Answer :
Let's walk through the solution step-by-step for calculating the mole fraction and partial pressure of each gas in the mixture.
### Step 1: Understanding the Composition
From the representative sample in the sketch, we observe that there are:
- 2 parts Nitrogen ([tex]$N_2$[/tex])
- 1 part Oxygen ([tex]$O_2$[/tex])
This gives us a total of [tex]\( 2 + 1 = 3 \)[/tex] parts.
### Step 2: Calculating Mole Fractions
The mole fraction for a gas is given by the ratio of the number of moles of that gas to the total number of moles in the mixture.
- For Nitrogen ([tex]$N_2$[/tex]):
[tex]\[ \text{Mole fraction of } N_2 = \frac{\text{Number of parts of } N_2}{\text{Total number of parts}} = \frac{2}{3} \approx 0.667 \][/tex]
- For Oxygen ([tex]$O_2$[/tex]):
[tex]\[ \text{Mole fraction of } O_2 = \frac{\text{Number of parts of } O_2}{\text{Total number of parts}} = \frac{1}{3} \approx 0.333 \][/tex]
### Step 3: Calculating Partial Pressures
The partial pressure of a gas in a mixture is given by multiplying the mole fraction of the gas by the total pressure of the mixture.
Given:
[tex]\[ \text{Total pressure} = 0.130 \text{ kPa} \][/tex]
- For Nitrogen ([tex]$N_2$[/tex]):
[tex]\[ \text{Partial pressure of } N_2 = \text{Mole fraction of } N_2 \times \text{Total pressure} = 0.667 \times 0.130 \approx 0.087 \text{ kPa} \][/tex]
- For Oxygen ([tex]$O_2$[/tex]):
[tex]\[ \text{Partial pressure of } O_2 = \text{Mole fraction of } O_2 \times \text{Total pressure} = 0.333 \times 0.130 \approx 0.043 \text{ kPa} \][/tex]
### Step 4: Presenting the Results
Now we can present the computed values in a tabular format.
[tex]\[ \begin{tabular}{|c|c|c|} \hline \text{Gas} & \text{Mole fraction} & \text{Partial pressure (kPa)} \\ \hline N_2 & 0.667 & 0.087 \\ \hline O_2 & 0.333 & 0.043 \\ \hline \end{tabular} \][/tex]
So, the solution is:
- The mole fraction of [tex]\(N_2\)[/tex] is [tex]\(0.667\)[/tex] and its partial pressure is [tex]\(0.087 \text{ kPa}\)[/tex].
- The mole fraction of [tex]\(O_2\)[/tex] is [tex]\(0.333\)[/tex] and its partial pressure is [tex]\(0.043 \text{ kPa}\)[/tex].
### Step 1: Understanding the Composition
From the representative sample in the sketch, we observe that there are:
- 2 parts Nitrogen ([tex]$N_2$[/tex])
- 1 part Oxygen ([tex]$O_2$[/tex])
This gives us a total of [tex]\( 2 + 1 = 3 \)[/tex] parts.
### Step 2: Calculating Mole Fractions
The mole fraction for a gas is given by the ratio of the number of moles of that gas to the total number of moles in the mixture.
- For Nitrogen ([tex]$N_2$[/tex]):
[tex]\[ \text{Mole fraction of } N_2 = \frac{\text{Number of parts of } N_2}{\text{Total number of parts}} = \frac{2}{3} \approx 0.667 \][/tex]
- For Oxygen ([tex]$O_2$[/tex]):
[tex]\[ \text{Mole fraction of } O_2 = \frac{\text{Number of parts of } O_2}{\text{Total number of parts}} = \frac{1}{3} \approx 0.333 \][/tex]
### Step 3: Calculating Partial Pressures
The partial pressure of a gas in a mixture is given by multiplying the mole fraction of the gas by the total pressure of the mixture.
Given:
[tex]\[ \text{Total pressure} = 0.130 \text{ kPa} \][/tex]
- For Nitrogen ([tex]$N_2$[/tex]):
[tex]\[ \text{Partial pressure of } N_2 = \text{Mole fraction of } N_2 \times \text{Total pressure} = 0.667 \times 0.130 \approx 0.087 \text{ kPa} \][/tex]
- For Oxygen ([tex]$O_2$[/tex]):
[tex]\[ \text{Partial pressure of } O_2 = \text{Mole fraction of } O_2 \times \text{Total pressure} = 0.333 \times 0.130 \approx 0.043 \text{ kPa} \][/tex]
### Step 4: Presenting the Results
Now we can present the computed values in a tabular format.
[tex]\[ \begin{tabular}{|c|c|c|} \hline \text{Gas} & \text{Mole fraction} & \text{Partial pressure (kPa)} \\ \hline N_2 & 0.667 & 0.087 \\ \hline O_2 & 0.333 & 0.043 \\ \hline \end{tabular} \][/tex]
So, the solution is:
- The mole fraction of [tex]\(N_2\)[/tex] is [tex]\(0.667\)[/tex] and its partial pressure is [tex]\(0.087 \text{ kPa}\)[/tex].
- The mole fraction of [tex]\(O_2\)[/tex] is [tex]\(0.333\)[/tex] and its partial pressure is [tex]\(0.043 \text{ kPa}\)[/tex].