To determine the partial pressure of helium gas in a mixture with oxygen, we can apply the ideal gas law, which is expressed as:
\( P = \frac{nRT}{V} \)
Where:
- P = pressure of the gas (in atmospheres)
- n = number of moles of the gas
- R = ideal gas constant, approximately \(0.08206 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K})\)
- T = temperature in Kelvin
- V = volume of the gas (in liters)
In this scenario, we have 12 grams of helium and 20 grams of oxygen in a 5-liter cylinder at a temperature of 30 degrees Celsius. First, we need to convert the mass of helium to moles:
Using the atomic mass of helium, which is approximately \(4.003 \, \text{g/mol}\), we calculate the moles of helium:
\( n = \frac{12 \, \text{g}}{4.003 \, \text{g/mol}} \approx 2.998 \, \text{mol} \)
Next, we convert the temperature from Celsius to Kelvin:
\( T = 30 \, \text{°C} + 273.15 = 303.15 \, \text{K} \)
Now, we can substitute the values into the ideal gas law formula to find the partial pressure of helium:
\( P = \frac{(2.998 \, \text{mol}) \times (0.08206 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K})) \times (303.15 \, \text{K})}{5 \, \text{L}} \)
Calculating this gives:
\( P \approx 14.9149 \, \text{atm} \)
When considering significant figures, we observe that the values provided have varying degrees of precision. To maintain consistency, we round the final answer to three significant figures:
\( P \approx 14.9 \, \text{atm} \)
This result indicates that the partial pressure of helium in the cylinder is approximately 14.9 atmospheres. Remember, the ideal gas law is a powerful tool for calculating the behavior of gases under various conditions, and understanding how to manipulate the equation is essential for solving related problems.