Chapter 14, Problem 14.2P

(a)

Interpretation Introduction

Interpretation:

The deepest $3\text{cm}$ diameter hole beneath the surface in Ann Arbor, Michigan that can be burrowed by a chipmunk is to be stated.

Concept introduction:

The rate law of the chemical reaction states that the rate of reaction is the function of the concentration of the reactants and the products present in that specific reaction. The rate is actually predicted by the slowest step of the reaction.

Answer to Problem 14.2P

The deepest $3\text{cm}$ diameter hole beneath the surface in Ann Arbor, Michigan that can be burrowed by a chipmunk is $\text{4}\text{.352}\text{m}$.

Explanation of Solution

The given minimum respiration rate of a chipmunk is $1.5\text{micromoles}$ or $1.5\times {10}^{-6}\text{moles}$ of ${\text{O}}_{\text{2}}\text{/min}$.

The given corresponding volumetric rate of gas intake is $0.05{\text{dm}}^{\text{3}}\text{/min}$ at STP.

The given value of diffusivity is $D_{\text{AB}}$ is $1.8\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}$.

The diameter of hole is $3\text{cm}$.

The hole is shown below.

Figure 1

There are two points at top and bottom of hole which are $0$ and L.

The corresponding concentration at $0$ is $C_{\text{A0}}$ and at L is $C_{\text{AL}}$.

The expression for the relation between molar flow rate of A and cross sectional area is given below.

    $F_{\text{AL}}=A_{\text{c}}{W}_{\text{AL}}$        (1)

Where,

$F_{\text{AL}}$ is the molar flow rate of A with length L as shown in figure 1.

$A_{\text{c}}$ is the cross-sectional area.

$W_{\text{AL}}$ is the flux.

The expression for the molar flux at the direction L is given below.

    $W_{\text{AL}}=\frac{D_{\text{AB}}}{\text{L}}\left(C_{\text{A0}}-C_{\text{AL}}\right)$        (2)

Where,

$D_{\text{AB}}$ is the diffusivity.

Substitute $W_{\text{AL}}=\frac{D_{\text{AB}}}{\text{L}}\left(C_{\text{A0}}-C_{\text{AL}}\right)$ in equation (1), where $F_{\text{AL}}$ is the molar flow rate.

    $\begin{array}{c}{F}_{\text{AL}}=A_{\text{c}}\times \frac{D_{\text{AB}}}{\text{L}}\left(C_{\text{A0}}-C_{\text{AL}}\right)\\ \text{L}=A_{\text{c}}\times \frac{D_{\text{AB}}}{F_{\text{AL}}}\left(C_{\text{A0}}-C_{\text{AL}}\right)\end{array}$        (3)

The area of the hole is expressed as follows.

    $A_{\text{c}}=\frac{\pi }{4}{D}^2$        (4)

Where,

$D$ is the diameter.

Substitute $D=3\text{cm}$ in equation (4).

    $\begin{array}{c}{A}_{\text{c}}=\frac{\pi }{4}\times {\left(\frac{3\text{cm}\times 1\text{m}}{100\text{cm}}\right)}^2\\ =\frac{3.14}{4}\times \left(\frac{9{\text{m}}^2}{1000}\right)\\ =7.065\times {10}^{-4}{\text{m}}^2\end{array}$

The relation between molar flow rate and volumetric flow rate is given below.

    $F_{\text{AL}}=C_{\text{AL}}{v}_0$

Where,

$v_0$ is the volumetric flow rate.

Substitute $1.5\times {10}^{-6}\text{moles}$ of ${\text{O}}_{\text{2}}\text{/min}$ as $F_{\text{AL}}$ and $0.05{\text{dm}}^{\text{3}}\text{/min}$ as $v_0$ in the above expression.

    $\begin{array}{c} 1.5\times {10}^{-6}\text{moles/min}=C_{\text{AL}}\times 0.05{\text{dm}}^{\text{3}}\text{/min}\\ \\ C_{\text{AL}}=\frac{1.5\times {10}^{-6}\text{moles/min}}{\frac{0.05{\text{dm}}^{\text{3}}\text{/min}}{1000{\text{dm}}^{\text{3}}}\times {\text{m}}^{\text{3}}}\\ \\ =0.03{\text{moles/m}}^{\text{3}}\end{array}$

The expression for $C_{\text{A0}}$ is given below.

