To determine the volume of 15 grams of oxygen gas at 25°C and 650 mmHg pressure, we can use the Ideal Gas Law, which is represented by the equation:
\(\[ PV = nRT \]\)
where:
- \(\( P \)\) is the pressure,
- \(\( V \)\) is the volume,
- \(\( n \)\) is the number of moles,
- \(\( R \)\) is the universal gas constant, and
- \(\( T \)\) is the temperature in Kelvin.
Here's the step-by-step solution:
1. Convert the temperature to Kelvin:
The temperature given is 25°C. We need to convert this to Kelvin:
\(\[ T = 25 + 273.15 = 298.15 \, \text{K} \]\)
2. Convert the pressure to atmospheres:
The pressure given is 650 mmHg. We need to convert this to atmospheres using the conversion factor \(\(1 \, \text{atm} = 760 \, \text{mmHg} \)\):
\(\[ P = \frac{650 \, \text{mmHg}}{760 \, \text{mmHg}} = 0.8553 \, \text{atm} \]\)
3. Calculate the number of moles of oxygen:
The molar mass of oxygen (O₂) is \(\(32 \, \text{g/mol}\)\). To find the number of moles, we use the formula:
\(\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{15 \, \text{g}}{32 \, \text{g/mol}} = 0.4688 \, \text{mol} \]\)
4. Use the Ideal Gas Law to find the volume:
The gas constant \(\( R \)\) is 0.0821 L·atm/(mol·K). Now, we can rearrange the Ideal Gas Law to solve for volume \(\( V \)\):
\(\[ V = \frac{nRT}{P} \]\)
Plugging in the values we have:
\(\[ V = \frac{0.4688 \, \text{mol} \times 0.0821 \, \frac{\text{L·atm}}{\text{mol·K}} \times 298.15 \, \text{K}}{0.8553 \, \text{atm}} \]\)
After calculating, we get:
\(\[ V = 13.4159 \, \text{L} \]\)
Thus, the volume of 15 grams of oxygen gas at 25°C and 650 mmHg pressure is approximately 13.42 liters.
