The Ideal Gas Equation (OCR A Level Chemistry A): Revision Note

Exam code: H432

Richard Boole

Written by: Richard Boole

Reviewed by: Philippa Platt

Updated on

Ideal gas equation & calculations

Kinetic theory of gases

  • The kinetic theory of gases explains the behaviour of gas molecules, which are in constant motion

  • It is based on these assumptions:

    • Gas molecules move rapidly and randomly

    • The volume of individual gas molecules is negligible

    • There are no attractive or repulsive forces between molecules

    • Collisions between molecules are elastic

      • This means that no kinetic energy is lost

    • Temperature is proportional to the average kinetic energy of the molecules

  • Gases that follow this model exactly are called ideal gases

  • Real gases do not follow all these assumptions perfectly

    • But, they often behave similarly under standard conditions

Ideal gases

  • The volume that an ideal gas occupies depends on:

    • Pressure

    • Temperature

  • When a gas is heated at constant pressure:

    • Particles gain kinetic energy

    • They collide more frequently with the container walls

    • To maintain constant pressure, the particles must spread out

      • So, the volume increases

  • Therefore, volume is directly proportional to temperature at constant pressure

Diagram showing gas molecules at low and high temperatures, with volume increasing to keep pressure constant. Graph of volume versus temperature, showing a direct relationship.
The volume of a gas increases upon heating to keep a constant pressure (a); volume is directly proportional to the temperature (b)

Limitations of the ideal gas law

  • At low temperatures and high pressures, real gases deviate from ideal behaviour

    • Molecules are closer together

    • Intermolecular forces (e.g. London forces, permanent dipoles) become significant

    • These forces reduce collisions with container walls

      • Therefore, the measured pressure is lower than predicted

    • The actual volume of gas particles is no longer negligible

      • Therefore, the measured volume is lower than predicted

  • Therefore, under these conditions:

    • The assumption of no attraction between particles is invalid

    • The assumption that particles have negligible volume is also invalid

Ideal gas equation

  • The ideal gas equation shows the relationship between pressure, volume, temperature, and number of moles:

pV = nRT

  • Where:

    • p = pressure (pascals, Pa)

    • V = volume (m3)

    • n = number of moles of gas (mol)

    • R = gas constant (8.314 J mol-1 K-1)

    • T = temperature (kelvin, K)

Worked Example

Calculating the volume of a gas

Calculate the volume occupied by 0.781 mol of oxygen at a pressure of 220 kPa and a temperature of 21 °C.

Answer:

  1. Rearrange the ideal gas equation to find volume of gas

Vfraction numerator n R T over denominator p end fraction

  1. Convert units:

    • p = 220 kPa = 220 000 Pa

    • n = 0.781 mol

    • R = 8.314 J mol-1 K-1 

    • T = 21 oC = 294 K

  2. Substitute and calculate V:

V = fraction numerator 0.781 space mol cross times 8.314 space straight J space straight K to the power of negative 1 end exponent space mol to the power of negative 1 end exponent cross times 294 space straight K over denominator 220000 space Pa end fraction 

V = 0.00867 m3

V = 8.67 dm3

Worked Example

Calculating the molar mass of a gas

A flask of volume 1000 cm3 contains 6.39 g of a gas. The pressure in the flask was 300 kPa and the temperature was 23 °C.

Calculate the relative molecular mass of the gas.

Answer

  1. Rearrange the ideal gas equation to find the number of moles of gas:

n = fraction numerator p V over denominator R T end fraction

  1. Convert units:

    • p = 300 kPa = 300 000 Pa

    • V = 1000 cm3 = 1 dm3 = 0.001 m3

    • R = 8.314 J mol-1 K-1

    • T = 23 oC = 296 K

  2. Substitute and calculate n:

n = fraction numerator 300000 space Pa space cross times 0.001 space straight m cubed over denominator 8.314 space straight J space straight K to the power of negative 1 end exponent space mol to the power of negative 1 end exponent cross times 296 space straight K end fraction

n = 0.1219 mol

  1. Use n and mass to calculate molar mass:

nfraction numerator mass over denominator molar space mass end fraction

molar mass = fraction numerator 6.39 space straight g over denominator 0.1219 space mol end fraction

molar mass = 52.42 g mol-1

Examiner Tips and Tricks

  • Always convert units:

    • Pressure to Pa

    • Volume to m3

    • Temperature to K (oC + 273)

  • Be confident rearranging pV = nRT to solve for any variable

  • Use the value of R from the Data Sheet: 8.314 J mol-1 K-1

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Richard Boole

Author: Richard Boole

Expertise: Chemistry Content Creator

Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.

Philippa Platt

Reviewer: Philippa Platt

Expertise: Chemistry Content Creator

Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener