Chapter 1.9, Problem 15E

a.

To determine

Find the amount of chemical in the lungs before breathing.

Answer to Problem 15E

The amount of chemical in the lungs before breathing is $9.0$ mmol

Explanation of Solution

Given :

It is given in the question that the volume of the lungs is V , the amount breathed in and out is W and the ambient concentration is $\gamma$ mmol/L and the values also given V= $1.0$ L, W= $0.9$ L, $\gamma$ = $5.0$ mmol/L, $c_0$ = $9.0$ mmol/L.

Concept Used:

In this we have to use the concept of a model of gas exchange in the lungs.

Calculation:

Initial concentration of chemical in lungs before breathing is:

  = volume of lungs (V) $\times$ concentration of chemical initial present in lungs ( $c_0$ ).

  = $V\times c_0$ mmol

  = $\text{1}.0\text{L}\times \text{9}.0$ mmol

  = $9.0$ mmol

Conclusion:

  $9.0$ mmol

b.

To determine

Find the amount of chemical breathed out.

Answer to Problem 15E

The amount of chemical breathed out is $8.1$ mmol.

Explanation of Solution

Given :

It is given in the question that the the volume of the lungs is V , the amount breathed in and out is W and the ambient concentration is $\gamma$ mmol/L and the values also given V= $1.0$ L, W= $0.9$ L, $\gamma$ = $5.0$ mmol/L, $c_0$ = $9.0$ mmol/L.

Concept Used:

In this we have to use the concept ofa model of gas exchange in the lungs.

Calculation:

The amount of chemical breathed out is given by,

  = the amount breathed out(W) $\times$ concentration of chemical initial present in lungs ( $c_0$ ).

  = $W\times c_0$

  = $0.9\times 0.9$ mmol

  = $8.1$ mmol

Conclusion:

  $8.1$ mmol

c.

To determine

Find the amount of chemical in the lungs after breathing out

Answer to Problem 15E

The amount of chemical in the lungs after breathing out is $0.\text{9}$ mmol.

Explanation of Solution

Given :

It is given in the question that the the volume of the lungs is V , the amount breathed in and out is W and the ambient concentration is $\gamma$ mmol/L and the values also given V= $1.0$ L, W= $0.9$ L, $\gamma$ = $5.0$ mmol/L, $c_0$ = $9.0$ mmol/L.

Concept Used:

In this we have to use the concept ofa model of gas exchange in the lungs.

Calculation:

The amount of chemical in the lungs after breathing out is given by,

  ={the volume of the lungs(V) - the amount breathed out(W)} $\times$ concentration of chemical initial present in lungs ( $c_0$ ).

  = $\left(V-W\right)\times c_0$

  = $\left(\text{1}.0-0.\text{9}\right)\times \text{9}.0$ mmol

  = $0.\text{1 }\times \text{ 9}.0$ mmol

  = $0.\text{9}$ mmol

Conclusion:

  $0.\text{9}$ mmol

d.

To determine

Find the amount of chemical breathed in .

Answer to Problem 15E

The amount of chemical breathed in is $0.\text{9}$ mmol.

Explanation of Solution

Given :

It is given in the question that the volume of the lungs is V , the amount breathed in and out is W and the ambient concentration is $\gamma$ mmol/L and the values also given V= $1.0$ L, W= $0.9$ L, $\gamma$ = $5.0$ mmol/L, $c_0$ = $9.0$ mmol/L.

Concept Used:

In this we have to use the concept ofa model of gas exchange in the lungs.

Calculation:

The amount of chemical in the lungs after breathing in is given by,

  ={the volume of the lungs(V)- the amount breathed in(W)} $\times$ concentration of chemical initial present in lungs ( $c_0$ ).

  = $\left(V-W\right)\times c_0$

  = $\left(\text{1}.0-0.\text{9}\right)\times \text{ 9}.0$ mmol

  = $0.\text{1 }\times \text{ 9}.0$ mmol

  = $0.\text{9}$ mmol

Conclusion:

  $0.\text{9}$ mmol

e.

To determine

Find the amount of chemical in the lungs after breathing in.

Answer to Problem 15E

The amount of chemical in the lungs after breathing in is $5.4$ mmol.

