Chapter 1.9, Problem 16E

a.

To determine

Find the amount of chemical in the lungs before breathing .

Answer to Problem 16E

The amount of chemical in the lungs before breathing is $90.0$ 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= $\text{1}0.0$ L, W= $0.2$ L, $\gamma$ = $1.0$ 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

  $=\text{ 1}0.0\text{L}\times \text{ 9}.0$ mmol

  = $90.0$ mmol

Conclusion:

  $90.0$ mmol

b.

To determine

Find the amount of chemical breathed out.

Answer to Problem 16E

The amount of chemical breathed out is $1.8$ 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= $\text{1}0.0$ L, W= $0.2$ L, $\gamma$ = $1.0$ 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:

The amount of chemical breathed out is given by,

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

  = $W\times c_0$

  = $0.\text{2 }\times \text{ 9}.0$ mmol

  = $1.8$ mmol

Conclusion:

  $1.8$ mmol

c.

To determine

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

Answer to Problem 16E

The amount of chemical in the lungs after breathing out is $7.2$ 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= $\text{1}0.0$ L, W= $0.2$ L, $\gamma$ = $1.0$ 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:

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 ( $c_0$ ).

  = $\left(V-W\right)\times c_0$

  = $\left\{\text{1}0.0-0.\text{2}\right\}\times \text{ 9}.0$ mmol

  $=0.\text{8 }\times \text{ 9}.0$ mmol

  = $7.2$ mmol

Conclusion:

  $7.2$ mmol

d.

To determine

Find the amount of chemical breathed in .

Answer to Problem 16E

The amount of chemical breathed in is $7.2$ 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= $\text{1}0.0$ L, W= $0.2$ L, $\gamma$ = $1.0$ 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:

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 ( $c_0$ ).

  = $\left(V-W\right)\times c_0$

  = $\left\{\text{1}0.0-0.\text{2}\right\}\times \text{ 9}.0$ mmol

  = $0.\text{8 }\times \text{ 9}.0$ mmol

  = $7.2$ mmol

Conclusion:

  $7.2$ mmol

e.

To determine

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

Answer to Problem 16E

The amount of chemical in the lungs after breathing in is $16.4$ 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= $\text{1}0.0$ L, W= $0.2$ L, $\gamma$ = $1.0$ 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:

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 ( $\gamma$ )

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

  = $\left[\left\{\text{2}.0-0.\text{2}\right\}\times \text{ 9}.0\right]+\left[0.\text{2 }\times \text{ 1}.0\right]$ mmol

  = $\left[\text{1}.\text{8 }\times \text{ 9}.0\right]+\left[0.\text{2}\right]$ mmol

  = $\left[\text{16}.\text{2}\right]+\left[0.\text{2}\right]$ mmol

  = $16.4$ mmol

Conclusion:

  $16.4$ mmol

f.

To determine

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

Answer to Problem 16E

The concentration of chemical in the lungs after breathing in is $8.94$ 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= $\text{1}0.0$ L, W= $0.2$ L, $\gamma$ = $1.0$ 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:

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 ( $\gamma$ )]

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

  = $\left[\left\{\text{1}0.0-0.\text{2}\right\}\times \text{ 9}.0\right]+\left[0.\text{2 }\times \text{ 1}.0\right]$ mmol

  $=\left[\text{9}.\text{8 }\times \text{ 9}.0\right]+\left[\text{1}.\text{2}\right]$ mmol

  $=\left[\text{88}.\text{2}\right]+\left[\text{1}.\text{2}\right]$ mmol

  = $8.94$ mmol

And, we know that total volume of lungs is = $10.0$ L

So, The concentration of chemical in the lungs after breathing in = $8.94/10.0$ mmol/L

  = $8.94$ mmol/L

Conclusion:

  $8.94$ 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 16E

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= $\text{1}0.0$ L, W= $0.2$ L, $\gamma$ = $1.0$ 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:

step	In this question	In discrete-time dynamical
a. The amount of chemical in the lungs before breathing	$90.0$ mmol	$12.0$ mmol
b. The amount of chemical breathed out	$1.8$ mmol	$1.2$ mmol
c. The amount of chemical in the lungs after breathing out	$7.2$ mmol	$10.8$ mmol
d. The amount of chemical breathed in	$7.2$ mmol	$3.0$ mmol
e. The amount of chemical in the lungs after breathing in	$16.4$ mmol	$13.8$ mmol
f. The concentration of chemical in the lungs after breathing in	$8.94$ mmol/L	$2.3$ mmol/L

Conclusion:

The comparison is shown above.