    $C_{\text{A0}}=\frac{P_{\text{T}}}{RT}{y}_{\text{A}}$        (5)

Where,

$P_{\text{T}}$ is the atmospheric pressure.

$R$ is the universal gas constant.

$T$ is the temperature.

If air occupies only $21\%$ oxygen and $79\%$ nitrogen then the value of $y_{\text{A}}$ is $0.21$.

The value of universal gas constant is $8.314\text{J/mol}\cdot \text{K}$ and temperature is $\text{298}\text{K}$ at STP.

The atmospheric pressure at STP is $1.013\times {10}^5{\text{N/m}}^{\text{2}}$.

Substitute the value $P_{\text{T}}$ as $1.013\times {10}^5{\text{N/m}}^{\text{2}}$, $y_{\text{A}}$ as $0.21$, $R$ as $8.314\text{J/mol}\cdot \text{K}$ and $T$ as $\text{298}\text{K}$ in equation (5).

    $\begin{array}{c}{C}_{\text{A0}}=\frac{1.013\times {10}^5{\text{N/m}}^{\text{2}}}{8.314\text{J/mol}\cdot \text{K}\times \text{298}\text{K}}\times 0.21\\ \\ =8.586{\text{mol/m}}^{\text{3}}\end{array}$

Substitute the values of $A_{\text{c}}$ as $7.065\times {10}^{-4}{\text{m}}^2$ $D_{\text{AB}}$ as $1.8\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}$ $F_{\text{AL}}$ as $1.5\times {10}^{-6}\text{moles/min}$ , $C_{\text{A0}}$ as $8.586{\text{mol/m}}^{\text{3}}$ and $C_{\text{AL}}$ as $0.03{\text{mol/m}}^{\text{3}}$ in equation (3).

    $\begin{array}{c}\text{L}=7.065\times {10}^{-4}{\text{m}}^2\times \frac{1.8\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}}{\left( 1.5\times {10}^{-6}\text{moles/min}\times \frac{\min }{\text{60 s}}\right)}\left(8.586{\text{mol/m}}^{\text{3}}-0.03{\text{mol/m}}^{\text{3}}\right)\\ =7.065\times {10}^{-4}{\text{m}}^2\times \frac{1.8\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}}{\left( 1.5\times {10}^{-6}\text{moles/min}\times \frac{\min }{\text{60 s}}\right)}\left(8.556{\text{mol/m}}^{\text{3}}\right)\\ =\text{4}\text{.352}\text{m}\end{array}$

Therefore, the deepest $3\text{cm}$ diameter hole beneath the surface in Ann Arbor, Michigan that can be burrowed by a chipmunk is $\text{4}\text{.352}\text{m}$.

(b)

Interpretation Introduction

Interpretation:

The deepest $3\text{cm}$ diameter hole beneath the surface in Boulder, Colorado that can be burrowed by a chipmunk is to be stated.

Concept introduction:

The rate law of the chemical reaction states that the rate of reaction is the function of the concentration of the reactants and the products present in that specific reaction. The rate is actually predicted by the slowest step of the reaction.

Answer to Problem 14.2P

The deepest $3\text{cm}$ diameter hole beneath the surface in Boulder, Colorado that can be burrowed by a chipmunk is $\text{3}\text{.559}\text{m}$.

Explanation of Solution

The given minimum respiration rate of a chipmunk is $1.5\text{micromoles}$ or $1.5\times {10}^{-6}\text{moles}$ of ${\text{O}}_{\text{2}}\text{/min}$.

The given corresponding volumetric rate of gas intake is $0.05{\text{dm}}^{\text{3}}\text{/min}$ at STP.

The given value of diffusivity is $D_{\text{AB}}$ is $1.8\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}$.

The diameter of hole is $3\text{cm}$.

The hole is shown below.

Figure 1

There are two points at top and bottom of hole which are $0$ and L.

The corresponding concentration at $0$ is $C_{\text{A0}}$ and at L is $C_{\text{AL}}$.

The expression for the relation between molar flow rate of A and cross sectional area is given below.

    $F_{\text{AL}}=A_{\text{c}}{W}_{\text{AL}}$        (1)

Where,

$F_{\text{AL}}$ is the molar flow rate.

