If a person exhales 25.5g of CO₂ in an hour, what volume (in L ) does this amount occupy at 1.01atm and 36.9°C ? Be sure your answer has the correct number of significant figures.
Note: Reference the Fundamental constants table for additional information.



Answer :

Answer: 14.6 L

Explanation:

To find the volume of CO₂ exhaled by a person, we can use the ideal gas law, which is given by:

[tex]$PV = nRT[/tex]

where:
[tex]$P$[/tex] is the pressure of the gas (in atmospheres, atm)

[tex]$V$[/tex] is the volume of the gas (in liters, L)

[tex]$n$[/tex] is the number of moles of the gas

[tex]$R$[/tex] is the ideal gas constant (0.0821 L·atm/mol·K)

[tex]$T$[/tex] is the temperature of the gas (in Kelvin, K)

First, we need to convert the mass of CO₂ to moles. The molar mass of CO₂ (carbon dioxide) is approximately 44.01 g/mol. Given the mass of CO₂ exhaled is 25.5 g, the number of moles, n, can be calculated as:

[tex]$n = \frac{m}{\text{molar mass}} = \frac{25.5 , \text{g}}{44.01 , \text{g/mol}} = 0.58 , \text{mol}$[/tex]

The temperature given is 36.9°C. To convert this to Kelvin:

[tex]$T = 36.9^\circ \text{C} + 273.15 = 310.05 , \text{K}$[/tex]

Now, substituting the values into the ideal gas law:

[tex]$PV = nRT[/tex]

We rearrange to solve for [tex]$V[/tex]:

[tex]$V = \frac{nRT}{P}[/tex]

[tex]$V = \frac{\left(\frac{25.5 , \text{g}}{44.01 , \text{g/mol}}\right) \times 0.0821 , \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}} \times 310.05 , \text{K}}{1.01 , \text{atm}}[/tex]

[tex]$V = 14.6 L $[/tex] (rounded to three significant figures)