College

If a person exhales 33.0 g of CO[tex]\(_2\)[/tex] in an hour, what volume (in L) does this amount occupy at 1.05 atm and 36.6 °C? Be sure your answer has the correct number of significant figures.

Note: Reference the Fundamental Constants table for additional information.

Answer :

To find the volume of carbon monoxide (CO) that a person exhales, we'll use the Ideal Gas Law, which is expressed as:

[tex]\[ PV = nRT \][/tex]

Where:
- [tex]\( P \)[/tex] is the pressure in atmospheres (atm),
- [tex]\( V \)[/tex] is the volume in liters (L),
- [tex]\( n \)[/tex] is the number of moles of gas,
- [tex]\( R \)[/tex] is the ideal gas constant, and
- [tex]\( T \)[/tex] is the temperature in Kelvin (K).

Here is a step-by-step solution:

1. Convert the Temperature to Kelvin:
- The temperature given is 36.6 °C. To convert this into Kelvin, use the formula:
[tex]\[
T(K) = T(°C) + 273.15
\][/tex]
[tex]\[
T(K) = 36.6 + 273.15 = 309.75 \, K
\][/tex]

2. Calculate the Number of Moles of CO:
- Given the mass of CO is 33.0 grams. We need the molar mass of CO, which is approximately 28.01 g/mol.
- Use the formula to find the number of moles ([tex]\( n \)[/tex]):
[tex]\[
n = \frac{\text{mass}}{\text{molar mass}} = \frac{33.0 \, \text{g}}{28.01 \, \text{g/mol}} = 1.178 \, \text{moles}
\][/tex]

3. Use the Ideal Gas Law to Find the Volume:
- The pressure [tex]\( P \)[/tex] is given as 1.05 atm.
- The ideal gas constant [tex]\( R \)[/tex] used here is 0.0821 atm·L/mol·K.
- Rearrange the Ideal Gas Law to solve for the volume:
[tex]\[
V = \frac{nRT}{P}
\][/tex]
Plug in the known values:
[tex]\[
V = \frac{(1.178 \, \text{moles}) \times (0.0821 \, \text{atm·L/mol·K}) \times (309.75 \, K)}{1.05 \, \text{atm}}
\][/tex]
[tex]\[
V \approx 28.534 \, \text{L}
\][/tex]

Thus, the volume of carbon monoxide exhaled is approximately 28.5 liters when rounded to three significant figures.

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