IBChemistryDPHLRevision NotesModels of the Particulate Nature of MatterIdeal GasesThe Ideal Gas Equation

The Ideal Gas Equation (DP IB Chemistry): Revision Note

Alexandra Brennan

Written by: Alexandra Brennan

Reviewed by: Philippa Platt

Updated on

DP Chemistry: HLDP Chemistry: SL

The Ideal Gas Equation

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

PV = nRT

P = pressure (pascals, Pa)

V = volume (m3)

n = number of moles of gas (mol)

R = gas constant (8.31 J K-1 mol-1)

T = temperature (Kelvin, K)

  • The ideal gas equation can also be used to calculate the molar mass (M) of a gas

Worked Example

Calculate the volume, in dm3, occupied by 0.781 mol of oxygen at a pressure of 220 kPa and a temperature of 21 °C.

Answer:

  • Step 1: Rearrange the ideal gas equation to find volume of the gas

V space equals space fraction numerator n R T over denominator P end fraction

  • Step 2: Convert into the correct units and calculate the volume the oxygen gas occupies

P = 220 kPa = 220 000 Pa

n = 0.781 mol

R = 8.31 J K-1 mol-1

T = 21 oC = 294 K

V space equals space fraction numerator 0.781 space mol space cross times 8.31 space straight J space straight K to the power of negative 1 end exponent mol cross times 294 straight K over denominator 220 space 000 space Pa end fraction

V = 0.00867 m3

V = 8.67 dm3

Examiner Tips and Tricks

Many students make mistakes converting between cm3 and m3 in gas law problems.

  • 1 m3 = 1,000,000 cm3 (because 100 × 100 × 100 = 106)

  • So:

    • To convert m3 to cm3, multiply by 106

    • To convert cm3 to m3, divide by 106 (or multiply by 10⁻6)

Always check units before using the ideal gas equation, it requires volume in m³

Worked Example

Calculate the pressure of a gas, in kPa, given that 0.20 moles of the gas occupy 10.1 dm3 at a temperature of 25 oC.

Answer:

  • Step 1: Rearrange the ideal gas equation to find the pressure of the gas

P space equals space fraction numerator n R T over denominator V end fraction

  • Step 2: Convert to the correct units and calculate the pressure

n = 0.20 mol

V = 10.1 dm3 = 0.0101 m3 = 10.1 x 10-3 m3 

R = 8.31 J K-1 mol-1

T = 25 oC = 298 K

P space equals space fraction numerator 0.20 space m o l space cross times 8.31 space J space K to the power of negative 1 space end exponent m o l to the power of negative 1 end exponent cross times 298 space K over denominator 10.1 space cross times 10 to the power of negative 3 end exponent m cubed end fraction

P = 49 037 Pa = 49 kPa (2 sig figs)

Worked Example

Calculate the temperature of a gas, in oC, if 0.047 moles of the gas occupy 1.2 dm3 at a pressure of 100 kPa.

Answer:

  • Step 1: Rearrange the ideal gas equation to find the temperature of the gas

T space equals space fraction numerator P V over denominator n R end fraction

  • Step 2: Convert to the correct units and calculate the pressure

n = 0.047 mol

V = 1.2 dm3 = 0.0012 m3 = 1.2 x 10-3 m3 

R = 8.31 J K-1 mol-1

P = 100 kPa = 100 000 Pa

T space equals space fraction numerator 100 space 000 space cross times 1.2 space cross times 10 to the power of negative 3 end exponent m cubed over denominator 0.047 space m o l space cross times 8.31 space J space K to the power of negative 1 end exponent m o l to the power of negative 1 end exponent end fraction

T = 307.24 K

T = 34.24 oC = 34 oC  (2 sig figs)

Worked Example

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 molar mass of the gas.

Answer:

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

n equals space fraction numerator P V over denominator R T end fraction

  • Step 2: Convert to the correct units and calculate the number of moles of gas

P = 300 kPa = 300 000 Pa

V = 1000 cm3 = 0.001 m3 = 1.0 x 10-3 m3 

R = 8.31 J K-1 mol-1

T = 23 oC = 296 K

      n = 0.12 mol

  • Step 3: Calculate the molar mass using the number of moles of gas

molar space mass space equals space mass over moles

M space equals space fraction numerator 6.39 space straight g over denominator 0.12 space mol end fraction equals bold space bold 53 bold space bold italic g bold space bold italic m bold italic o bold italic l to the power of bold minus bold 1 end exponent bold space stretchy left parenthesis 2 space sig space figs stretchy right parenthesis

Examiner Tips and Tricks

To calculate the temperature in Kelvin, add 273 to the Celsius temperature, eg. 100 oC is 373 Kelvin.

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Author
Reviewer
Alexandra Brennan

Author: Alexandra Brennan

Expertise: Chemistry Content Creator

Alex studied Biochemistry at Newcastle University before embarking upon a career in teaching. With nearly 10 years of teaching experience, Alex has had several roles including Chemistry/Science Teacher, Head of Science and Examiner for AQA and Edexcel. Alex’s passion for creating engaging content that enables students to succeed in exams drove her to pursue a career outside of the classroom 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

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