Sure, let's solve this problem step-by-step.
Given:
- The total pressure of the gas mixture is 0.130 kPa.
- The mixture is composed of nitrogen ([tex]\(N_2\)[/tex]) and oxygen ([tex]\(O_2\)[/tex]) gas.
- We assume that the gases are present in equal amounts in the mixture.
### Step 1: Determine the mole fractions of [tex]\(N_2\)[/tex] and [tex]\(O_2\)[/tex]
Since it's given that nitrogen ([tex]\(N_2\)[/tex]) and oxygen ([tex]\(O_2\)[/tex]) gases are present in equal amounts, each gas will constitute half of the total number of moles. Hence, the mole fraction for each gas is:
[tex]\[
\text{Mole fraction of } N_2 = \frac{\text{Moles of } N_2}{\text{Total moles}} = 0.5
\][/tex]
[tex]\[
\text{Mole fraction of } O_2 = \frac{\text{Moles of } O_2}{\text{Total moles}} = 0.5
\][/tex]
### Step 2: Calculate the partial pressures of [tex]\(N_2\)[/tex] and [tex]\(O_2\)[/tex]
The partial pressure of a gas in a mixture can be calculated using its mole fraction multiplied by the total pressure of the mixture.
[tex]\[
\text{Partial pressure of } N_2 = \text{Total pressure} \times \text{Mole fraction of } N_2
\][/tex]
[tex]\[
\text{Partial pressure of } O_2 = \text{Total pressure} \times \text{Mole fraction of } O_2
\][/tex]
Given the total pressure is 0.130 kPa:
[tex]\[
\text{Partial pressure of } N_2 = 0.130 \, \text{kPa} \times 0.5 = 0.065 \, \text{kPa}
\][/tex]
[tex]\[
\text{Partial pressure of } O_2 = 0.130 \, \text{kPa} \times 0.5 = 0.065 \, \text{kPa}
\][/tex]
### Final Results:
[tex]\[
\begin{array}{|c|c|c|}
\hline
\text{Gas} & \text{Mole Fraction} & \text{Partial Pressure (kPa)} \\
\hline
N_2 & 0.500 & 0.065 \\
\hline
O_2 & 0.500 & 0.065 \\
\hline
\end{array}
\][/tex]
Here, we have calculated the mole fraction and partial pressure for each gas in the mixture:
- The mole fraction of [tex]\(N_2\)[/tex] is 0.500 and its partial pressure is 0.065 kPa.
- The mole fraction of [tex]\(O_2\)[/tex] is 0.500 and its partial pressure is 0.065 kPa.
