To solve a mixture question, you have to integrate with respect to mass because mass doesn’t change with volume, only time. Concentration changes with both so that’s hard to deal with.
Unfortunately, most of our measurements are given in concentrations. We need to do some converstions. For simplicity, let
x = mass inside tank
V = volume inside tank
t = time
It’s clear that
dx/dt = rate of mass in – rate of mass out
***This is a really important concept, you have to start here no matter what. You have to start with something you know is true, then convert that into the stuff we are given.
Mass = concentration * volume, so let’s start the conversions.
dx/dt = concentration in * rate of volume in – concentration out * rate of volume out
We know concentration in, rate of volume in and rate of volume out. Concentration out is our problem.
Luckily, it’s just same as the concentration already inside the tank at that moment. So
dx/dt = concentration in * rate of volume in – x/V * rate of volume out.
V may not be a constant. You need to express it in terms of time.
Now we have expressed dx/dt in everything that’s given to us in the question, we can plug those in and do the integration.
V = 200 (in this case, since inflow rate = outflow rate, V is constant)
Concentration in = 0
Rate of volume out = 2
So
dx/dt = -x/200 * 2
dx/dt=-x/100
x’=-x/100
x’/x=-100 (You can do this by the rules of linear ODE as well but it’s simpler to do this as we recognize x’/x is integrated to log x)
log x = -100t + c
Now we put in the initial values, x=200 (g), t=0 (s)
log 200 = c
log x = -100t + log 200
When the concentration is 1% of the original, i.e. only 2g left in the tank
log 2 = -100t + log 200
100t = log 200 – log 2
100t = log (200/2)
t=log(100)/100
There done!
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