Molar Volume, Avogadro's Law

The molar volume is the volume occupied by one mole of a substance (chemical element or chemical compound) at a given temperature and pressure.

There are two standards, commonly used in schools:

• STP (standard temperature and pressure) which is 0 ºC and 1 atmosphere.
• RTP (room temperature and pressure) which is 20 ºC and 1 atmosphere.

Avogadro’s Law states that:

1 mole of every gas occupies the same volume, at the same temperature and pressure.

At STP (standard temperature and pressure), this volume is 22.4 liters

At RTP (room temperature and pressure), this volume is 24 dm³ (liters)

We can also say:

The molar volume of a gas is 22.4 liters at STP (standard temperature and pressure).

The molar volume of gas is 24 dm³ at RTP (room temperature and pressure).

Gas volumes from moles and grams

Example:

Calculate the volume of carbon dioxide gas, CO₂, occupied by (a) 5 moles and (b) 0.5 moles of the gas occupied at STP.

Solution:

a) Volume of CO₂

= number of moles of CO₂ × 22.4 L

= 5 × 22.4

= 112 L

b) Volume of CO₂

= number of moles of CO₂ × 22.4 L

= 0.5 × 22.4

= 11.2 L

The following video shows an example of grams to liters conversion.

Moles from Gas Volume

Example:

Calculate the number of moles of ammonia gas, NH₃, in a volume of 80 L of the gas measured at STP.

Solution:

Volume of gas = number of moles × 22.414 L/mol

The following video shows an example of liters to moles conversion.

Gas volume from equations

From the equation for a reaction, we can tell how many moles of a gas take part. Using Avogadro’s Law, we can also work out its volume.

Example:

What volume of hydrogen will react with 22.4 liters of oxygen to form water? (All volumes are measured at STP)

Solution:

Step 1: Write a balanced equation for the reaction.

2H₂ (g) + O₂ (g) → 2H₂O (l)

Step 2: Calculate the volume.

From the equation, 2 volumes of hydrogen react with 1 of oxygen or

2 × 22.4 liters of hydrogen react with 22.4 liters of oxygen.

The volume of hydrogen that will react is 44.8 liters.

Example:

When sulfur burns in air it forms sulfur dioxide. What volume of this gas is produced when 1 g of sulfur burns? (A_r : S = 32) (All volumes are measured at STP)

Solution:

Step 1: Write a balanced equation for the reaction.

S (s) + O₂ (g) → SO₂ (g)

Step 2: Get the number of moles from the grams.

32 g of sulfur atoms = 1 mole of sulfur atoms

So, 1 g = 1 ÷ 32 mole or 0.03125 moles of sulfur atoms

1 mole of sulfur atoms gives 1 mole of sulfur dioxide molecules

So, 0.03125 moles of sulfur atoms gives 0.03125 moles of sulfur dioxide

Step 3: Get the volume.

1 mole of sulfur dioxide molecules has a volume of 22.4 at STP

So, 0,03125 moles has a volume of 0.03125 × 22.4 = 0.7 liters at STP

So, 0.7 liters of sulfur dioxide are produced.

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