Question

What is the mass present in a 10.0L container of oxygen at a pressure of 105kPa and 20 degrees Celsius

Answer 1

1.31 × 10⁴ grams.

Explanation

Assume that oxygen acts like an ideal gas. In other words, assume that the oxygen here satisfies the ideal gas law:

$P \cdot V = n \cdot R\cdot T$,

where

• $P$ the pressure on the gas, $\bf P = 10^{5}\;\textbf{kPa}=10^{8}\;\textbf{Pa}$;
• $V$ the volume of the gas, $V = 10.0 \;\text{L} = 10.0\times 10^{-3}\;\text{m}^{3}=10^{-2}\;\text{m}^{3}$;
• $n$ the number of moles of the gas, which needs to be found;
• $T$ the absolute temperature of the gas, $T=20\;\textdegree{}\text{C} = (20 + 273.15)\;\text{K} = 293.15\;\text{K}$.
• $R$ the ideal gas constant, $R = 8.314$ if P, V, and T are in their corresponding SI units: Pa, m³, and K.

Apply the ideal gas law to find $n$:

$n = \dfrac{P\cdot V}{R\cdot T} = \dfrac{{\bf 10^{8}\;\textbf{Pa}}\times 10^{-2}\;\text{m}^{3}}{8.314 \;\text{Pa}\cdot\text{m}^{3}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}\times 293.15\;\text{K}} = 410.3\;\text{mol}$.

In other words, there are 410.3 moles of O₂ molecules in that container.

There are two oxygen atoms in each O₂ molecules. The mass of mole of O₂ molecules will be ${\bf 2}\times 16.00 = 32.00\;\text{g}$. The mass of 410.3 moles of O₂ will be:

$410.3 \times 32.00 = 1.31\times10^{4}\;\text{g}$.

What would be the mass of oxygen in the container if the pressure is approximately the same as STP at $10^{5}\;\textbf{Pa}$ or  $10^{2}\;\text{kPa}$ instead?