To determine the number of moles of gas in a human breath that occupies 56.0 L at a pressure of 753 mmHg and a temperature of 37 °C, we will use the Ideal Gas Law:
[tex]\[ PV = nRT \][/tex]
where:
- \( P \) is the pressure of the gas
- \( V \) is the volume of the gas
- \( n \) is the number of moles of the gas
- \( R \) is the ideal gas constant
- \( T \) is the temperature of the gas in Kelvin
Given the values:
- Volume (\( V \)) = 56.0 L
- Pressure (\( P \)) = 753 mmHg
- Temperature (\( T \)) = 37 °C
First, we need to convert the pressure from mmHg to atmospheres (atm):
[tex]\[ 1 \text{ atm} = 760 \text{ mmHg} \][/tex]
[tex]\[ P = \frac{753 \text{ mmHg}}{760 \text{ mmHg/atm}} = 0.990789 \text{ atm} \][/tex]
Next, we convert the temperature from Celsius to Kelvin:
[tex]\[ T(\text{K}) = T(\text{°C}) + 273.15 \][/tex]
[tex]\[ T = 37 + 273.15 = 310.15 \text{ K} \][/tex]
The ideal gas constant \( R \) in units of L·atm/(K·mol) is:
[tex]\[ R = 0.0821 \text{ L·atm/(K·mol)} \][/tex]
Now we can rearrange the Ideal Gas Law to solve for the number of moles (\( n \)):
[tex]\[ n = \frac{PV}{RT} \][/tex]
Substitute in the known values:
[tex]\[ n = \frac{(0.990789 \text{ atm}) \times (56.0 \text{ L})}{(0.0821 \text{ L·atm/(K·mol)}) \times (310.15 \text{ K})} \][/tex]
[tex]\[ n \approx 2.179 \][/tex]
The number of moles of gas contained in the breath, to the correct number of significant figures (three, based on the given data), is:
[tex]\[ n = 2.18 \text{ moles} \][/tex]
So, there are approximately 2.18 moles of gas in the human breath.
