Chapter 1.9, Problem 13E

a.

To determine

Find the amount of chemical in the lungs before breathing .

Answer to Problem 13E

The amount of chemical in the lungs before breathing is $2.0$ 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 = $2.0$ L, W = $0.5$ L, $\gamma$ = $5.0$ mmol/L, $c_0$ = $1.0$ 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

  = $2.0\times 1.0$ mmol

  = $2.0$ mmol

Conclusion:

  $2.0$ mmol .

b.

To determine

Find the amount of chemical breathed out.

Answer to Problem 13E

The amount of chemical breathed outis $0.5$ 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 = $2.0$ L, W = $0.5$ L, $\gamma$ = $5.0$ mmol/L, $c_0$ = $1.0$ 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 ( $c_0$ ).

  = $W\times c_0$

  = $0.5\times 1.0$ mmol

  = $0.5$ mmol

Conclusion:

  $0.5$ mmol .

c.

To determine

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

Answer to Problem 13E

The amount of chemical in the lungs after breathing outis $1.5$ 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 = $2.0$ L, W = $0.5$ L, $\gamma$ = $5.0$ mmol/L, $c_0$ = $1.0$ 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 ( $c_0$ ).

  $\begin{array}{l}=\left(V-W\right)\times c_0\\ =\left(2.0-0.5\right)\times 1.0\text{mmol}\\ =1.5\times 1.0\text{mmol}\\ \text{=1}\text{.5}\text{mmol}\end{array}$

Conclusion:

  $1.5$ mmol.

d.

To determine

Find the amount of chemical breathed in .

Answer to Problem 13E

The amount of chemical breathed in is $\text{1}.\text{5}$ 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 = $2.0$ L, W = $0.5$ L, $\gamma$ = $5.0$ mmol/L, $c_0$ = $1.0$ 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 ( $c_0$ ).

  $\begin{array}{l}=\left(V-W\right)\times c_0\\ =\left(2.0-0.5\right)\times 1.0\text{mmol}\\ =1.5\times 1.0\text{mmol}\\ \text{=1}\text{.5}\text{mmol}\end{array}$

Conclusion:

  $\text{1}.\text{5}$ mmol

e.

To determine

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

Answer to Problem 13E

The amount of chemical in the lungs after breathing in is $4.0\text{ mmol}$ .

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 = $2.0$ L, W = $0.5$ L, $\gamma$ = $5.0$ mmol/L, $c_0$ = $1.0$ 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)} $\times$ the ambient concentration ( $\gamma$ )]

  $\begin{array}{l}=\left[\left(V-W\right)\times c_0\right]+\left[W\times \gamma \right]\\ =\left[\left\{\text{2}.0-0.\text{5}\right\}\times \text{ 1}.0\right]+\left[0.\text{5 }\times \text{ 5}.0\right]\\ =\left[\text{1}.\text{5 }\times \text{ 1}.0\right]+\left[\text{2}.\text{5}\right]\\ =4.0\text{ mmol}\end{array}$

Conclusion:

  $4.0\text{ mmol}$

f.

To determine

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

Answer to Problem 13E

The concentration of chemical in the lungs after breathing in is 2 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 = $2.0$ L, W = $0.5$ L, $\gamma$ = $5.0$ mmol/L, $c_0$ = $1.0$ 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 ( $\gamma$ )]

  $\begin{array}{l}=\left[\left(V-W\right)\times c_0\right]+\left[W\times \gamma \right]\\ =\left[\left\{\text{2}.0-0.\text{5}\right\}\times \text{ 1}.0\right]+\left[0.\text{5 }\times \text{ 5}.0\right]\\ =\left[\text{1}.\text{5 }\times \text{ 1}.0\right]+\left[\text{2}.\text{5}\right]\\ =4.0\text{ mmol}\end{array}$

And, we know that total volume of lungs is = $2.0$ L

So, The concentration of chemical in the lungs after breathing in = $4.0/2.0$ mmol/L

  = 2 mmol/L

Conclusion:

2 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 $\text{q}=\text{W}/\text{V}$ .

Answer to Problem 13E

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 = $2.0$ L, W = $0.5$ L, $\gamma$ = $5.0$ mmol/L, $c_0$ = $1.0$ 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	$2.0$ mmol	$12.0$ mmol
b. The amount of chemical breathed out	$0.5$ mmol	$1.2$ mmol
c. The amount of chemical in the lungs after breathing out	$1.5$ mmol	$10.8$ mmol
d. The amount of chemical breathed in	$1.5$ mmol	$3.0$ mmol
e. The amount of chemical in the lungs after breathing in	$3.5$ mmol	$13.8$ mmol
f. The concentration of chemical in the lungs after breathing in	$1.75$ mmol/L	$2.3$ mmol/L

Conclusion :

The comparison is shown above.