01 Introduction to Chemistry
02 Matter and the Atomic Structure
03 The Mole Concept, Chemical Formula and Equation
04 The Periodic Table of Elements
05 Chemical Bond
06 Acids, Bases and Salts
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3.2.5 Number of Mole and Volume of Gas

Molar Volume of Gas

  1. Equal amount of gas occupies same volume of space.
  2. Molar volume of a gas is defined as the volume occupied by one mol the gas at certain condition
  3. At room temperature(25ºC) and pressure (1 atm), one mole of any gas occupies a volume of 24 dm³ (24 000 cm³ )
  4. At standard temperature(0ºC) and pressure (1 atm) one mole of any gas occupies a volume of 22.4 dm³ (22400cm³).
Number of mole of gassesVolume at s.t.p. (dm3)
1 mol of O2 gas 1 x 22.4 = 22.4
1 mol of H2 gas 1 x 22.4 = 22.4
2 mol of He gas 2 x 22.4 = 44.8
3.5 mol of N2 gas 3.5 x 22.4 = 78.4
Number of mole of gassesVolume at r.t.p. (dm3)
1 mol of O2  gas 1 x 24 = 24
1 mol of H2 gas 1 x 24 = 24
2 mol of He gas 2 x 24 = 48
3.5 mol of N2 gas 3.5 x 24 = 84

Example:

What volume (at room temperature and pressure r.t.p.) would 2 moles of oxygen gas occupy? (Molar Gas Volume at r.t.p. = 24 dm³)

Answer:
Volume of gas = 2 x 24 = 48 dm³

Example:
A sample of ozone gas has a volume of 960cm³ at room temperature and pressure. Find the number of mole of the ozone. [Molar volume at r.t.p. = 24.0dm³]

Answer:
Number of mole 

= 960 cm3 / 24000 cm3
=0.04 mol

Volume of Gas, Mass of Substance and Number of Particles

Figure below shows the relationship between the number of mole of particles in a substance with the mass, number of particles and volume of gas.

Example

Find the number of particle in 1200cm³ of nitrogen gas at room temperature. (Molar Gas Volume at r.t.p. = 24 dm³)

Answer:

Number of mole of nitrogen gas
= 1200 cm3 / 24000 cm3
=0.05 mol

Number of nitrogen molecules
= 0.05 mol × 6.02 × 1 023
= 3.01 × 1 022

 

Example:

A gas jar containing 200cm³ of oxygen gas at s.t.p. Find the mass of the oxygen gas. [Relative atomic mass: Oxygen = 16; Molar volume of gas at s.t.p. = 22.4 dm³]

Answer:

Number of mole of oxygen gas, 
n = 200  cm3 / 22400  cm3
= 1/112  mol

Relative molecular mass of oxygen 
= 2×16 = 32
Mass of oxygen
= 1/112  × 32 gmol-1
=0.286g