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:
[tex]\[ PV = nRT \][/tex]
where:
- [tex]\( P \)[/tex] is the pressure,
- [tex]\( V \)[/tex] is the volume,
- [tex]\( n \)[/tex] is the number of moles,
- [tex]\( R \)[/tex] is the universal gas constant, and
- [tex]\( T \)[/tex] 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:
[tex]\[ T = 25 + 273.15 = 298.15 \, \text{K} \][/tex]
2. Convert the pressure to atmospheres:
The pressure given is 650 mmHg. We need to convert this to atmospheres using the conversion factor [tex]\(1 \, \text{atm} = 760 \, \text{mmHg} \)[/tex]:
[tex]\[ P = \frac{650 \, \text{mmHg}}{760 \, \text{mmHg}} = 0.8553 \, \text{atm} \][/tex]
3. Calculate the number of moles of oxygen:
The molar mass of oxygen (O₂) is [tex]\(32 \, \text{g/mol}\)[/tex]. To find the number of moles, we use the formula:
[tex]\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{15 \, \text{g}}{32 \, \text{g/mol}} = 0.4688 \, \text{mol} \][/tex]
4. Use the Ideal Gas Law to find the volume:
The gas constant [tex]\( R \)[/tex] is 0.0821 L·atm/(mol·K). Now, we can rearrange the Ideal Gas Law to solve for volume [tex]\( V \)[/tex]:
[tex]\[ V = \frac{nRT}{P} \][/tex]
Plugging in the values we have:
[tex]\[ V = \frac{0.4688 \, \text{mol} \times 0.0821 \, \frac{\text{L·atm}}{\text{mol·K}} \times 298.15 \, \text{K}}{0.8553 \, \text{atm}} \][/tex]
After calculating, we get:
[tex]\[ V = 13.4159 \, \text{L} \][/tex]
Thus, the volume of 15 grams of oxygen gas at 25°C and 650 mmHg pressure is approximately 13.42 liters.
