Chapter 2.6, Problem 21P

(A)

Interpretation Introduction

Interpretation:

The pressure of the air at $0°\text{F}$.

Concept Introduction:

Write the ideal gas law for states 1 and 2.

$\frac{P_1{V}_1}{P_2{V}_2}=\frac{N_{1}R{T}_1}{N_{2}R{T}_2}$

Here, initial and final volume is $V_1\text{and}{V}_2$, initial and final temperature is $T_1\text{and}{T}_2$, number of moles in states 1 and 2 is $N_1$ and $N_2$, initial and final pressure is $P_1\text{and}{P}_2$ respectively.

Explanation of Solution

Given information:

Initial Temperature is $T_{\text{i}}=0°\text{F}$.

Final Temperature is $T_{\text{f}}=70°\text{F}$.

Pressure of the air is $P=1\text{ atm}$.

Convert the unit of initial temperature from degree Fahrenheit to Kelvin.

$T_K=\left(T_{°F}+459.67\right)\left(5/9\right)$        (1)

Here, temperature in Kelvin is $T_K$, and temperature in degree Fahrenheit is $T_{°F}$.

Substitute $\text{0°F}$ for $T_{°F}$ in Equation (1).

$\begin{array}{l}{T}_K=\left(0+459.67\right)\left(5/9\right)\\ T_{\text{i}}=255\text{K}\end{array}$

Here, initial temperature is $T_{\text{i}}$.

Convert the unit of initial temperature from degree Fahrenheit to Kelvin.

$T_K=\left(T_{°F}+459.67\right)\left(5/9\right)$        (2)

Substitute $\text{70°F}$ for $T_{°F}$ in Equation (2).

$\begin{array}{l}{T}_K=\left(70+459.67\right)\left(5/9\right)\\ T_{\text{f}}=294\text{K}\end{array}$

Here, final temperature is $T_{\text{f}}$.

Write the ideal gas law for states 1 and 2.

$\frac{P_1{V}_1}{P_2{V}_2}=\frac{N_{1}R{T}_1}{N_{2}R{T}_2}$        (3)

The volume is constant for an isochoric system. For a closed system, number of moles must be constant.

The Equation (3) is reduces as follows.

$\begin{array}{l}\frac{P_1}{P_2}=\frac{T_1}{T_2}\\ P_2=P_1\frac{T_2}{T_1}\end{array}$        (4)

Substitute $\text{1}\text{atm}$ for $P_1$, $255\text{ K}$ for $T_2$, and $294\text{ K}$ for $T_1$ in Equation (4).

$\begin{array}{c}{P}_2=\left(\text{1}\text{atm}\right)\left(\frac{255\text{ K}}{294\text{ K}}\right)\\ =0.8678\text{atm}\end{array}$

Thus, the final pressure is $\text{0}\text{.8678}\text{atm}$.

(B)

Interpretation Introduction

Interpretation:

The pressure of the air at $105°\text{F}$.

Explanation of Solution

Given information:

Initial Temperature is $T_{\text{i}}=70°\text{F}$.

Final Temperature is $T_{\text{f}}=105°\text{F}$.

Convert the unit of initial temperature from degree Fahrenheit to Kelvin.

$T_K=\left(T_{°F}+459.67\right)\left(5/9\right)$        (5)

Substitute $\text{70°F}$ for $T_{°F}$ in Equation (5).

$\begin{array}{l}{T}_K=\left(70+459.67\right)\left(5/9\right)\\ T_{\text{i}}=294\text{K}\end{array}$

Convert the unit of initial temperature from degree Fahrenheit to Kelvin.

$T_K=\left(T_{°F}+459.67\right)\left(5/9\right)$        (6)

Substitute $\text{105°F}$ for $T_{°F}$ in Equation (6).

$\begin{array}{c}{T}_K=\left(105+459.67\right)\left(5/9\right)\\ T_{\text{f}}=313\text{K}\end{array}$

Substitute $\text{1}\text{atm}$ for $P_1$, $313\text{ K}$ for $T_2$, and $294\text{ K}$ for $T_1$ in Equation (4).

$\begin{array}{c}{P}_2=P_1\frac{T_2}{T_1}\\ =\left(\text{1}\text{atm}\right)\left(\frac{294\text{ K}}{313\text{ K}}\right)\\ =1.0661\text{atm}\end{array}$

Thus, the final pressure is $1.0661\text{atm}$.

(C)

Interpretation Introduction

Interpretation:

The volume of air that the lungs must hold when fully expanded.

Concept Introduction:

Write the expression to obtain the minimum number of moles $\left(N_{\min }\right)$.

