Question

Air is basically 80% of nitrogen and 20% of oxygen. A 2.0 Mol sample of air is found to occupy 6.0L at 27%. What is the partial pressure of oxygen in the sample?

Answer 1

`\large \boxed{\text{1.64 atm}}`

Step-by-step explanation:

1. Moles of O₂

`\text{Moles } = \text{2.0 mol air} * \frac{\text{20 mol O}_(2)}{\text{100 mol air }} = \text{0.40 mol O}_(2)`

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

`\begin{array}{rcl}p * \text{6.00 L} & = & \text{0.40 mol} * \text{0.082 06 L$\cdot$ atm$\cdot$K$^(-1)$mol$^(-1)*$ 300.15K}\\6.00p & = & \text{9.85 atm}\\p & = & \textbf{1.64 atm}\\\end{array}\\\text{The partial pressure of the oxygen is $\large \boxed{\textbf{1.642 atm}}$}`