A 1.00-L sample of dry air at 25 °C contains 0.0319 mol N2, 0.00856 mol O2, ... The main components of dry air, by volume, are N2, 78.08%; O2, 20.95%; Ar, ... by volume, of the following main gaseous components: N2, 74.1%; O2, ... What is the partial pressure of each gas in the expired air at 37 °C and 1.000 atm?
To find the partial pressure of each gas in the expired air at 37 °C and 1.000 atm, we first need to calculate the number of moles of each gas based on the given percentages by volume.
Let's assume the total volume of the expired air is 1 L, which means it also contains 1 mol of gas (according to Avogadro's law).
First, calculate the moles of N2:
moles of N2 = (74.1 / 100) * 1 mol = 0.741 mol
Next, calculate the moles of O2:
moles of O2 = (20.95 / 100) * 1 mol = 0.2095 mol
Now, we can determine the moles of the other remaining gases (Ar, CO2, etc.) by subtracting the sums of the moles of N2 and O2 from 1 mol:
moles of other gases = 1 mol - (0.741 mol + 0.2095 mol) = 0.0495 mol
Next, we need to convert the moles of each gas to partial pressures using the ideal gas law:
PV = nRT
Where:
P = Partial pressure
V = Volume
n = Moles of gas
R = Ideal gas constant
T = Temperature in Kelvin
First, convert the temperature from Celsius to Kelvin:
T = 37 °C + 273.15 = 310.15 K
Now, we can calculate the partial pressures of each gas:
Partial pressure of N2:
PN2 = (moles of N2 / total moles) * total pressure
= (0.741 mol / 1 mol) * 1.000 atm
= 0.741 atm
Partial pressure of O2:
PO2 = (moles of O2 / total moles) * total pressure
= (0.2095 mol / 1 mol) * 1.000 atm
= 0.2095 atm
Partial pressure of other gases:
Pother gases = (moles of other gases / total moles) * total pressure
= (0.0495 mol / 1 mol) * 1.000 atm
= 0.0495 atm
Therefore, the partial pressure of each gas in the expired air at 37 °C and 1.000 atm is:
PN2 = 0.741 atm
PO2 = 0.2095 atm
Pother gases = 0.0495 atm