$N_{\min }=\frac{P{V}_{\text{collapsed}}}{R{T}_{\min }}$

Here, gas constant is $R$, volume of air in lungs when it is fully collapsed is $V_{\text{collapsed}}$.

Write the expression to obtain the maximum volume at maximum temperature.

$V_{\text{maximum}}=\frac{N_{\min }R{T}_{\max }}{P}$

Here, maximum temperature is $T_{\max }$.

Write the expression to obtain the volume of air must the lungs hold when fully expanded.

$V_{\text{lung}}=V_{\text{maximum}}-V_{\text{collapsed}}$

Here, volume of air must the lungs hold when fully expanded is $V_{\text{lung}}$.

Explanation of Solution

Since the system is closed, the number of moles is constant and can be found from minimum volume. The minimum volume occurs at the minimum temperature.

Write the expression to obtain the minimum number of moles $\left(N_{\min }\right)$.

$N_{\min }=\frac{P{V}_{\text{collapsed}}}{R{T}_{\min }}$        (7)

Substitute $\text{1}\text{atm}$ for $P$, $\text{125,000}{\text{m}}^{\text{3}}$ for $V_{\text{collapsed}}$, $\text{8}\text{.314}\text{Pa}\cdot {\text{m}}^{\text{3}}/\text{mol}\cdot \text{K}$ for $R$, and $\text{255}\text{K}$ for $T_{\min }$ in Equation (7).

$\begin{array}{c}{N}_{\min }=\frac{\left(\text{1}\text{atm}\right)\left(\text{125,000}{\text{m}}^{\text{3}}\right)}{\left(\text{8}\text{.314}\text{Pa}\cdot {\text{m}}^{\text{3}}/\text{mol}\cdot \text{K}\right)\left(\text{255}\text{K}\right)}\\ =\frac{\left[\left(\text{1}\text{atm}\right)\left(0.1MPa/\text{1}\text{atm}\right)\right]\left(\text{125,000}{\text{m}}^{\text{3}}\right)}{\left(\text{8}\text{.314}\text{Pa}\cdot {\text{m}}^{\text{3}}/\text{mol}\cdot \text{K}\right)\left(\text{255}\text{K}\right)}\\ =\frac{\left(0.1MPa\right)\left(\text{125,000}{\text{m}}^{\text{3}}\right)}{\left(\text{8}\text{.314}\text{Pa}\cdot {\text{m}}^{\text{3}}/\text{mol}\cdot \text{K}\right)\left(\text{255}\text{K}\right)}\\ =5.90\text{Mmol}\end{array}$

Write the expression to obtain the maximum volume at maximum temperature.

$V_{\text{maximum}}=\frac{N_{\min }R{T}_{\max }}{P}$        (8)

Substitute $\text{1}\text{atm}$ for $P$, $\text{5}\text{.90}\text{Mmol}$ for $N_{\min }$, $\text{8}\text{.314}\text{Pa}\cdot {\text{m}}^{\text{3}}/\text{mol}\cdot \text{K}$ for $R$, and $\text{313}\text{K}$ for $T_{\max }$ in Equation (8).

$\begin{array}{c}{V}_{\text{maximum}}=\frac{\left(\text{5}\text{.90}\text{Mmol}\right)\left(\text{8}\text{.314}\text{Pa}\cdot {\text{m}}^{\text{3}}/\text{mol}\cdot \text{K}\right)\left(\text{313}\text{K}\right)}{\text{1}\text{atm}\left(0.1MPa/\text{1}\text{atm}\right)}\\ =\frac{\left(\text{5}\text{.90}\text{Mmol}\right)\left(\text{8}\text{.314}\text{Pa}\cdot {\text{m}}^{\text{3}}/\text{mol}\cdot \text{K}\right)\left(\text{313}\text{K}\right)}{0.1MPa}\\ =153,534.63{\text{m}}^{\text{3}}\end{array}$

Write the expression to obtain the volume of air must the lungs hold when fully expanded.

$V_{\text{lung}}=V_{\text{maximum}}-V_{\text{collapsed}}$        (9)

Substitute $153,534.63{\text{m}}^{\text{3}}$ for $V_{\text{maximum}}$, and $\text{125,000}{\text{m}}^{\text{3}}$ for $V_{\text{collapsed}}$ in Equation (9).

$\begin{array}{c}{V}_{\text{lung}}=153,534.63{\text{m}}^{\text{3}}-\text{125,000}{\text{m}}^{\text{3}}\\ =28,534.63{\text{m}}^{\text{3}}\end{array}$

Thus, the volume of air must the lungs hold when fully expanded is $28,534.63{\text{m}}^{\text{3}}$.

