Syllabus Edition

First teaching 2023

First exams 2025

|

Molar Gas Volume (HL IB Chemistry)

Revision Note

Test Yourself
Stewart

Author

Stewart

Expertise

Chemistry Lead

Molar Gas Volume

  • Gases in a container exert a pressure as the gas molecules are constantly colliding with the walls of the container

A particle model of a gas in a container
states-of-matter-pressure

Gas particles exert a pressure by constantly colliding with the walls of the container

Changing gas volume

  • Decreasing the volume (at constant temperature) of the container causes the molecules to be squashed together which results in more frequent collisions with the container wall
  • The pressure of the gas increases

Gas Molecule Collision Frequency with Increasing Pressure Diagram

Diagram showing how volume of a gas varies with pressure

Decreasing the volume of a gas causes an increased collision frequency of the gas particles with the container wall

  • The pressure is therefore inversely proportional to the volume (at constant temperature)
  • This is known as Boyle's Law
  • Mathematically, we say P ∝ 1 / V or PV = a constant
  • We can show a graphical representation of Boyle's Law in three different ways:
    • A graph of pressure of gas plotted against 1 / volume gives a straight line
    • A graph of pressure against volume gives a curve
    • A graph of PV versus P gives a straight line

Sketch graphs of Boyles' Law

Sketch graphs representing Boyles' Law

Three graphs that show Boyle’s Law

Changing gas temperature

  • When a gas is heated (at constant pressure) the particles gain more kinetic energy and undergo more frequent collisions with the container walls
  • To keep the pressure constant, the molecules must get further apart and therefore the volume increases
  • The volume is therefore directly proportional to the temperature in Kelvin (at constant pressure)
  • This is known as Charles' Law
  • Mathematically, V T or V / T = a constant
  • A graph of volume against temperature in Kelvin gives a straight line

Gas Molecule Collision Frequency with Increasing Temperature Diagram

states-of-matter-volume-and-temperature

Increasing the temperature of a gas causes an increased collision frequency of the gas particles with the container wall (a); volume is directly proportional to the temperature in Kelvin (b)

Changing gas pressure

  • Increasing the temperature (at constant volume) of the gas causes the molecules to gain more kinetic energy
  • This means that the particles will move faster and collide with the container walls more frequently
  • The pressure of the gas increases
  • The temperature is therefore directly proportional to the pressure (at constant volume)
  • Mathematically, we say that PT or P / T = a constant 
  • A graph of temperature in Kelvin of a gas plotted against pressure gives a straight line

Gas Molecule Collision Frequency with Increasing Pressure at Constant Volume Diagram

states-of-matter-temperature-and-pressure

Increasing the temperature of a gas causes an increased collision frequency of the gas particles with the container wall (a); temperature is directly proportional to the pressure (b)

Pressure, volume and temperature

  • Combining these three relationships together:
    • P / V = a constant
    • V / T = a constant
    • P / T = a constant
  • We can see how the ideal gas equation is constructed
    • PV / T = a constant
    • PV = a constant x T
  • This constant is made from two components, the number of moles, n, and the gas constant, R, resulting in the overall equation:
    • PV = nRT

Changing the conditions of a fixed amount of gas

  • For a fixed amount of gas, n and R will be constant, so if you change the conditions of a gas we can ignore n and R in the ideal gas equation
  • This leads to a very useful expression for problem solving

fraction numerator bold italic P subscript bold 1 bold italic V subscript bold 1 over denominator bold italic T subscript bold 1 end fraction bold equals bold space fraction numerator bold italic P subscript bold 2 bold italic V subscript bold 2 over denominator bold italic T subscript bold 2 end fraction

  • Where P1, V1 and T1 are the initial conditions of the gas and P2, V2 and T2 are the final conditions

Worked example

At 25 oC and 100 kPa a gas occupies a volume of 20 dm3. Calculate the new temperature, in oC, of the gas if the volume is decreased to 10 dm3 at constant pressure.

Answer:

Step 1: Rearrange the formula to change the conditions of a fixed amount of gas. Pressure is constant so it is left out of the formula

italic T subscript bold italic 2 bold equals bold space fraction numerator V subscript bold 2 italic T subscript bold italic 1 over denominator V subscript bold italic 1 end fraction

Step 2: Convert the temperature to Kelvin. There is no need to convert the volume to mbecause the formula is using a ratio of the two volumes

 V1 = 20 dm

V2 = 10 dm

T1 = 25 + 273 = 298 K

Step 3: Calculate the new temperature

bold italic T subscript bold italic 2 bold equals bold space fraction numerator bold 10 bold space bold dm to the power of bold 3 bold italic space bold italic cross times bold space bold 298 bold italic space bold K over denominator bold 20 bold space bold dm to the power of bold 3 end fraction bold space bold equals bold space bold 149 bold space bold K bold space bold equals bold space bold minus bold 124 bold space to the power of bold o bold C

Worked example

A 2.00 dm3 container of oxygen at a pressure of 80 kPa was heated from 20 oC to 70 oC The volume expanded to 2.25 dm3 . What was the final pressure of the gas?

Answer:

Step 1: Rearrange the formula to change the conditions of a fixed amount of gas

bold italic P subscript bold 2 bold equals bold space fraction numerator bold italic P subscript bold 1 bold italic V subscript bold 1 bold italic T subscript bold italic 2 over denominator bold italic V subscript bold italic 2 bold italic T subscript 1 end fraction

Step 2: Substitute in the values and calculate the final pressure

P1 = 80 kPa

V= 2.00 dm

V= 2.25 dm

T= 20 + 273 = 293 K

T2 = 70 + 273 = 343 K

bold italic P subscript bold 2 bold equals bold space fraction numerator bold 80 bold space bold kPa bold space bold cross times bold space bold 2 bold. bold 00 bold space bold dm to the power of bold 3 bold space bold x bold space bold 343 bold space bold K over denominator bold 293 bold space bold K bold space bold cross times bold space bold 2 bold. bold 25 bold space bold dm to the power of bold 3 end fraction bold space bold equals bold space bold 83 bold space bold kPa

You've read 0 of your 0 free revision notes

Get unlimited access

to absolutely everything:

  • Downloadable PDFs
  • Unlimited Revision Notes
  • Topic Questions
  • Past Papers
  • Model Answers
  • Videos (Maths and Science)

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Stewart

Author: Stewart

Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.

Join over 500 thousand students
getting better grades