To determine the number of moles of oxygen in the mixture, we first need to find the partial pressure of oxygen using the total pressure and the partial pressures of nitrogen and argon.
Given:
Total pressure (P_total) = 50.0 kPa
Partial pressure of nitrogen (P_N2) = 22.0 kPa
Partial pressure of argon (P_Ar) = 22.0 kPa
Partial pressure of oxygen (P_O2) can be calculated by subtracting the partial pressures of nitrogen and argon from the total pressure:
\[ P_{O2} = P_{total} - (P_{N2} + P_{Ar}) \]
\[ P_{O2} = 50.0 \, \text{kPa} - (22.0 \, \text{kPa} + 22.0 \, \text{kPa}) \]
\[ P_{O2} = 50.0 \, \text{kPa} - 44.0 \, \text{kPa} \]
\[ P_{O2} = 6.0 \, \text{kPa} \]
Now, we can use the ideal gas law to find the number of moles of oxygen:
\[ PV = nRT \]
Where:
- P is the pressure (in atmospheres or kilopascals),
- V is the volume (in liters),
- n is the number of moles,
- R is the ideal gas constant (0.0821 L·atm/mol·K or 8.31 kPa·L/mol·K), and
- T is the temperature (in kelvin).
We need to convert the temperature from Celsius to Kelvin:
\[ T(K) = T(C) + 273.15 \]
\[ T(K) = 25^\circ \text{C} + 273.15 = 298.15 \text{ K} \]
Now, we can rearrange the ideal gas law to solve for the number of moles:
\[ n = \frac{PV}{RT} \]
Substituting the given values:
\[ n_{O2} = \frac{(6.0 \, \text{kPa}) \times (3.5 \, \text{L})}{(8.31 \, \text{kPa} \cdot \text{L/mol} \cdot \text{K}) \times (298.15 \, \text{K})} \]
\[ n_{O2} = \frac{21.0 \, \text{L} \cdot \text{kPa}}{2468.4865 \, \text{L} \cdot \text{kPa/mol} \cdot \text{K}} \]
\[ n_{O2} \approx 0.0085 \, \text{mol} \]
Therefore, the number of moles of oxygen in the mixture is approximately 0.0085 mol.