Explanation of Solution

Given :

It is given in the question that the the volume of the lungs is V , the amount breathed in and out is W and the ambient concentration is $\gamma$ mmol/L and the values also given V= $1.0$ L, W= $0.9$ L, $\gamma$ = $5.0$ mmol/L, $c_0$ = $9.0$ mmol/L.

Concept Used:

In this we have to use the concept ofa model of gas exchange in the lungs.

Calculation:

The amount of chemical in the lungs after breathing in,

  =[ {the volume of the lungs(V) - the amount breathed out(W) } $\times$ concentration of chemical initial present in lungs ( $c_0$ )] + [{the amount breathed in(W)} * the ambient concentration ( $\gamma$ )]

  = $\left[\left(V-W\right)\times c_0\right]+\left[W\times \gamma \right]$

  = $[\{\text{1}.0-0.\text{9}\}\times \text{ 9}.0]+[0.\text{9}\times \text{ 5}.0]$ mmol

  $=\left[0.\text{1}\times \text{ 9}.0\right]+\left[\text{4}.\text{5}\right]$ mmol

  $=\left[0.\text{9}\right]+\left[\text{4}.\text{5}\right]$ mmol

  = $5.4$ mmol

Conclusion:

  $5.4$ mmol

f.

To determine

Find the concentration of chemical in the lungs after breathing in.

Answer to Problem 15E

The concentration of chemical in the lungs after breathing in is $5.4$ mmol/L.

Explanation of Solution

Given :

It is given in the question that the the volume of the lungs is V , the amount breathed in and out is W and the ambient concentration is $\gamma$ mmol/L and the values also given V= $1.0$ L, W= $0.9$ L, $\gamma$ = $5.0$ mmol/L, $c_0$ = $9.0$ mmol/L.

Concept Used:

In this we have to use the concept ofa model of gas exchange in the lungs.

Calculation:

The concentration of chemical in the lungs after breathing in= (total amount / total volume) So, The amount of chemical in the lungs after breathing in,

  =[ {the volume of the lungs(V)- the amount breathed out(W) } $\times$ concentration of chemical initial present in lungs ( $c_0$ )] + [{the amount breathed in(W)} $\times$ the ambient concentration ( $\gamma$ )]

  = $\left[\left(V-W\right)\times c_0\right]+\left[W\times \gamma \right]$

  = $\left[\left\{\text{1}.0-0.\text{9}\right\}\times \text{ 9}.0\right]+\left[0.\text{9 }\times \text{ 5}.0\right]$ mmol

  $=\left[0.\text{1 }\times \text{ 9}.0\right]+\left[\text{4}.\text{5}\right]$ mmol

  $=\left[0.\text{9}\right]+\left[\text{4}.\text{5}\right]$ mmol

  = $5.4$ mmol

And, we know that total volume of lungs is = $1.0$ L

So, the concentration of chemical in the lungs after breathing in = $5.4/1.0$ mmol/L

  = $5.4$ mmol/L

Conclusion:

  $5.4$ mmol/L

g.

To determine

Compare this result with the result of using the general lungs discrete-time dynamical system (equation $\mathrm{1.9.1}$ ). Remember that q=W/V .

Answer to Problem 15E

The comparison is shown below and we easily compare it by seeing general lungs discrete − time system to the result.

Explanation of Solution

Given :

It is given in the question that the the volume of the lungs is V , the amount breathed in and out is W and the ambient concentration is $\gamma$ mmol/L and the values also given V= $1.0$ L, W= $0.9$ L, $\gamma$ = $5.0$ mmol/L, $c_0$ = $9.0$ mmol/L.

Concept Used:

In this we have to use the concept ofa model of gas exchange in the lungs.

Calculation:

Step	In this question	In discrete-time dynamical
a. The amount of chemical in the lungs before breathing	$9.0$ mmol	$12.0$ mmol
b. The amount of chemical breathed out	$8.1$ mmol	$1.2$ mmol
c. The amount of chemical in the lungs after breathing out	$0.9$ mmol	$10.8$ mmol
d. The amount of chemical breathed in	$0.9$ mmol	$3.0$ mmol
e. The amount of chemical in the lungs after breathing in	$5.4$ mmol	$13.8$ mmol
f. The concentration of chemical in the lungs after breathing in	$5.4$ mmol/L	$2.3$ mmol/L

Conclusion:

The comparison is shown above.