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. Since we are only interested in the partial pressure of helium, we first need to convert the mass of helium to moles. The atomic mass of helium is approximately 4.003 grams per mole, so we calculate the moles of helium as follows:
\( n_{\text{He}} = \frac{12 \, \text{g}}{4.003 \, \text{g/mol}} \approx 2.998 \, \text{mol} \)
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:
\( P_{\text{He}} = \frac{(2.998 \, \text{mol}) \times (0.08206 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K})) \times (303.15 \, \text{K})}{5 \, \text{L}} \)
Calculating this gives:
\( P_{\text{He}} \approx 14.9149 \, \text{atm} \)
When considering significant figures, we observe that the values provided have varying degrees of precision. The most appropriate representation of the pressure, maintaining three significant figures, is:
\( P_{\text{He}} \approx 14.9 \, \text{atm} \)
This calculation illustrates how the ideal gas law can be utilized to find the partial pressure of a specific gas in a mixture, emphasizing the importance of moles, temperature, and volume in the process.