Chapter 5, Problem 5.96P

(a)

Interpretation Introduction

Interpretation:

The partial pressure of each gas in the atmosphere is to be calculated

Concept introduction:

The ideal gas equation can be expressed as follows,

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

The expression to calculate the partial pressure of the gas is as follows:

$P_{\text{gas}}=X_{\text{gas}}\cdot P_{\text{total}}$

Here, $P_{\text{gas}}$ is the partial pressure of the gas, $X_{\text{gas}}$ mole fraction of the gas and $P_{\text{total}}$ is the total pressure.

The expression to calculate the mole fraction of the gas is as follows,

$X_{\text{gas}}=\frac{\text{Moles of gas}}{\text{Total moles}}$

Here, $X_{\text{gas}}$ is the mole fraction of the gas.

Answer to Problem 5.96P

The partial pressure of ${\text{N}}_{\text{2}}$ is $\text{597 torr}$, the partial pressure of ${\text{O}}_{\text{2}}$ is $\text{159 torr}$, The partial pressure of ${\text{CO}}_{\text{2}}$ is $\text{0}\text{.3 torr}$ and the partial pressure of water is $\text{3}\text{.5 torr}$.

Explanation of Solution

The expression to calculatethe mole fraction of nitrogen is as follows:

$P_{\text{nitrogen}}=X_{\text{nitrogen}}\cdot P_{\text{total}}$ (1)

Substitute the value $0.786$ for $X_{\text{nitrogen}}$ and $1\text{ atm}$ for $P_{\text{Total}}$ in the equation (1)

$\begin{array}{c}{P}_{\text{nitrogen}}=\left(0.786\right)\left(1\text{ atm}\left(\frac{760\text{ torr}}{1\text{ atm}}\right)\right)\\ =597.36\text{ torr}\\ \text{= 597 torr}\end{array}$

The expression to calculatethe mole fraction of oxygen is as follows:

$P_{\text{oxygen}}=X_{\text{oxygen}}\cdot P_{\text{total}}$ (2)

Substitute the value $0.209$ for $X_{\text{oxygen}}$ and $1\text{ atm}$ for $P_{\text{Total}}$ in the equation (2)

$\begin{array}{c}{P}_{\text{oxygen}}=\left(0.209\right)\left(1\text{ atm}\left(\frac{760\text{ torr}}{1\text{ atm}}\right)\right)\\ =158.84\text{ torr}\\ \text{= 597 torr}\end{array}$

The expression to calculatethe mole fraction of carbon dioxide is as follows:

$P_{\text{carbon dioxide}}=X_{\text{carbon dioxide}}\cdot P_{\text{total}}$ (3)

Substitute the value $0.0004$ for $X_{\text{carbon dioxide}}$ and $1\text{ atm}$ for $P_{\text{Total}}$ in the equation (3)

$\begin{array}{c}{P}_{\text{carbon dioxide}}=\left(0.0004\right)\left(1\text{ atm}\left(\frac{760\text{ torr}}{1\text{ atm}}\right)\right)\\ =0.304\text{ torr}\\ \text{= 0}\text{.3 torr}\end{array}$

The expression tocalculatethe mole fraction of water is as follows:

$P_{\text{water}}=X_{\text{water}}\cdot P_{\text{total}}$ (4)

Substitute the value $0.0046$ for $X_{\text{water}}$ and $1\text{ atm}$ for $P_{\text{Total}}$ in the equation (4)

$\begin{array}{c}{P}_{\text{water}}=\left(0.0046\right)\left(1\text{ atm}\left(\frac{760\text{ torr}}{1\text{ atm}}\right)\right)\\ =3.496\text{ torr}\\ \text{= 3}\text{.5 torr}\end{array}$

Conclusion

The partial pressure of ${\text{N}}_{\text{2}}$ is $\text{597 torr}$, the partial pressure of ${\text{O}}_{\text{2}}$ is $\text{159 torr}$, The partial pressure of ${\text{CO}}_{\text{2}}$ is $\text{0}\text{.3 torr}$ and the partial pressure of water is $\text{3}\text{.5 torr}$.

(b)

Interpretation Introduction

Interpretation:

The mole percent of each gas in the alveoli is to be calculated.