(D)

Interpretation Introduction

Interpretation:

The change in internal energy that the air undergoes.

Concept Introduction:

Write the expression to obtain the change in specific internal energy $\left(d\widehat{U}\right)$.

$d\widehat{U}={\widehat{C}}_{V}dT$

Here, constant volume heat capacity is ${\widehat{C}}_V$, and change in temperature is $dT$.

Write the expression to obtain the constant pressure heat capacity of nitrogen for ideal gas $\left(C_{P,N_2}^{\ast }\right)$.

$\begin{array}{l}\frac{C_{P,N_2}^{\ast }}{R}=A+BT+C{T}^2+D{T}^3+E{T}^4\\ C_{P,N_2}^{\ast }=R\left(A+BT+C{T}^2+D{T}^3+E{T}^4\right)\end{array}$

Here, gas constant is $R$, temperature is $T$, and constant are $A,B,C,D\text{and}E$ respectively.

Write the expression to obtain the constant pressure heat capacity of oxygen for ideal gas $\left(C_{P,O_2}^{\ast }\right)$.

$\begin{array}{l}\frac{C_{P,O_2}^{\ast }}{R}=A+BT+C{T}^2+D{T}^3+E{T}^4\\ C_{P,O_2}^{\ast }=R\left(A+BT+C{T}^2+D{T}^3+E{T}^4\right)\end{array}$

Write the expression to obtain the parameter $R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}dT}$.

$R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}dT}=R\left(T_2-T_1\right)$

Write the expression to obtain the change in internal energy $\left(\Delta U\right)$.

$\Delta U=\Delta \widehat{U}\left(N_{\min }\right)$

Explanation of Solution

Since this is being modeled as an ideal gas, internal energy is depends only on temperature.

Write the expression to obtain the change in specific internal energy $\left(d\widehat{U}\right)$.

$d\widehat{U}={\widehat{C}}_{V}dT$        (10)

As we do not have the heat capacity at constant volume for the air, use constant pressure heat capacity in Equation (10).

Rewrite Equation (10).

$\begin{array}{c}d\widehat{U}=C_{V,air}^{\ast }dT\\ =\left(C_{P,air}^{\ast }-R\right)dT\end{array}$        (11)

Here, constant pressure heat capacity of air for ideal gas is $C_{P,air}^{\ast }$, and constant volume heat capacity of air for ideal gas is $C_{V,air}^{\ast }$.

Substitute $0.79C_{P,N_2}^{\ast }+0.21C_{P,O_2}^{\ast }$ for $C_{P,air}^{\ast }$ in Equation (11).

$d\widehat{U}=\left(0.79C_{P,N_2}^{\ast }+0.21C_{P,O_2}^{\ast }-R\right)dT$        (12)

Here, constant pressure heat capacity of nitrogen for ideal gas is $C_{P,N_2}^{\ast }$, and constant pressure heat capacity of oxygen for ideal gas is $C_{P,O_2}^{\ast }$.

Integrate Equation (12).

$\begin{array}{c}{\displaystyle \int d\widehat{U}}={\displaystyle \underset{T_1}{\overset{T_2}{\int }}\left(0.79C_{P,N_2}^{\ast }+0.21C_{P,O_2}^{\ast }-R\right)}dT\\ =0.79{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\left(C_{P,N_2}^{\ast }\right)}dT+0.21{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\left(C_{P,O_2}^{\ast }\right)}dT-R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}dT}\end{array}$        (13)

Write the expression to obtain the constant pressure heat capacity of nitrogen for ideal gas $\left(C_{P,N_2}^{\ast }\right)$.

$C_{P,N_2}^{\ast }=R\left(A+BT+C{T}^2+D{T}^3+E{T}^4\right)$        (14)

Integrate Equation (14).

$\begin{array}{c}{\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_{P,N_2}^{\ast }}={\displaystyle \underset{T_1}{\overset{T_2}{\int }}R\left(A+BT+C{T}^2+D{T}^3+E{T}^4\right)}dT\\ =R\left[\begin{array}{c}A\left(T_2-T_1\right)+\frac{B}{2}\left(T_2^2-T_1^2\right)+\\ \frac{C}{3}\left(T_2^3-T_1^3\right)+\frac{D}{4}\left(T_2^4-T_1^4\right)+\frac{E}{5}\left(T_2^5-T_1^5\right)\end{array}\right]\end{array}$        (15)

From Appendix D.1, “Ideal Gas Heat Capacity”, obtain and write the constant values of nitrogen and oxygen as in Table (1).

