Calculate the mass density of a pure sample of oxygen gas at a pressure of 1.0 atm, a volume of 7.8 L at 197.2C?

Calculate the mass density of a pure sample of oxygen gas at a pressure of 1.0 atm, a volume of 7.8 L at 197.2C.

May 11, 2016

The density of ${\text{O}}_{2}$ under these conditions is is 0.83 g/L.

Explanation:

We can use the Ideal Gas Law to solve this problem.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

Since $\text{moles" = "mass"/"molar mass}$ or $n = \frac{m}{M}$, we can write

$P V = \frac{m}{M} R T$

We can rearrange this to

$P M = \frac{m}{V} R T$

But $\text{density"= "mass"/"volume}$ or color(brown)(|bar(ul(color(white)(a/a)ρ = m/Vcolor(white)(a/a)|)))" "

PM = ρRT and

color(blue)(|bar(ul(color(white)(a/a)ρ = (PM)/(RT)color(white)(a/a)|)))" "

$P = \text{1.0 atm"; M = "32.00 g/mol"; R = "0.082 06 L·atm·K"^"-1""mol"^"-1"; T = "197.2 °C" = "470.35 K}$

ρ = (1.0 color(red)(cancel(color(black)("atm"))) × "32.00 g"·color(red)(cancel(color(black)("mol"^"-1"))))/("0.082 06" color(red)(cancel(color(black)("atm")))·"L"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 470.35 color(red)(cancel(color(black)("K")))) = "0.83 g/L"