# The air in a small room 12 ft x 8 ft x 8 ft is 3% carbon monoxide. Starting at t = 0, fresh air...

## Question:

The air in a small room 12 ft x 8 ft x 8 ft is 3% carbon monoxide. Starting at t = 0, fresh air (no carbon monoxide) is blown into the room at a rate of 100 {eq}ft^{3}/min {/eq}. If the air in the room flows out through a vent at the same rate, when will the air in the room be 0.01% carbon monoxide?

## Volume:

A quantity that represents the space occupy by the coordinates or the variable on the plane by the closed surface is known as volume. Its measurable unit is meter cube or feet cube. It essential quantity that shows the ability of the system to store.

Given Data

• The dimension of the room is: {eq}{V_o} = 12\;{\rm{ft}} \times {\rm{8}}\;{\rm{ft}} \times {\rm{8}}\;{\rm{ft}} {/eq}
• The percentage volume of carbon -monoxide at t = 0 time is: {eq}{V_1} = 3\% {/eq}
• The volume of fresh rate with no carbon monoxide is: {eq}F = 100\;{\rm{f}}{{\rm{t}}^3}/{\rm{min}} {/eq}
• The percentage volume of carbon -monoxide at time is: {eq}{V_2} = 0.01\% {/eq}

The percentage of volume of the carbon monoxide in room is: {eq}{V_m} {/eq}

The volume of monoxide at {eq}t = 0 {/eq}

{eq}\begin{align*} {V_o} &= 12\;{\rm{ft}} \times {\rm{8}}\;{\rm{ft}} \times {\rm{8}}\;{\rm{ft}}\\ &{\rm{ = 768}}\;{\rm{f}}{{\rm{t}}^3} \end{align*} {/eq}

The expression for percentage of the room air changes in one minute {eq}T = 1\;{\rm{minutes}} {/eq} is

{eq}K = \left( {F{V_o}} \right)T {/eq}

Substitute the value and solve the above expression

{eq}\begin{align*} K &= \left( {\dfrac{{100\;{\rm{f}}{{\rm{t}}^3}{\rm{/min}}}}{{{\rm{768}}\;{\rm{f}}{{\rm{t}}^3}}} \times 1\;{\rm{min}}} \right)\\ &= {\rm{0}}{\rm{.1302}} \end{align*} {/eq}

The expression for percentage of the carbon monoxide in room in time tis

{eq}\begin{align*} \dfrac{{d{V_m}}}{{dt}}& = - K{V_m}\\ \dfrac{{d{V_m}}}{{{V_m}}} &= - Kdt \end{align*} {/eq}

Integrate the above expression

{eq}\begin{align*} \int_{{V_1}}^{{V_2}} {\dfrac{{d{V_m}}}{{{V_m}}}} &= \int_0^t { - Kdt} \\ \ln \left( {{V_2}} \right) - \ln {V_1} &= - K\left( {t - 0} \right)\\ \ln \dfrac{{{V_2}}}{{{V_1}}} &= - Kt \end{align*} {/eq}

Substitute the value and solve the above expression

{eq}\begin{align*} \ln \dfrac{{0.01}}{3} &= - 0.1302 \times t\\ - 5.7037 &= - 0.1302 \times t\\ t &= 43.80\;{\rm{min}} \end{align*} {/eq}

Thus the air in the room be 0.01% carbon monoxide is {eq}43.80\;{\rm{min}} {/eq}