$A_{\text{c}}$ is the cross-sectional area.

$W_{\text{AL}}$ is the flux.

The expression for the molar flux at the direction L is given below.

    $W_{\text{AL}}=\frac{D_{\text{AB}}}{\text{L}}\left(C_{\text{A0}}-C_{\text{AL}}\right)$        (2)

Where,

$D_{\text{AB}}$ is the diffusivity.

Substitute $W_{\text{AL}}=\frac{D_{\text{AB}}}{\text{L}}\left(C_{\text{A0}}-C_{\text{AL}}\right)$ in equation (1).

    $\begin{array}{c}{F}_{\text{AL}}=A_{\text{c}}\times \frac{D_{\text{AB}}}{\text{L}}\left(C_{\text{A0}}-C_{\text{AL}}\right)\\ \text{L}=A_{\text{c}}\times \frac{D_{\text{AB}}}{F_{\text{AL}}}\left(C_{\text{A0}}-C_{\text{AL}}\right)\end{array}$        (3)

The area of the hole is expressed as follows.

    $A_{\text{c}}=\frac{\pi }{4}{D}^2$        (4)

Where,

$D$ is the diameter.

Substitute $D=3\text{cm}$ in equation (4).

    $\begin{array}{c}{A}_{\text{c}}=\frac{\pi }{4}\times {\left(\frac{3\text{cm}\times \text{m}}{100\text{cm}}\right)}^2\\ =\frac{3.14}{4}\times \left(\frac{9{\text{m}}^2}{1000}\right)\\ =7.065\times {10}^{-4}{\text{m}}^2\end{array}$

The relation between molar flow rate and volumetric flow rate is given below.

    $F_{\text{AL}}=C_{\text{AL}}{v}_0$

Where,

$v_0$ is the volumetric flow rate.

Substitute $1.5\times {10}^{-6}\text{moles}$ of ${\text{O}}_{\text{2}}\text{/min}$ as $F_{\text{AL}}$ and $0.05{\text{dm}}^{\text{3}}\text{/min}$ as $v_0$ in the above expression.

    $\begin{array}{c} 1.5\times {10}^{-6}\text{moles/min}=C_{\text{AL}}\times 0.05{\text{dm}}^{\text{3}}\text{/min}\\ \\ C_{\text{AL}}=\frac{1.5\times {10}^{-6}\text{moles/min}}{\frac{0.05{\text{dm}}^{\text{3}}\text{/min}}{1000{\text{dm}}^{\text{3}}}\times {\text{m}}^{\text{3}}}\\ \\ =0.03{\text{moles/m}}^{\text{3}}\end{array}$

The expression for $C_{\text{A0}}$ is given below.

    $C_{\text{A0}}=\frac{P_{\text{T}}}{RT}{y}_{\text{A}}$        (5)

Where,

$P_{\text{T}}$ is the atmospheric pressure.

$R$ is the universal gas constant.

$T$ is the temperature.

If air occupies only $21\%$ oxygen and $79\%$ nitrogen then the value of $y_{\text{A}}$ is $0.21$.

The value of universal gas constant is $8.314\text{J/mol}\cdot \text{K}$ and temperature is $\text{298}\text{K}$ at STP.

The atmospheric pressure in Colorado at STP is $0.829\times {10}^5{\text{N/m}}^{\text{2}}$.

Substitute the value $P_{\text{T}}$ as $0.829\times {10}^5{\text{N/m}}^{\text{2}}$, $y_{\text{A}}$ as $0.21$, $R$ as $8.314\text{J/mol}\cdot \text{K}$ and $T$ as $\text{298}\text{K}$ in equation (5).

    $\begin{array}{c}{C}_{\text{A0}}=\frac{0.829\times {10}^5{\text{N/m}}^{\text{2}}}{8.314\text{J/mol}\cdot \text{K}\times \text{298}\text{K}}\times 0.21\\ \\ =7.0266{\text{mol/m}}^{\text{3}}\end{array}$

Substitute the values of $A_{\text{c}}$ as $7.065\times {10}^{-4}{\text{m}}^2$ $D_{\text{AB}}$ as $1.8\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}$ $F_{\text{AL}}$ as $1.5\times {10}^{-6}\text{moles/min}$ , $C_{\text{A0}}$ as $7.0266{\text{mol/m}}^{\text{3}}$ and $C_{\text{AL}}$ as $0.03{\text{mol/m}}^{\text{3}}$ in equation (3).