Concept introduction:

The ideal gas equation can be expressed as follows,

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

The expression to calculate the partial pressure of the gas is as follows:

$P_{\text{gas}}=X_{\text{gas}}\cdot P_{\text{total}}$

Here, $P_{\text{gas}}$ is the partial pressure of the gas, $X_{\text{gas}}$ mole fraction of the gas and $P_{\text{total}}$ is the total pressure.

The expression to calculate the mole fraction of the gas is as follows,

$X_{\text{gas}}=\frac{\text{Moles of gas}}{\text{Total moles}}$

Here, $X_{\text{gas}}$ is the mole fraction of the gas.

Answer to Problem 5.96P

The mole percent of ${\text{N}}_{\text{2}}$ is $74.9\text{ \%}$, the mole percent of ${\text{O}}_{\text{2}}$ is $13.7\text{ \%}$, the mole percent of ${\text{CO}}_{\text{2}}$ is $5.3\text{ \%}$ and the mole percent of water is $6.2\text{ \%}$.

Explanation of Solution

The expression to calculate the mole fraction percent of nitrogen is as follows:

$P_{\text{nitrogen}}=X_{\text{nitrogen}}\cdot P_{\text{total}}$ (5)

Rearrange the equation (5) to calculate the percent of $X_{\text{nitrogen}}$ as follows:

$X_{\text{nitrogen}}=\left(\frac{P_{\text{nitrogen}}}{P_{\text{total}}}\right)\left(100\text{ \%}\right)$ (6)

Substitute the value $\text{569 torr}$ for $P_{\text{nitrogen}}$ and $760\text{ torr}$ for $P_{\text{Total}}$ in the equation (6)

$\begin{array}{c}{X}_{\text{nitrogen}}=\left(\frac{569\text{ torr}}{760\text{ torr}}\right)\left(100\text{ \%}\right)\\ =74.8684\%\\ =74.9\%\end{array}$

The expression to calculate the mole fraction percent of oxygen as follows:

$P_{\text{oxygen}}=X_{\text{oxygen}}\cdot P_{\text{total}}$ (7)

Rearrange the equation (7) to calculate the percent of $X_{\text{oxygen}}$ as follows:

$X_{\text{oxygen}}=\left(\frac{P_{\text{oxygen}}}{P_{\text{total}}}\right)\left(100\text{ \%}\right)$ (8)

Substitute the value $\text{104 torr}$ for $P_{\text{oxygen}}$ and $760\text{ torr}$ for $P_{\text{Total}}$ in the equation (8)

$\begin{array}{c}{X}_{\text{oxygen}}=\left(\frac{\text{104 torr}}{760\text{ torr}}\right)\left(100\text{ \%}\right)\\ =13.6842\%\\ =13.7\%\end{array}$

The expression to calculate the mole fraction percent of carbon dioxide as follows:

$P_{\text{carbon dioxide}}=X_{\text{carbon dioxide}}\cdot P_{\text{total}}$ (9)

Rearrange the equation (5) to calculate the percent of $X_{\text{carbon dioxide}}$ as follows:

$X_{\text{carbon dioxide}}=\left(\frac{P_{\text{carbon dioxide}}}{P_{\text{total}}}\right)\left(100\text{ \%}\right)$ (10)

Substitute the value $\text{40 torr}$ for $P_{\text{carbon dioxide}}$ and $760\text{ torr}$ for $P_{\text{Total}}$ in the equation (10)

$\begin{array}{c}{X}_{\text{carbon dioxide}}=\left(\frac{\text{40 torr}}{760\text{ torr}}\right)\left(100\text{ \%}\right)\\ =5.263\%\\ =5.3\%\end{array}$

The expression to calculate the mole fraction percent of water as follows:

$P_{\text{water}}=X_{\text{water}}\cdot P_{\text{total}}$ (11)

Rearrange the equation (11) to calculate $X_{\text{water}}$ as follows:

$X_{\text{water}}=\left(\frac{P_{\text{water}}}{P_{\text{total}}}\right)$ (12)

