The mole fraction of oxygen gas (\(O_2\)) is approximately \(0.380\) in the given gas mixture.
To find the mole fraction (\(X\)) of oxygen gas (\(O_2\)), you can use the formula:
\[ X_{O_2} = \frac{n_{O_2}}{n_{\text{total}}} \]
where
\[ n_{O_2} = \frac{P_{O_2}V}{RT} \]
Given:
- \(P_{O_2} = 2.31 \, \text{atm}\)
- \(P_{H_2} = 3.75 \, \text{atm}\)
- Total pressure (\(P_{\text{total}}\)) is the sum of partial pressures: \(P_{\text{total}} = P_{O_2} + P_{H_2}\)
- Volume (\(V\)), temperature (\(T\)), and the ideal gas constant (\(R\)) are constant.
1. Find \(n_{O_2}\) and \(n_{\text{total}}\):
\[ n_{O_2} = \frac{P_{O_2}V}{RT} \]
\[ n_{\text{total}} = \frac{P_{\text{total}}V}{RT} \]
2. Calculate \(X_{O_2}\):
\[ X_{O_2} = \frac{n_{O_2}}{n_{\text{total}}} \]
Substitute the given values and calculate the mole fraction of oxygen gas (\(X_{O_2}\)).
\[ X_{O_2} = \frac{n_{O_2}}{n_{\text{total}}} = \frac{\frac{P_{O_2}V}{RT}}{\frac{P_{\text{total}}V}{RT}} = \frac{P_{O_2}}{P_{\text{total}}} \]
\[ X_{O_2} = \frac{2.31}{2.31 + 3.75} \]
\[ X_{O_2} \approx 0.380 \]
Therefore, the mole fraction of oxygen gas (\(O_2\)) is approximately \(0.380\) in the given gas mixture.
The probable question may be:
Assuming that only the listed gases are present, what would be the mole fraction of oxygen gas be for each of the following situations? A gas sample of 2.31 atm of oxygen gas and 3.75 atm of hydrogen gas that react to form water vapor. Assume the volume of the container and the temperature inside the container does not change.