Question

What volume would be occupied by 4.8 g of oxygen gas at 0.50 atm and 133 °c? (o: 15.999)

Answer 1

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 = $273 +133=406\;K$

• Molar mass of oxygen gas = $2 \times 15.999$ = 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;

$Number\;of\;moles = \frac{Mass}{molar\;mass}\\\\Number\;of\;moles = \frac{4.8}{31.998 }$

Number of moles = 0.15 moles.

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

$V = \frac{nRT}{P}$

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;

$V =\frac{0.15 \times 0.0821 \times 406}{0.50} \\\\V =\frac{4.99989}{0.50}$

Volume, V = 9.9998 ≈ 10.0 Liters.

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