The gaseous state: ideal and real gases and pV = nRT

AS Level Chemistry · States of matter. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

Gas molecules move quickly and randomly. They hit the walls of the container, and each collision pushes on the wall. The total force of these collisions on each unit of area is the pressure of the gas.

An ideal gas is a model. Its particles have zero volume, and there are no forces of attraction between them. Such a gas obeys the equation pV = nRT exactly.

Real gases are close to ideal when the molecules are far apart and moving fast, which means low pressure and high temperature. They deviate at high pressure and low temperature, where the size of the molecules and the attractions between them begin to matter.

Rules to remember

Common mistake

Students put cm3, kPa or °C straight into pV = nRT. Convert to m3, Pa and K before you calculate.

Worked example

0.880 g of a gas occupies 500 cm3 at 100 kPa and 301 K. What is the relative molecular mass of the gas? (R = 8.31 J K−1 mol−1)

  1. Convert: p = 100 000 Pa; V = 500 × 10−6 = 5.00 × 10−4 m3.
  2. n = pV ÷ (RT) = (100 000 × 5.00 × 10−4) ÷ (8.31 × 301) = 50.0 ÷ 2501 = 0.0200 mol.
  3. Mr = mass ÷ n = 0.880 ÷ 0.0200 = 44.0.

Answer: Mr = 44.0

Practice questions

Try each one, then open the answer.

1. Which statement is an assumption of the ideal gas model?

  1. A
    the gas molecules have zero volume and no forces of attraction between them
  2. B
    the gas molecules are weakly attracted to each other
  3. C
    collisions between the gas molecules lose energy
  4. D
    all the gas molecules move at the same speed
Show answer

Answer: A. An ideal gas is treated as point particles with no intermolecular forces, so pV = nRT applies exactly. Real gases deviate most at high pressure (molecular volume matters) and low temperature (attractions matter).

2. 0.270 g of a gas occupies 240 cm3 at 100 kPa and 300 K. What is the relative molecular mass of the gas?

  1. A
    14.0
  2. B
    28.0
  3. C
    44.0
  4. D
    56.0
Show answer

Answer: B. n = pV/RT = (100 000 × 2.40 × 10−4)/(8.31 × 300) = 24.0/2493 = 9.63 × 10−3 mol. Mr = 0.270/9.63 × 10−3 = 28.0. Convert to Pa, m3 and K before substituting.

3. Why does a gas exert a pressure on the walls of its container?

  1. A
    the molecules repel each other and are pushed outwards
  2. B
    the molecules collide with the walls and exert a force on them
  3. C
    the molecules are attracted to the walls
  4. D
    the gas has weight that presses on the walls
Show answer

Answer: B. Gas molecules are in constant random motion; every collision with a wall applies a force, and pressure is that force per unit area. More frequent or harder collisions mean a higher pressure.

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