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If a person inhales 82.3 g of O₂ in an hour, what volume (in L) does this amount occupy at 1.03 atm 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 :

To find the volume of oxygen (O₂) that a person inhales under given conditions, we can use the Ideal Gas Law, which is expressed by the formula:

[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 (approximately 0.0821 L·atm/(mol·K))
- [tex]\( T \)[/tex] is the temperature in Kelvin (K)

Here are the steps to solve the problem:

1. Convert the temperature from Celsius to Kelvin:

The temperature given is 36.9 °C. To convert Celsius to Kelvin, use the formula:

[tex]\[ T(K) = T(°C) + 273.15 \][/tex]

Thus, [tex]\( T = 36.9 + 273.15 = 310.05 \)[/tex] K

2. Calculate the number of moles of O₂:

Given the mass of inhaled oxygen is 82.3 g and the molar mass of oxygen (O₂) is approximately 32.00 g/mol, the number of moles of oxygen can be calculated using:

[tex]\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{82.3\, \text{g}}{32.00\, \text{g/mol}} = 2.57 \, \text{moles} \][/tex]

3. Calculate the volume using the Ideal Gas Law:

Rearrange the Ideal Gas Law to solve for volume ([tex]\( V \)[/tex]):

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

Plugging in the values:
- [tex]\( n = 2.57 \)[/tex] moles
- [tex]\( R = 0.0821 \)[/tex] L·atm/(mol·K)
- [tex]\( T = 310.05 \)[/tex] K
- [tex]\( P = 1.03 \)[/tex] atm

[tex]\[ V = \frac{2.57 \, \text{moles} \times 0.0821 \, \text{L·atm/(mol·K)} \times 310.05 \, \text{K}}{1.03 \, \text{atm}} \][/tex]

[tex]\[ V \approx 63.6 \, \text{L} \][/tex]

The volume of oxygen inhaled is approximately 63.6 liters, ensuring the result has the appropriate number of significant figures based on the given data.

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