Changing Atmosphere

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September 7th 2008, 07:19 AM
ThePerfectHacker

Changing Atmosphere

I came up with my own problem. (Emo)
There is nothing complicated.

Suppose a room (air tight) is filled with $100\%$ oxygen.
In the room there is a machine which takes in $5$ liters/minute of air and gives up pure nitrogen.

Find a way to model the amount of oxygen at any given time.
(Of course use obvious assumption necessary).
September 7th 2008, 08:12 AM
Laurent

Nice problem indeed, thank you. I guess we assume that nitrogen and oxygen mix immediately in the whole volume.

The whole volume is $V$ , the proportion of oxygen at time $t$ (in minutes) is $x(t)$ , so $x(0)=1$ and probably $x(t)\to_{t\to\infty} 0$ .

During an infinitesimal time interval $\Delta t$ , the initial volume of oxygen $Vx(t)$ becomes $Vx(t+\Delta t)=Vx(t)-5\Delta t x(t)$ (since $5\Delta t$ is the volume that was taken away, among which a proportion $x(t)$ was composed of oxygen). So:
$\frac{x(t+\Delta t)-x(t)}{\Delta t}=-\frac{5}{V}x(t)$ .

Stated differently, $\frac{dx}{dt}=-\frac{5}{V}x(t)$ , so that $x(t)=e^{-\frac{5}{V}t}$ . The proportion hence decreases exponentially fast.

Laurent.
September 11th 2008, 09:41 AM
ThePerfectHacker

Here is how I would do it.
Let $y(t)$ be a function which represents amount of oxygen at time $t$ in minutes.
Notice that $y'(t)$ is the rate of change.
Thus, $y'(t) = \text{ rate out } - \text{ rate in }$ .
The rate in is amount of oxygen coming in per minute is $5\cdot \tfrac{y}{V}$ .
The rate out is $0$ .
Thus, $y'(t) = - \tfrac{5}{V}y$ with $y(0) = V$ .