Q53E - University Physics with Modern Physics

Question

The density of air under standard laboratory conditions is$\mathbf{1}\mathbf{.}\mathbf{29} \mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$ , and about 20% of that air consists of oxygen. Typically, people breathe about 1/2 L of air per breath. (a) How many grams of oxygen does a person breathe in a day? (b) If this air is stored uncompressed in a cubical tank, how long is each side of the tank?

Step-by-Step Solution

Answer

1. The mass of the oxygen in grams a person breathes in a day is $2200 \text{g}.$
2. The length of each side of the tank is $2.1 \text{m}.$

Step 1: Identification of given data

The given data can be listed below,

• The density of the air in standard lab conditions is,.${\mathrm{p}}_{\mathrm{A}}=1.29\mathrm{kg}/{\mathrm{m}}^3.$
• The amount of oxygen in the air is, $20\%$
• Oxygen intake of people per breath is, $\frac{1}{2}\mathrm{L}.$

Step 2: Concept/Significance of mass and density

Physically, the dimensions of any given object of arbitrary shape are provided clearly by the mass and density of the object. Density and mass have a directly proportional relation.

The density is more when the mass is packed more densely.

Step 3: (a) Determination of the mass of oxygen in grams a person breathes in a day

By unit conversion, the breathing intake of a person in one day is given by,

$\left(12 \text{breathe/min}\right)\left(\frac{60 \text{min}}{1 \text{hr}}\right)\left(\frac{24 \text{hr}}{1 \text{day}}\right)=17,280 \text{breathe/day}$

The volume of air breathed in one day is given by,

  

$$\begin{gathered} V_A=\left(\frac{1}{2}\text{L/breath}\right)\left(17280 \text{breaths/day}\right) \\ =8640 \text{L}\left(\frac{1 {\text{m}}^3}{1000 \text{L}}\right) \\ =8.64 {\text{m}}^3 \end{gathered}$$

The mass of air breathed in one day is the density of air times the volume of air breathed.

So the mass of air is given by,

$m_A={\rho }_A\times V_A$

Here, ${\rho }_A$ is the density of the air and $V_A$ is the volume of the air.

Substitute values in the above,

$$\begin{gathered} m_A=1.29 {\text{kg/m}}^3\times 8.64 {\text{m}}^3 \\ =11.1 \text{kg} \end{gathered}$$

The mass of oxygen in the air can be expressed as,

$m_o=20\%\left(m_A\right)$

Substitute value in the above equation.

  

$$\begin{gathered} m_o=20\%\left(11.1 \text{kg}\right) \\ =\left(0.20\right)\left(11.1 \text{kg}\right) \\ =2.2 \text{kg}\left(\frac{1000 \text{g}}{1 \text{kg}}\right) \\ =2200 \text{g} \end{gathered}$$

Thus, the mass of the oxygen in grams a person breathes in a day is $2200 \text{g}.$

Step 4: (b) Determination of the length of each side of the tank when air is stored uncompressed in a cubical tank

The volume of a cube is given by,

$$\begin{gathered} V=l^3 \\ l={\left(V\right)}^{1/3} \end{gathered}$$

Here, l is the length of the faces of the cubical tank.

Substitute values in the above,

$$\begin{gathered} l={\left(8.64 {\text{m}}^3\right)}^{1/3} \\ =2.1 \text{m} \end{gathered}$$

Thus, the length of each side of the tank is $2.1 \text{m}.$