Substitute the value $\text{47 torr}$ for $P_{\text{water}}$ and $760\text{ torr}$ for $P_{\text{Total}}$ in the equation (12)

$\begin{array}{c}{X}_{\text{water}}=\left(\frac{\text{47 torr}}{760\text{ torr}}\right)\left(100\text{ \%}\right)\\ =6.184\%\\ =6.2\%\end{array}$

Conclusion

The mole percent of ${\text{N}}_{\text{2}}$ is $74.9\text{ \%}$, the mole percent of ${\text{O}}_{\text{2}}$ is $13.7\text{ \%}$, the mole percent of ${\text{CO}}_{\text{2}}$ is $5.3\text{ \%}$ and the mole percent of water is $6.2\text{ \%}$.

(c)

Interpretation Introduction

Interpretation:

The number of ${\text{O}}_{\text{2}}$ molecules in $0.50\text{ L}$ of alveoli air at $37°\text{C}$ is to be calculated.

Concept introduction:

The relationship between pressure and volume can be expressed as follows,

$PV=\text{constant}$

Here, $P$ is the pressure and $V$ is the volume.

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

$V\alpha T$

Here, $T$ is the temperature and $V$ is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

$\text{V }\alpha n$

Here, $n$ is the mole of the gas and $V$ is the volume.

The ideal gas equation can be expressed as follows,

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

Answer to Problem 5.96P

The number of ${\text{O}}_{\text{2}}$ molecules in $0.50\text{ L}$ of alveoli air at $37°\text{C}$ is $1.6\times {10}^{21}\text{ molecules}$.

Explanation of Solution

The formula to convert $°\text{C}$ to Kelvin is:

$T_{\text{K}}=T_{\text{°C}}+273.15$ (12)

Substitute $\text{37 }°\text{C}$ for $T_{\text{°C}}$ in equation (12).

$\begin{array}{c}{T}_{\text{K}}=37+273.15\\ =310.15\text{ K}\end{array}$

The pressure in $\text{torr}$ is converted into $\text{atm}$ is as follows,

$\begin{array}{c}P=104\text{ torr}\\ \text{=}\left(104\text{ torr}\right)\left(\frac{1\text{ atm}}{\text{760 torr }}\right)\\ =0.136842105\text{ atm}\end{array}$

The expression to calculate the moles of the ${\text{O}}_{\text{2}}$ is as follows,

$PV=nRT$ (13)

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the $\text{NO}$ and $R$ is the gas constant.

Rearrange the equation (13) to calculate $n$ as follows,

$n=\frac{PV}{RT}$                                                                                                                     (14)

Substitute the value $0.136842105\text{ atm}$ for $P$, $\text{310}\text{.15 K}$ for $T$, $\text{0}\text{.50 L}$ for $V$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (14).

$\begin{array}{c}n=\frac{\left(0.50\text{ L}\right)\left(\text{0}\text{.136842105 atm}\right)}{\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(310.15\text{ K}\right)}\\ =0.0026870\text{ mol}\end{array}$

The expression to calculate the molecules of ${\text{O}}_{\text{2}}$ is as follows:

${\text{Moles of O}}_{\text{2}}=\frac{{\text{Molecules of O}}_{\text{2}}}{\text{Avogadro's number}}$ (15)

Rearrange the equation (15) to calculate the molecules of ${\text{O}}_{\text{2}}$.

$\left({\text{Molecules of O}}_{\text{2}}\right)=\left({\text{Moles of O}}_{\text{2}}\right)\left(\text{Avogadro's number}\right)$ (16)

Substitute the value $0.0026870\text{ mol}$ for moles of ${\text{O}}_{\text{2}}$ and $6.022\times {10}^{23}\text{ molecules}$ for the Avogadro's number in the equation (16).

$\begin{array}{c}\left({\text{Molecules of O}}_{\text{2}}\right)=\left(0.0026870\text{ mol}\right)\left(6.022\times {10}^{23}\right)\\ =1.681\times {10}^{21}\text{ molecules}\end{array}$

Conclusion

The number of ${\text{O}}_{\text{2}}$ molecules in $0.50\text{ L}$ of alveoli air at $37°\text{C}$ is $1.6\times {10}^{21}\text{ molecules}$.