Question

When people smoke, carbon monoxide is released into the air. In a room of volume 70 m3, air containing 5% carbon monoxide is introduced at a rate of 0.002 m3/ min. The carbon monoxide mixes immediately with the rest of the air in the room and the mixture leaves the room at the same rate as it enters. (a) Which of the following can be used to model the amount of carbon monoxide, Q, in the room at time t, in minutes? dQ dt = 0.002 70 Q − 0.0001 dQ dt = 0.0001 − 0.002Q dQ dt = 0.0001 − 70 0.002 Q dQ dt = 0.002 − 0.002Q dQ dt = 0.0001 − 0.002 70 Q

Answer 1

`\frac {dQ}{dt}=0.0001-\frac {Q}{70}* 0.002`

Step-by-step explanation:

Given that the volume is
`70\ m^3`

The air flows (F) at the rate of
`0.002\ m^3/min`. The concentration of carbon monoxide in the air is 5%.

Let Q be the amount of the carbon monoxide present in the room at time t. then,

`\frac {dQ}{dt}=Rate\ of\ Q_(in)-Rate\ of\ Q_(out)`

Also,

`Rate\ of\ Q_(in)=5\%\ of\ 0.002=0.0001`

`Rate\ of\ Q_(out)=\frac {Q}{70}* 0.002`

Thus,

`\frac {dQ}{dt}=0.0001-\frac {Q}{70}* 0.002`