Chapter 13, Problem 84P

(a)

To determine

The number of ${\text{O}}_2$ molecules per cubic meter in the surface air at a temperature $20.0\text{°C}$ and pressure $1.00\text{atm}$.

Answer to Problem 84P

The number of ${\text{O}}_2$ molecules per cubic meter in the surface air at a temperature $20.0\text{°C}$ and pressure $1.00\text{atm}$ is $5.2\times {10}^{24}{\text{m}}^{\text{-3}}$.

Explanation of Solution

The percentage of oxygen in dry surface air is $21\%$. The percentage of nitrogen is $78\%$. The percentage of argon is $1\%$.

Write the formula for the number of oxygen molecule for cubic meter of air.

$\frac{N_O}{V}=\frac{0.21}{m_a/{\rho }_a}$ (I)

Here, $N_O$ is the number of oxygen molecule, $V$ is the volume, $m_a$ is the mass of air, ${\rho }_a$ is the density of air.

Write the formula for the mass of air.

$m_a=\frac{M}{N_A}$ (II)

Here, $M$ is the molar mass, $N_A$ is the Avogadro’s number.

Substitute equation (II) in equation (I).

$\frac{N_O}{V}=\frac{0.21{\rho }_a{N}_A}{M}$ (III)

Write the formula for the molar mass of air.

$M=\left(0.78\right)\left(2\times M_N\right)+\left(0.21\right)\left(2\times M_O\right)+\left(0.01\right)\left(M_A\right)$ (IV)

Here, $M_N$ is the molar mass of nitrogen, $M_O$ is the molar mass of oxygen, $M_A$ is the molar mass of argon.

Substitute equation (IV) in equation (III).

$\frac{N_O}{V}=\frac{0.21{\rho }_a{N}_A}{\left(0.78\right)\left(2\times M_N\right)+\left(0.21\right)\left(2\times M_O\right)+\left(0.01\right)\left(M_A\right)}$ (V)

Conclusion:

Substitute $1.20\times {10}^3{\text{g/m}}^{\text{3}}$ for ${\rho }_a$, $6.022\times {10}^{23}{\text{mol}}^{\text{-1}}$ for $N_A$, $14.00674\text{g/mol}$ for $M_N$, $15.9994\text{g/mol}$ for $M_O$, $39.948\text{g/mol}$ for $M_A$,$\begin{array}{c}\frac{N_O}{V}=\frac{0.21\left(1.20\times {10}^3{\text{g/m}}^{\text{3}}\right)\left(6.022\times {10}^{23}{\text{mol}}^{\text{-1}}\right)}{\left(0.78\right)\left(2\times \left(14.00674\text{g/mol}\right)\right)+\left(0.21\right)\left(2\times \left(15.9994\text{g/mol}\right)\right)+\left(0.01\right)\left(39.948\text{g/mol}\right)}\\ =5.2\times {10}^{24}{\text{m}}^{\text{-3}}\end{array}$

The number of ${\text{O}}_2$ molecules per cubic meter in the surface air at a temperature $20.0\text{°C}$ and pressure $1.00\text{atm}$ is $5.2\times {10}^{24}{\text{m}}^{\text{-3}}$.

(b)

To determine

The percentage of oxygen molecule in the tank of the diver who goes to a depth of $100.0\text{m}$.

Answer to Problem 84P

The percentage of oxygen molecule in the tank of the diver who goes to a depth of $100.0\text{m}$ is $1.9\%$

Explanation of Solution

The percentage of oxygen in dry surface air is $21\%$. The percentage of nitrogen is $78\%$. The percentage of argon is $1\%$. The density of sea water is $1025{\text{kg/m}}^{\text{3}}$.

Write the ideal gas equation.

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the universal gas constant, $T$ is the temperature.

When number of moles and temperature is constant, the volume is inversely proportional to the pressure.

Write the equation for the ratio of initial and final volume.

$\frac{V_f}{V_i}=\frac{P_i}{P_f}$

Here, $V_i$ is the initial volume, $P_i$ is the initial pressure, $P_f$ is the pressure of the pump as air starts to flow into the tire.

Write the equation for the percentage of oxygen molecules in the tank.

$\frac{V_f}{V_i}\left(21\%\right)=\frac{P_i}{P_f}\left(21\%\right)$ (V)

The initial pressure is at atmospheric pressure.

$P_i=P_{\text{atm}}$ (VI)

Here, $P_{\text{atm}}$ is the atmospheric pressure.

Write the formula for the final pressure.

$P_f=P_{\text{atm}}+\rho gh$ . (VII)

Here, $\rho$ is the density, $g$ is the acceleration due to gravity, $h$ is the depth.

Substitute equation (VI) and (VII) in equation (V).

$\frac{V_f}{V_i}\left(21\%\right)=\frac{P_{\text{atm}}}{P_{\text{atm}}+\rho gh}\left(21\%\right)$ (VIII)

Conclusion:

Substitute $1025{\text{kg/m}}^{\text{3}}$ for $\rho$, $1.013\times {10}^5\text{Pa}$ for $P_{\text{atm}}$, $9.80{\text{m/s}}^{\text{2}}$ for $g$, $100.0\text{m}$ for $h$.

$\begin{array}{c}\frac{V_f}{V_i}\left(21\%\right)=\frac{1.013\times {10}^5\text{Pa}}{1.013\times {10}^5\text{Pa}+\left(1025{\text{kg/m}}^{\text{3}}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)\left(100.0\text{m}\right)}\left(21\%\right)\\ =1.9\%\end{array}$

The percentage of oxygen molecule in the tank of the diver who goes to a depth of  $100.0\text{m}$ is $1.9\%$