# How many kilograms of O_2 gas are needed to give an oxygen pressure of 3.48 atm at 22°C?

Feb 6, 2017

Just to retire this question, let us specify a volume of $1 \cdot {m}^{3}$; and I get a mass of dioxygen gas of under $5 \cdot k g$.

#### Explanation:

We use the Ideal Gas Equation, with $R = 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1$.

And thus $n = \frac{P V}{R T} = \frac{3.48 \cdot \cancel{a t m} \times 1 \cdot \cancel{{m}^{3}} \times {10}^{3} \cdot \cancel{L} \cdot \cancel{{m}^{-} 3}}{0.0821 \cdot \cancel{L \cdot a t m} \cdot \cancel{{K}^{-} 1} \cdot m o {l}^{-} 1 \times 295 \cdot \cancel{K}}$

$= 143.7 \cdot m o l$ (note that the answer has units $\frac{1}{m o {l}^{-} 1} = \frac{1}{\frac{1}{m o l}} = m o l$ $\text{as required}$

And we convert this molar quantity into a mass:

$143.7 \cdot \cancel{m o l} \times 32.00 \cdot \cancel{g \cdot m o {l}^{-} 1} \times {10}^{-} 3 \cdot k g \cdot \cancel{{g}^{-} 1}$

$= 4.60 \cdot k g .$

Please review my arithmetic. There are no money-back guarantees.

Note that $1 \cdot {m}^{3}$ is a very large volume, and is equal to $1000 \cdot L$. Chemists tend to deal with $\text{litres}$ and $m L$ and $c {m}^{3}$ ($1 \cdot m L \equiv 1 \cdot c {m}^{3} \equiv {10}^{-} 3 \cdot L \equiv {10}^{-} 6 \cdot {m}^{3}$). Of course all the measurements should be equivalent dimensionally, but it is so easy to divide instead of multiply and vice versa.