(0)+(RV_{in}-RV_{out})t \; .$$
For the mass of solvant in the system, we have the equation
$$\frac{dM_S}{dt}=RM_{in}-\frac{dM_{out}}{dt} \; .$$
Of course, the rate of outflow of mass of solvant will depend on the concentration of the solute in the system and the rate of outflow so that
$$\frac{dM_{out}}{dt}=RV_{out} \cdot C_S \; ,$$
leading to the mass balance equation 
$$\frac{dM_S}{dt}=RM_{in}-RV_{out} \cdot C_S \; .$$
The concentration of solute in the system is then $C_S=M_S/V_S$ which changes in time as
$$\frac{dC_S}{dt}=\frac{d}{dt}\left(\frac{M_S}{V_S}\right)= \frac{1}{V_S}\frac{dM_S}{dt}-\frac{M_S}{V_S^2}\frac{dV_S}{dt} \; .$$
Combining our equations, we get
$$\frac{dC_S}{dt} = \frac{1}{V_S}\left(\frac{dM_S}{dt} -C_S\frac{dV_S}{dt}\right) $$
$$\frac{dC_S}{dt} = \frac{1}{V_S}\left(RM_{in}-RV_{out} \cdot C_S-C_S(RV_{in}-RV_{out})\right) $$  
$$\frac{dC_S}{dt} =  \frac{1}{V_S}\left(RM_{in}-C_S\cdot RV_{in}\right) $$
Remembering our solution for the volume
$$\frac{dC_S}{dt} =  \frac{RM_{in}-C_S\cdot RV_{in}}{V_S(0)+(RV_{in}-RV_{out})t} $$
which can be solved by separation of variables
$$\frac{dC_S}{RM_{in}-C_S\cdot RV_{in}} =  \frac{dt}{V_S(0)+(RV_{in}-RV_{out})t} $$
Integrating gives
$$\frac{-1}{RV_{in}}\log(RM_{in}-C_S\cdot RV_{in}) =  \frac{1}{(RV_{in}-RV_{out})}\log(V_S(0)+(RV_{in}-RV_{out})t) + K $$
where we introduced some integration constant $K$ which we'll specify later. Working further out
$$RM_{in}-C_S\cdot RV_{in} =  A (V_S(0)+(RV_{in}-RV_{out})t)^{\frac{RV_{in}}{(RV_{out}-RV_{in})}} $$
in which $\exp(K)=A$. The constant $A$ should be chosen in such a way that
$$RM_{in}-C_S(0)\cdot RV_{in} =  A (V_S(0))^{\frac{RV_{in}}{(RV_{out}-RV_{in})}} \; .$$
Finally,
$$C_S =  \frac{RM_{in} - A (V_S(0)+(RV_{in}-RV_{out})t)^{\frac{RV_{in}}{(RV_{out}-RV_{in})}}}{RV_{in}} \; ,$$
or with the formula for $A$ substituted in
$$C_S(t) =  \frac{RM_{in} - (RM_{in}-C_S(0)\cdot RV_{in}) (1+(\frac{RV_{in}-RV_{out}}{V_S(0)})t)^{\frac{RV_{in}}{(RV_{out}-RV_{in})}}}{RV_{in}} \; .$$
Hope this helps. Sorry for the longwinded derivation with little text and too many formulas. Feel free to ask questions if something is unclear.
