Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules. Several other gases make up the remaining 1% of air molecules.
A)What is the partial pressure of nitrogen in air at atmospheric pressure (1 atm)? Assume ideal behavior.
Wouldn't it be = X*Po where X is the mole fraction of nitrogen. No mass is given but you can assume something convenient like 100 g, calculate the mole N2, O2, and other gases and from that obtain mole fraction nitrogen. Check my thinking.
.78*760
To calculate the partial pressure of nitrogen in air at atmospheric pressure (1 atm), we can use the concept of mole fraction and the ideal gas law.
1) First, we need to convert the percentages of nitrogen and oxygen to mole fractions. Mole fraction (X) is the ratio of the number of moles of a specific gas to the total number of moles of all gases present.
The mole fraction of nitrogen (Xn2) can be calculated as follows:
Xn2 = (percentage of nitrogen / 100) = 78.0 / 100 = 0.780
Similarly, the mole fraction of oxygen (Xo2) is calculated as:
Xo2 = 21.0 / 100 = 0.210
2) Next, we need to determine the total pressure exerted by air in the system, which is given as atmospheric pressure (1 atm).
3) Now, we can use the ideal gas law equation to calculate the partial pressure of nitrogen. The ideal gas law states:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of moles
R is the ideal gas constant (0.0821 L * atm / mol * K)
T is the temperature in Kelvin (assumed to be constant)
However, in this case, we are interested in partial pressure, so we rearrange the equation as follows:
Pn2 = Xn2 * P
Where:
Pn2 is the partial pressure of nitrogen
Xn2 is the mole fraction of nitrogen
P is the total pressure (atmospheric pressure)
4) Plugging in the values:
Pn2 = 0.780 * 1 atm
Pn2 = 0.780 atm
Therefore, the partial pressure of nitrogen in air at atmospheric pressure (1 atm) is 0.780 atm.