Name	Formula	A	$B\times {10}^3$	$C\times {10}^5$	$D\times {10}^8$	$E\times {10}^{11}$
Nitrogen	${\text{N}}_{\text{2}}$	3.539	$-0.261$	0.007	0.157	$-0.099$
oxygen	${\text{O}}_{\text{2}}$	3.630	$-1.794$	0.658	$-0.601$	$0.179$

Substitute 3.539 for A, $-0.261\times {10}^{-3}$ for B, $0.007\times {10}^{-5}$ for C, $0.157\times {10}^{-8}$ for D, $-0.099\times {10}^{-11}$ for E, $\text{8}\text{.314}\text{Pa}\cdot {\text{m}}^{\text{3}}/\text{mol}\cdot \text{K}$ for R, $\text{313}\text{K}$ for $T_2$, and $\text{255}\text{K}$ for $T_1$ in Equation (15).

$\begin{array}{c}{\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_{P,N_2}^{\ast }}=R\left[\begin{array}{c}A\left(T_2-T_1\right)+\frac{B}{2}\left(T_2^2-T_1^2\right)+\\ \frac{C}{3}\left(T_2^3-T_1^3\right)+\frac{D}{4}\left(T_2^4-T_1^4\right)+\frac{E}{5}\left(T_2^5-T_1^5\right)\end{array}\right]\\ =\left(\text{8}\text{.314}\text{Pa}\cdot {\text{m}}^{\text{3}}/\text{mol}\cdot \text{K}\right)\left[\begin{array}{l}3.539\left(313\text{K}-255\text{K}\right)\\ +\frac{-0.261\times {10}^{-3}}{2}\left[{\left(313\text{K}\right)}^2-{\left(255\text{K}\right)}^2\right]+\\ \frac{0.007\times {10}^{-5}}{3}\left[{\left(313\text{K}\right)}^3-{\left(255\text{K}\right)}^3\right]+\\ \frac{0.157\times {10}^{-8}}{4}\left[{\left(313\text{K}\right)}^4-{\left(255\text{K}\right)}^4\right]+\\ \frac{-0.099\times {10}^{-11}}{5}\left[{\left(313\text{K}\right)}^5-{\left(255\text{K}\right)}^5\right]\end{array}\right]\\ =1,341.26\text{J}/\text{mol}\end{array}$

Write the expression to obtain the constant pressure heat capacity of oxygen for ideal gas $\left(C_{P,O_2}^{\ast }\right)$.

$C_{P,O_2}^{\ast }=R\left(A+BT+C{T}^2+D{T}^3+E{T}^4\right)$        (16)

Integrate Equation (16).

$\begin{array}{c}{\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_{P,O_2}^{\ast }}={\displaystyle \underset{T_1}{\overset{T_2}{\int }}R\left(A+BT+C{T}^2+D{T}^3+E{T}^4\right)}dT\\ =R\left[\begin{array}{c}A\left(T_2-T_1\right)+\frac{B}{2}\left(T_2^2-T_1^2\right)+\\ \frac{C}{3}\left(T_2^3-T_1^3\right)+\frac{D}{4}\left(T_2^4-T_1^4\right)+\frac{E}{5}\left(T_2^5-T_1^5\right)\end{array}\right]\end{array}$        (17)

Substitute 3.630 for A, $-1.794\times {10}^{-3}$ for B, $0.658\times {10}^{-5}$ for C, $-0.601\times {10}^{-8}$ for D, $-0.099\times {10}^{-11}$ for E, $8.314\text{J}/\text{mol}\cdot \text{K}$ for R, $\text{313}\text{K}$ for $T_2$, and $\text{255}\text{K}$ for $T_1$ in Equation (17).

$\begin{array}{c}{\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_{P,O_2}^{\ast }}=8.314\text{J}/\text{mol}\cdot \text{K}\left[\begin{array}{l}3.630\left(313\text{K}-255\text{K}\right)\\ +\frac{-1.794\times {10}^{-3}}{2}\left[{\left(313\text{K}\right)}^2-{\left(255\text{K}\right)}^2\right]+\\ \frac{0.658\times {10}^{-5}}{3}\left[{\left(313\text{K}\right)}^3-{\left(255\text{K}\right)}^3\right]+\\ \frac{-0.601\times {10}^{-8}}{4}\left[{\left(313\text{K}\right)}^4-{\left(255\text{K}\right)}^4\right]+\\ \frac{0.179\times {10}^{-11}}{5}\left[{\left(313\text{K}\right)}^5-{\left(255\text{K}\right)}^5\right]\end{array}\right]\\ =359.28\text{J}/\text{mol}\end{array}$

Write the expression to obtain the parameter $R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}dT}$.

