Question

Air containing 0.06% carbon dioxide is pumped into a room whose volume is $8000\mathit{}\mathit{f}{\mathit{t}}^3$ . The air is pumped in at a rate of $2000\mathit{}\mathit{f}{\mathit{t}}^3/\mathit{m}\mathit{i}\mathit{n}$ , and the circulated air is then pumped out at the same rate. If there is an initial concentration of $0.2\mathit{\%}$carbon dioxide in the room, determine the subsequent amount in the room at time t. What is the concentration of carbon dioxide at 10 minutes? What is the steady-state, or equilibrium, concentration of carbon dioxide?

Short answer

The steady state equilibrium concentration of carbon dioxide in the room is$0.06\%$

Step-by-step solution

Concept of modelling with first order differential equations:

A first-order differential equation is defined by an equation: $\frac{dy}{dx}=f(x,y)$of two variables x and y with its function $\mathit{f}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$ defined on a region in the xy plane.

Given data.

• Let the amount of carbon dioxide in the room be$x\left(t\right)f{t}^3$.
• Volume of the room is $8000f{t}^3$.
• Air containing carbon dioxide of concentration $0.06\%$ is pumped in the room.
• Rate of flow of air in the room is $2000f{t}^3/min$.
• Initial concentration of carbon dioxide in the room is $0.2\%$.

For the room $x\left(0\right)=16$.

Find the  x value:

$$\begin{gathered} Rateofinflow=2000\times 0.06\% \\ =1.2f{t}^3/min \\ Rateofoutflow=\frac{2000.x\left(t\right)}{8000} \\ =0.25x\left(t\right)f{t}^3/min \end{gathered}$$

$$\begin{gathered} Rateofchangeofx=Rateofinflow-Rateofoutflow \\ {D}_x=\left[1.2\right]-\left[0.25x\right] \\ (D+0.25)x=1.2 \\ \frac{dx}{dt}+0.25x=1.2 \\ {e}^{0.25t}\frac{dx}{dt}+0.25e^{0.25t}=1.2e^{0.25t} \\ \frac{d\left[x{e}^{0.25t}\right]}{dt}=1.2e^{0.25t} \\ x{e}^{0.25t}=\int 1.2e^{0.25t}dt \\ =\frac{1.2t{e}^{0.25t}}{0.25}+c \\ =\frac{24}{5}{e}^{0.25t}+c \end{gathered}$$

role="math" localid="1668609983651" $$\begin{gathered} x{e}^{0.25t}=4.8e^{0.25t}+c \\ x=4.8+c{e}^{-0.25t} \\ x\left(0\right)=4.8+c{e}^0 \\ 16=4.8+c \\ c=11.2 \\ x\left(t\right)=4.8+11.2e^{-0.25t} \end{gathered}$$

Find the value of c(10) :

The percentage concentration of carbon dioxide in the room after time is given by:

$$\begin{gathered} \%x\left(t\right)=\frac{x\left(t\right)}{8000}.100\% \\ c\left(t\right)=\frac{4.8+11.2e^{-0.25t}}{80} \\ =0.06+0.14e^{-0.25t} \end{gathered}$$

The concentration of carbon dioxide in the room after 10 minutes is given by:

$$\begin{gathered} C\left(10\right)=0.06+0.14e^{-0.25\times 10} \\ =0.06+0.14e^{-2.5} \\ =0.06+0.14\times 0.082 \\ =0.06+0.0115 \\ =0.0715 \end{gathered}$$

Find the steady state.

The concentration of carbon dioxide after 10 minutes is $0.0715\%$. The steady state equilibrium concentration is given by

$$\begin{gathered} \underset{t\to \infty }{lim}c\left(t\right)=\underset{t\to \infty }{lim}(0.06+0.14e^{-0.025t}) \\ =0.06+0.14e^{-\infty } \\ =0.06+0 \\ =0.06 \end{gathered}$$

So, the steady state equilibrium concentration of carbon dioxide in the room is $0.06\%$.