    $\begin{array}{c}\text{L}=7.065\times {10}^{-4}{\text{m}}^2\times \frac{1.8\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}}{\left( 1.5\times {10}^{-6}\text{moles/min}\times \frac{\min }{\text{60 s}}\right)}\left(7.0266{\text{mol/m}}^{\text{3}}-0.03{\text{mol/m}}^{\text{3}}\right)\\ =7.065\times {10}^{-4}{\text{m}}^2\times \frac{1.8\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}}{\left( 1.5\times {10}^{-6}\text{moles/min}\times \frac{\min }{\text{60 s}}\right)}\left(6.997{\text{mol/m}}^{\text{3}}\right)\\ =\text{3}\text{.559}\text{m}\end{array}$

Therefore, the deepest $3\text{cm}$ diameter hole beneath the surface in Boulder, Colorado that can be burrowed by a chipmunk is $\text{3}\text{.559}\text{m}$.

(c)

Interpretation Introduction

Interpretation:

The deepest $3\text{cm}$ diameter hole beneath the surface in Ann Arbor, Michigan and Boulder, Colorado that can be burrowed by a chipmunk if temperature is $0°\text{F}$ is to be stated.

Concept introduction:

The rate law of the chemical reaction states that the rate of reaction is the function of the concentration of the reactants and the products present in that specific reaction. The rate is actually predicted by the slowest step of the reaction.

Answer to Problem 14.2P

The deepest $3\text{cm}$ diameter hole beneath the surface in Ann Arbor, Michigan and Boulder, Colorado that can be burrowed by a chipmunk if temperature is $0°\text{F}$ is $\text{4}\text{.0}\text{m}$ and $3.\text{564}\text{m}$ respectively.

Explanation of Solution

The given minimum respiration rate of a chipmunk is $1.5\text{micromoles}$ or $1.5\times {10}^{-6}\text{moles}$ of ${\text{O}}_{\text{2}}\text{/min}$.

The given corresponding volumetric rate of gas intake is $0.05{\text{dm}}^{\text{3}}\text{/min}$ at STP.

The given value of diffusivity is $D_{\text{AB}}$ is $1.8\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}$.

The diameter of hole is $3\text{cm}$.

There are two points at top and bottom of hole which are $0$ and L.

The corresponding concentration at $0$ is $C_{\text{A0}}$ and at L is $C_{\text{AL}}$.

The standard temperature is $\text{0°F}$ instead of $\text{298}\text{K}$.

The conversion of $\text{0°F}$ into Kelvin is given below.

    $\begin{array}{c}\text{0°F}=\left(\frac{0-32}{1.8}\right)+273\text{K}\\ =255\text{K}\end{array}$

The expression for the relation between diffusivity, two temperatures and two pressures is given below.

    $D_{\text{AB}}=D_{\text{AB}}\left(T_2,P_2\right)\frac{P_1}{P_2}{\left(\frac{T_2}{T_1}\right)}^{1.75}$        (6)

Substitute the value $T_{\text{1}}$ as $2\text{98}\text{K}$, $T_{\text{2}}$ as $255\text{K}$, and $D_{\text{AB}}$ as $1.8\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}$ in equation (6).

    $\begin{array}{c}{D}_{\text{AB}}=1.8\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}{\left(\frac{255}{298}\right)}^{1.75}\\ =1.516\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}\end{array}$

If air occupies only $21\%$ oxygen and $79\%$ nitrogen then the value of $y_{\text{A}}$ is $0.21$.

The value of universal gas constant is $8.314\text{J/mol}\cdot \text{K}$ and temperature is $\text{298}\text{K}$ at STP.

The atmospheric pressure in Colorado at STP is $0.829\times {10}^5{\text{N/m}}^{\text{2}}$.

For Ann arbor, Michigan, substitute the value $P_{\text{T}}$ as $1.013\times {10}^5{\text{N/m}}^{\text{2}}$, $y_{\text{A}}$ as $0.21$, $R$ as $8.314\text{J/mol}\cdot \text{K}$ and $T$ as $\text{255}\text{K}$ in equation (5).

