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.

Step-by-step 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* 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 = (P\cdot V)/(R\cdot T) = \frac{{\bf 10^(8)\;\textbf{Pa}}* 10^(-2)\;\text{m}^(3)}{8.314 \;\text{Pa}\cdot\text{m}^(3)\cdot\text{K}^(-1)\cdot\text{mol}^(-1)* 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}* 16.00 = 32.00\;\text{g}`. The mass of 410.3 moles of O₂ will be:

`410.3 * 32.00 = 1.31*10^(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?