Solved: Dry air near sea level has the following

Step 1 of 2

Given:

\(\begin{aligned}&\mathrm{V} \text { of } \mathrm{N}_{2}=78.08 \% \\&\mathrm{~V} \text { of } \mathrm{O}_{2}=20.94 \% \\&\mathrm{~V} \text { of } \mathrm{Ar}=0.93 \% \\&\mathrm{~V} \text { of } \mathrm{CO}_{2}=0.05 \% \\&\mathrm{P}=1.00 \mathrm{~atm}\end{aligned}\)

From the ideal gas equation, we can say that volume is proportional to the number of moles of gas present. Therefore, the volume percents can be directly converted to mole fractions.

Mole fraction of \(\mathrm{N}_{2}=\frac{78.08}{100}=0.7808\)

Mole fraction of \(\mathrm{O}_{2}=\frac{20.94}{100}=0.2097\)

Mole fraction of \(A r=\frac{0.93}{100}=0.0093\)

Mole fraction of \(\mathrm{CO}_{2}=\frac{0.05}{100}=0.0005\)

(a) Here, we are asked to calculate the partial pressure of each gas in atm.

Now, let's determine the partial pressure of gases:

\(p_{i}=X i P T\) i.e, \(\quad\) Partial pressure \(=\) mole fraction \(\times\) total pressure

The partial pressure of \(\mathrm{N}_{2}\) will be:

\(\begin{aligned}&=0.7808 \times 1 \mathrm{~atm} \\&=0.7808 \mathrm{~atm}\end{aligned}\)

Partial pressure of \(\mathrm{O}_{2}\) will be:

\(\begin{aligned}&=0.2097 \times 1 \mathrm{~atm} \\&=0.2097 \mathrm{~atm}\end{aligned}\)

Partial pressure of Ar will be:

\(\begin{aligned}&=0.0093 \times 1 \mathrm{~atm} \\&=0.0093 \mathrm{~atm}=9.3 \times 10^{-3} \mathrm{~atm}\end{aligned}\)

Partial pressure of \(\mathrm{CO}_{2}\) will be:

\(\begin{aligned}&=0.0005 \times 1 \mathrm{~atm} \\&=0.0005 \mathrm{~atm}=5 \times 10^{-4} \mathrm{~atm}\end{aligned}\)