    $\begin{array}{c}{C}_{\text{A0}}=\frac{1.013\times {10}^5{\text{N/m}}^{\text{2}}}{8.314\text{J/mol}\cdot \text{K}\times \text{255}\text{K}}\times 0.21\\ =10.034{\text{mol/m}}^{\text{3}}\end{array}$

For, Ann, Arbor Substitute the values of $A_{\text{c}}$ as $7.065\times {10}^{-4}{\text{m}}^2$ $D_{\text{AB}}$ as $1.516\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}$ $F_{\text{AL}}$ as $1.5\times {10}^{-6}\text{moles/min}$ , $C_{\text{A0}}$ as $10.034{\text{mol/m}}^{\text{3}}$ and $C_{\text{AL}}$ as $0.03{\text{mol/m}}^{\text{3}}$ in equation (3).

    $\begin{array}{c}\text{L}=7.065\times {10}^{-4}{\text{m}}^2\times \frac{1.516\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}}{\left( 1.5\times {10}^{-6}\text{moles/min}\times \frac{\min }{\text{60 s}}\right)}\left(10.034{\text{mol/m}}^{\text{3}}-0.03{\text{mol/m}}^{\text{3}}\right)\\ =7.065\times {10}^{-4}{\text{m}}^2\times \frac{1.516\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}}{\left( 1.5\times {10}^{-6}\text{moles/min}\times \frac{\min }{\text{60 s}}\right)}\left(10.004{\text{mol/m}}^{\text{3}}\right)\\ =\text{4}\text{.0}\text{m}\end{array}$

For Boulder, Colorado, substitute the value $P_{\text{T}}$ as $0.829\times {10}^5{\text{N/m}}^{\text{2}}$, $y_{\text{A}}$ as $0.21$, $R$ as $8.314\text{J/mol}\cdot \text{K}$ and $T$ as $\text{255}\text{K}$ in equation (5).

    $\begin{array}{c}{C}_{\text{A0}}=\frac{0.829\times {10}^5{\text{N/m}}^{\text{2}}}{8.314\text{J/mol}\cdot \text{K}\times \text{255}\text{K}}\times 0.21\\ =8.211{\text{mol/m}}^{\text{3}}\end{array}$

For, Boulder, Colorado, Substitute the values of $A_{\text{c}}$ as $7.065\times {10}^{-4}{\text{m}}^2$ $D_{\text{AB}}$ as $1.516\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}$, $F_{\text{AL}}$ as $1.5\times {10}^{-6}\text{moles/min}$ , $C_{\text{A0}}$ as $8.211{\text{mol/m}}^{\text{3}}$ and $C_{\text{AL}}$ as $0.03{\text{mol/m}}^{\text{3}}$ in equation (3).

    $\begin{array}{c}\text{L}=7.065\times {10}^{-4}{\text{m}}^2\times \frac{1.516\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}}{\left( 1.5\times {10}^{-6}\text{moles/min}\times \frac{\min }{\text{60 s}}\right)}\left(8.211{\text{mol/m}}^{\text{3}}-0.03{\text{mol/m}}^{\text{3}}\right)\\ =7.065\times {10}^{-4}{\text{m}}^2\times \frac{1.516\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}}{\left( 1.5\times {10}^{-6}\text{moles/min}\times \frac{\min }{\text{60 s}}\right)}\left(8.181{\text{mol/m}}^{\text{3}}\right)\\ =3.\text{564}\text{m}\end{array}$

Therefore, the deepest $3\text{cm}$ diameter hole beneath the surface in Ann Arbor, Michigan and Boulder, Colorado that can be burrowed by a chipmunk if temperature is $0°\text{F}$ is $\text{4}\text{.0}\text{m}$ and $3.\text{564}\text{m}$ respectively

(d)

Interpretation Introduction

Interpretation:

The criticism and the extension of the given problem is to be stated.

Concept introduction:

The rate law of the chemical reaction states that the rate of reaction is the function of the concentration of the reactants and the products present in that specific reaction. The rate is actually predicted by the slowest step of the reaction.

Explanation of Solution

The most important base of the given problem is the composition of air that is air occupies only $21\%$ oxygen and $79\%$ nitrogen. If some other gas is present in the air then the above calculations must be wrong.

If carbon dioxide is added in the air, then the extension of the above calculations takes place by adding the percentage of carbon dioxide in the problem.