Molar Gas Volume (DP IB Chemistry): Revision Note

Stewart Hird

Written by: Stewart Hird

Reviewed by: Philippa Platt

Updated on

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

Diagram of a cube showing gas molecules as blue circles with arrows indicating random motion in different directions within the cube.
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 increased gas pressure by compressing molecules, leading to more frequent collisions with container walls.
Decreasing the volume of a gas causes an increased collision frequency of the gas particles with the container wall
  • Pressure is inversely proportional to volume (when temperature is constant):

P1 over V

PV = a constant

  • This is known as Boyle's Law

  • 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

Three graphs: Pressure vs 1/Volume (linear increase), Pressure vs Volume (hyperbolic decrease), and Pressure x Volume vs Pressure (constant line).
Three graphs that show Boyle’s Law

Changing gas temperature

  • When a gas is heated (at constant pressure) the particles gain more kinetic energy

  • This leads to more frequent collisions with the container walls

  • To keep the pressure constant, the gas must expand, so the volume increases

  • The volume is therefore directly proportional to the temperature in Kelvin (at constant pressure)

V T

V over T = a constant

  • This is known as Charles' Law

  • A graph of volume against temperature in Kelvin gives a straight line

Gas molecule collision frequency with increasing temperature diagram

Diagram of gas expansion: low temperature with slow molecules, high temperature with faster molecules. Volume-temperature graph illustrates a direct relationship.
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

  • The particles move faster and collide more frequently with the container walls

  • This leads to an increase in pressure

  • Therefore, pressure is directly proportional to temperature (in Kelvin) at constant volume

PT

P over 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

Two diagrams: A shows gas molecules in a container with increased collisions when temperature rises. B shows a graph of temperature vs pressure, forming a straight line.
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:

    • PV = 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

  • When n (moles) and R (gas constant) are constant, the ideal gas law simplifies to:

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

    • P2, V2 and T2 are the final conditions of the gas

  • The temperature must be in Kelvin

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 in oC

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

Convert to oC = -124 oC

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

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Stewart Hird

Author: Stewart Hird

Expertise: Chemistry Content Creator

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.

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