The partial pressure of oxygen is approximately 0.306 atm, the partial pressure of nitrogen is approximately 0.491 atm, and the total pressure in the tank is approximately 0.797 atm.
To find the partial pressures of oxygen (O2) and nitrogen (N2) in the gaseous mixture, we can use the ideal gas law:
\[ PV = nRT \]
Where:
- \( P \) is the pressure
- \( V \) is the volume
- \( n \) is the number of moles
- \( R \) is the ideal gas constant
- \( T \) is the temperature in Kelvin
We first need to find the number of moles (\( n \)) for each gas using the given masses and molar masses:
1. For oxygen (O2):
\[ n_{\text{O2}} = \frac{\text{mass}_{\text{O2}}}{\text{molar mass}_{\text{O2}}} \]
\[ n_{\text{O2}} = \frac{5.00 \, \text{g}}{32.00 \, \text{g/mol}} \]
\[ n_{\text{O2}} = 0.15625 \, \text{mol} \]
2. For nitrogen (N2):
\[ n_{\text{N2}} = \frac{\text{mass}_{\text{N2}}}{\text{molar mass}_{\text{N2}}} \]
\[ n_{\text{N2}} = \frac{7.00 \, \text{g}}{28.02 \, \text{g/mol}} \]
\[ n_{\text{N2}} = 0.24982 \, \text{mol} \]
Now, we can find the partial pressures using the ideal gas law:
\[ P_{\text{O2}} = \frac{n_{\text{O2}}RT}{V} \]
\[ P_{\text{N2}} = \frac{n_{\text{N2}}RT}{V} \]
Substitute the values:
\[ P_{\text{O2}} = \frac{(0.15625 \, \text{mol})(0.0821 \, \text{L}\cdot\text{atm/mol}\cdot\text{K})(298 \, \text{K})}{12.0 \, \text{L}} \]
\[ P_{\text{O2}} \approx 0.306 \, \text{atm} \]
\[ P_{\text{N2}} = \frac{(0.24982 \, \text{mol})(0.0821 \, \text{L}\cdot\text{atm/mol}\cdot\text{K})(298 \, \text{K})}{12.0 \, \text{L}} \]
\[ P_{\text{N2}} \approx 0.491 \, \text{atm} \]
The total pressure (\( P_{\text{total}} \)) is the sum of the partial pressures:
\[ P_{\text{total}} = P_{\text{O2}} + P_{\text{N2}} \]
\[ P_{\text{total}} = 0.306 \, \text{atm} + 0.491 \, \text{atm} \]
\[ P_{\text{total}} \approx 0.797 \, \text{atm} \]
Therefore, the partial pressure of oxygen is approximately 0.306 atm, the partial pressure of nitrogen is approximately 0.491 atm, and the total pressure in the tank is approximately 0.797 atm.