$R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}dT}=R\left(T_2-T_1\right)$        (18)

Substitute $8.314\text{J}/\text{mol}\cdot \text{K}$ for R, $\text{313}\text{K}$ for $T_2$, and $\text{255}\text{K}$ for $T_1$ in Equation (18).

$\begin{array}{c}R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}dT}=\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(\text{313}\text{K}-\text{255}{\text{K}}_1\right)\\ =485\text{J}/\text{mol}\end{array}$

Substitute $1,341.26\text{J}/\text{mol}$ for $C_{P,N_2}^{\ast }$, $359.28\text{J}/\text{mol}$ for $C_{P,O_2}^{\ast }$, and $485\text{J}/\text{mol}$ for $R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}dT}$ in Equation (13).

$\begin{array}{c}{\displaystyle \int d\widehat{U}}=0.79{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\left(C_{P,N_2}^{\ast }\right)}dT+0.21{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\left(C_{P,O_2}^{\ast }\right)}dT-R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}dT}\\ \Delta \widehat{U}=0.79{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\left(1,341.26\text{J}/\text{mol}\right)}dT+0.21{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\left(359.28\text{J}/\text{mol}\right)}dT-485\text{J}/\text{mol}\\ =\left\{\begin{array}{c}\left[0.79\left(1,341.26\text{J}/\text{mol}\right)\left(T_2-T_1\right)\right]+0.21\\ \left[\left(359.28\text{J}/\text{mol}\right)\left(T_2-T_1\right)\right]-485\text{J}/\text{mol}\end{array}\right\}\end{array}$        (19)

Substitute $\text{313}\text{K}$ for $T_2$, and $\text{255}\text{K}$ for $T_1$ in Equation (19).

$\begin{array}{c}\Delta \widehat{U}=\left\{\begin{array}{c}\left[0.79\left(1,341.26\text{J}/\text{mol}\right)\left(\text{313}\text{K}-\text{255}\text{K}\right)\right]+0.21\\ \left[\left(359.28\text{J}/\text{mol}\right)\left(\text{313}\text{K}-\text{255}\text{K}\right)\right]-485\text{J}/\text{mol}\end{array}\right\}\\ =1,341.26\text{J}/\text{mol}+359.28\text{J}/\text{mol}-485\text{J}/\text{mol}\\ =1215.5\text{J}/\text{mol}\end{array}$

Write the expression to obtain the change in internal energy $\left(\Delta U\right)$.

$\Delta U=\Delta \widehat{U}\left(N_{\min }\right)$        (20)

Substitute $5.90\text{Mmol}$ for $N_{\min }$, and $1215.5\text{J}/\text{mol}$ for $\Delta \widehat{U}$ in Equation (20).

$\begin{array}{c}\Delta U=1215.5\text{J}/\text{mol}\left(5.90\text{Mmol}\right)\\ =7,171.45\text{MJ}\end{array}$

Thus, the change in internal energy does the air undergo is $7,171.45\text{MJ}$.

(E)

Interpretation Introduction

Interpretation:

The work done by the lungs on the surroundings.

Concept Introduction:

Write the expression to obtain the work done $\left(W_{EC}\right)$ by the lungs.

$\begin{array}{c}{W}_{EC}=-{\displaystyle \int Pd{V}_{lung}}\\ =-P{\displaystyle \int d{V}_{lung}}\\ =-P\Delta V_{lung}\\ =-P\left(V_{\text{maximum}}-V_{\text{collapsed}}\right)\end{array}$

Explanation of Solution

Write the expression to obtain the work done $\left(W_{EC}\right)$ by the lungs.

$W_{EC}=-P\left(V_{\text{maximum}}-V_{\text{collapsed}}\right)$        (21)

Substitute $\text{1}\text{atm}$ for $P$, $153,534.63{\text{m}}^{\text{3}}$ for $V_{\text{maximum}}$, and $\text{125,000}{\text{m}}^{\text{3}}$ for $V_{\text{collapsed}}$ in Equation (21).

$\begin{array}{c}{W}_{EC}=-\text{1}\text{atm}\left(153,534.63{\text{m}}^{\text{3}}-\text{125,000}{\text{m}}^{\text{3}}\right)\\ =-\left[\left(\text{1}\text{atm}\right)\left(0.1MPa/\text{1}\text{atm}\right)\right]\\ =-\left(0.1MPa\right)\left(28,534.63{\text{m}}^{\text{3}}\right)\left(\text{J}/\text{Pa}\cdot {\text{m}}^{\text{3}}\right)\\ =-2,853.46\text{MJ}\end{array}$

Thus, the work done by the lungs is $-2,853.46\text{MJ}$.