To solve the problem, we will use the ideal gas law, which states that \( PV = nRT \). In this case, we are interested in the number of moles of oxygen gas (\( O_2 \)) before and after the leakage, as well as the corresponding pressures.
### Step-by-Step Solution:
1. **Identify Initial Conditions:**
- Initial volume of \( O_2 \) (\( V \)) = 5 L
- Initial pressure (\( P_1 \)) = 8 atm
- Temperature (\( T \)) = 25°C = 298 K (convert to Kelvin by adding 273)
- Ideal gas constant (\( R \)) = 0.0821 L·atm/(K·mol)
2. **Calculate Initial Moles of \( O_2 \) (Before Leakage):**
Using the ideal gas equation:
\[
n_1 = \frac{P_1 V}{RT}
\]
Substituting the values:
\[
n_1 = \frac{8 \, \text{atm} \times 5 \, \text{L}}{0.0821 \, \text{L·atm/(K·mol)} \times 298 \, \text{K}}
\]
\[
n_1 = \frac{40}{24.4758} \approx 1.63 \, \text{mol}
\]
3. **Identify Final Conditions After Leakage:**
- Final pressure (\( P_2 \)) = 2 atm
- Volume remains the same (\( V \)) = 5 L
4. **Calculate Final Moles of \( O_2 \) (After Leakage):**
Using the ideal gas equation again:
\[
n_2 = \frac{P_2 V}{RT}
\]
Substituting the values:
\[
n_2 = \frac{2 \, \text{atm} \times 5 \, \text{L}}{0.0821 \, \text{L·atm/(K·mol)} \times 298 \, \text{K}}
\]
\[
n_2 = \frac{10}{24.4758} \approx 0.41 \, \text{mol}
\]
5. **Calculate the Ratio of Masses:**
The mass of \( O_2 \) can be calculated using the formula:
\[
\text{mass} = n \times M
\]
where \( M \) (molar mass of \( O_2 \)) = 32 g/mol.
- Initial mass (\( m_1 \)):
\[
m_1 = n_1 \times M = 1.63 \, \text{mol} \times 32 \, \text{g/mol} \approx 52.16 \, \text{g}
\]
- Final mass (\( m_2 \)):
\[
m_2 = n_2 \times M = 0.41 \, \text{mol} \times 32 \, \text{g/mol} \approx 13.12 \, \text{g}
\]
6. **Calculate the Ratio of Masses:**
\[
\text{Ratio} = \frac{m_1}{m_2} = \frac{52.16 \, \text{g}}{13.12 \, \text{g}} \approx 4
\]
### Final Answer:
The ratio of the mass of \( O_2 \) initially present to that left after leakage is equal to **4**.
