Answer:
[tex]\large \boxed{\text{1.64 atm}}[/tex]
Explanation:
1. Moles of O₂
[tex]\text{Moles } = \text{2.0 mol air} \times \dfrac{\text{20 mol O}_{2}}{\text{100 mol air }} = \text{0.40 mol O}_{2}[/tex]
2. Partial pressure of O₂
To calculate the partial pressure of the oxygen, we can use the Ideal Gas Law:
pV = nRT
Data:
V = 6.0 L
n = 0.40 mol
T = 27 °C
Calculations:
T = (27 + 273.15) K = 300.15 K
[tex]\begin{array}{rcl}p \times \text{6.00 L} & = & \text{0.40 mol} \times \text{0.082 06 L$\cdot$ atm$\cdot$K$^{-1}$mol$^{-1}\times$ 300.15K}\\6.00p & = & \text{9.85 atm}\\p & = & \textbf{1.64 atm}\\\end{array}\\\text{The partial pressure of the oxygen is $\large \boxed{\textbf{1.642 atm}}$}[/tex]
