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The volume occupied by this oxygen gas is equal to 10.0 Liters.

Given the following data:

  • Mass of oxygen gas = 4.8 grams
  • Pressure = 0.50 atm
  • Temperature = 133°C to Kelvin = [tex]273 +133=406\;K[/tex]
  • Molar mass of oxygen gas = [tex]2 \times 15.999[/tex] = 31.998 g/mol.

Scientific data:

  • Ideal gas constant, R = 0.0821 L⋅atm/mol⋅K

To determine the volume occupied by this oxygen gas:

First of all, we would calculate the number of moles of oxygen gas;

[tex]Number\;of\;moles = \frac{Mass}{molar\;mass}\\\\Number\;of\;moles = \frac{4.8}{31.998 }[/tex]

Number of moles = 0.15 moles.

Now, we can determine the volume by using the ideal gas law equation;

[tex]V = \frac{nRT}{P}[/tex]

Where;

  • P is the pressure.
  • V is the volume.
  • n is the number of moles of a gas.
  • R is the ideal gas constant.
  • T is the temperature.

Substituting the given parameters into the formula, we have;

[tex]V =\frac{0.15 \times 0.0821 \times 406}{0.50} \\\\V =\frac{4.99989}{0.50}[/tex]

Volume, V = 9.9998 ≈ 10.0 Liters.

Read more on volume here: https://brainly.com/question/25290815

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