Question

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

Answer 1

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Step-by-step explanation:

The rate-in;

`R_(in)`
`= (0.04)/(100)*2000`

`R_(in)` = 0.8

The rate-out

`R_(out)` =
`(A)/(6000)*2000`

`R_(out)` =
`(A)/(3)`

We can say that:

`(dA)/(dt)=`
`0.8-(A)/(3)`

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

`(dA)/(dt) +(A)/(3) =0.8`

Integration of the above linear equation =

`e^{\int\limits \frac {1}{3}dt } =`
`e^{(1)/(3)t`

so we have:

`e^{(1)/(3)t}(dA)/(dt)} +(1)/(3)e^{(1)/(3)t}A`
`= 0.8e^{(1)/(3)t`

`(d)/(dt)[e^{(1)/(3)t}A]`
`= 0.8e^{(1)/(3)t`

`Ae^{(1)/(3)t} =2.4e(1)/(3)t +C`

∴
`A(t) = 2.4 +Ce^{-(1)/(3)t`

Since A(0) = 12

Then;

`12 =2.4 + Ce^{-(1)/(3)}(0)`

`C= 12-2.4`

`C =9.6`

Hence;

`A(t) = 2.4 +9.6e^{-(t)/(3)}`

`A(0) = 2.4 +9.6e^{-(10)/(3)}`

`A(t) = 2.74`

∴ the concentration at 10 minutes is ;

=
`(2.74)/(6000)*100`%

= 0.0456667 %

= 0.046% to three decimal places