A

Step 1: Calculate the number of moles of each gas.
The molar mass of oxygen (O₂) is approximately 32 g/mol. Therefore, the number of moles of oxygen is:
$$ n(O_2) = \frac{16 \text{ g}}{32 \text{ g/mol}} = 0.5 \text{ mol} $$
Step 2: Calculate the number of moles of nitrogen.
The molar mass of nitrogen (N₂) is approximately 28 g/mol. Therefore, the number of moles of nitrogen is:
$$ n(N_2) = \frac{14 \text{ g}}{28 \text{ g/mol}} = 0.5 \text{ mol} $$
Step 3: Calculate the total number of moles of the gas mixture.
$$ n_{total} = n(O_2) + n(N_2) = 0.5 + 0.5 = 1.0 \text{ mol} $$
Step 4: Use the Ideal Gas Law to calculate the pressure of the mixture.
The Ideal Gas Law is given by:
$$ PV = nRT $$
where
- P is the pressure,
- V is the volume in cubic meters,
- n is the number of moles,
- R is the universal gas constant (approximately 8.314 J/(mol K)), and
- T is the temperature in Kelvin.
Step 5: Substitute the values into the Ideal Gas Law.
First, convert the volume from cm³ to m³:
$$ V = 5000 \text{ cm}^3 = 5000 \times 10^{-6} \text{ m}^3 = 0.005 \text{ m}^3 $$
Now substituting the values:
$$ P \cdot 0.005 = 1.0 \cdot 8.314 \cdot 300 $$
Step 6: Solve for P.
$$ P = \frac{1.0 \cdot 8.314 \cdot 300}{0.005} \approx \frac{2494.2}{0.005} = 498840 \text{ Pa} \approx 5 \times 10^5 \text{ Pa} $$
Therefore, the resulting pressure of the mixture is nearly:
A. 5 × 10⁵ Pa.