Answer: The amount of air that could be held in lungs is 6.873g
Explanation:
To calculate the amount of air, we use the equation given by Ideal gas equation, which follows:
[tex]PV=nRT[/tex]
where,
P = Pressure of the gas = 102 kPa
V = volume of the gas = 6 L
n = number of moles of gas = ? mol
R = Gas constant = [tex]8.31\text{L kPa }mol^{-1}K^{-1}[/tex]
T = temperature of the gas = 37°C = 310 K (Conversion factor: [tex]T(K)=T(^oC)+273)[/tex]
Putting values in above equation, we get:
[tex]102kPa\times 6L=n\times 8.31\text{L kPa }\times 310K\\\\n=0.237mol[/tex]
To calcultae the mass of the air, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We are given:
Moles of air entered = 0.237 mol
Molar mass of air = 29 g/mol
Putting values in above equation, we get:
[tex]0.237mol=\frac{\text{Mass of air}}{29g/mol}\\\\\text{Mass of air}=6.873g[/tex]
Hence, the amount of air that could be held in lungs is 6.873g
