STATES OF MATTER - properties of gases and liquids (fluids) and solids

19. The kinetic particle theory of gases and Charles' Law and Gay-Lussac's Law Calculations - practise exam questions

involving the ideal gas equations V₁/V₂ = T₁/T₂  (V T) and  P₁/T₁ = P₂/T₂  (P T) and exam practice questions

 Also helpful for UK advanced level chemistry students aged ~16-18, IB chemistry courses and US grades 11-12 K12 AP honors courses

INDEX of all my notes on the states of matter

GCSE (~US grades 8-10) level multiple choice QUIZ on the states of matter: gases, liquids & solids

19. Charles's Law and Gay-Lussac's Law for pressure/volume and temperature calculations

• The particle theory of gas pressure was explained in Part 1 so this section concentrates on the gas law calculations involving pressure and volume and their variation with temperature.

• (i) Charles's Law states for a fixed mass of gas at constant pressure:

  • The volume of a gas is directly proportional to the absolute temperature (K) at constant pressure

  • V = constant x T (right graph), or

  • V/T = constant, or

  • V₁/V₂ = T₁/T₂  for conditions changing from 1 (initial) to 2 (final)

  • or V₁/T₁ = V₂/T₂  (for constant pressure)

  • V₁ x T₂ = V₂ x T₁

  • V₂ = V₁ x T₂/T₁

  • or T₂ = T₁ x V₂/V₁  (for constant pressure)

  • Kinetic particle model reasoning - increasing the temperature increases the kinetic energy of the molecules giving more forceful collisions which 'push out' to expand the gas at constant pressure (major factor) and the chance of particle collision with the sides of the container is also increased (minor factor).

  • So both effects contribute to an increase in volume at constant pressure with increase in temperature (above right graph.

  • Note that the graphs extrapolate back to 0K (absolute zero, Kelvin scale) or -273^oC (Celsius scale).

• (ii) Gay-Lussac's Law states that for a fixed mass of gas at constant volume:

• The pressure of a gas is directly proportional to the absolute temperature (K) at constant volume,

  • p = constant x T (right graph), or

  • p/T = constant, or

  • p₁/p₂ = T₁/T2  for conditions changing from 1 (initial) to 2 (final),

  • or p₁/T₁ = p₂/T₂  (for constant volume)

  • p₁ x T₂ = T₁ x p₂

  • p₂ = p₁ x T₂/T₁

  • or T₂ = T₁ x p₂/p₁ (for constant volume)

  • Kinetic particle model reasoning - increasing the temperature increases the kinetic energy of the molecules giving more forceful collisions (major factor) and greater chance of collision (minor factor), both of which contribute to an increase in the pressure if the volume is constrained (kept constant).

  • Note again that the graphs extrapolate back to 0K (absolute zero, Kelvin scale) or -273^oC (Celsius scale).

• In all calculations, the absolute or Kelvin scale of temperature must be used for T (K = ^oC + 273).

Some exam practice questions based on Charles's Law and Gay Lussac's Law

ANSWERS

• Q1

  • The pressure exerted by a gas in sealed container is 100kPa at 17^oC. It was found that the container might leak if the internal pressure exceeds 120kPa. Assuming constant volume, at what temperature in ^oC will the container start to leak?

• Q2

  • A cylinder of propane gas at 20^oC exerted a pressure of 8.5 atmospheres. When exposed to sunlight it warmed up to 28^oC. What pressure does the container side now experience?

• Q3

  • 12.0 dm³ of gas in a cylinder and piston system is heated from 290 K to 340 K. If the pressure remains constant, calculate the final volume of gas in the cylinder.

• Q4

• The fuel and air gases in the cylinders of a 1200 cm³ car engine go from 25^oC before combustion and rise to a peak temperature of 2100^oC after combustion. If normal atmospheric pressure is 101 kPa, calculate the peak pressure reached after combustion assuming the volume is constant at both ends of the cycle.

ANSWERS

If you think there are any errors, email details to chem55555@hotmail.com

Key points about the  Kinetic Particle Theory & Gas Laws

Kinetic Particle Theory of Gases

• Gas particles are in constant random motion.
• Temperature is a measure of the average kinetic energy of particles.
• As temperature increases:
  • Particles move faster.
  • Collisions with container walls are more frequent and more forceful.
  • So with change in temperature, this explains both Charles’ Law (volume change at constant pressure) and Gay-Lussac’s Law (pressure change at constant volume).

Charles’ Law

• Statement: At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature (Kelvin).
• (V T)
• Equation:
  V₁/V₂ = T₁/T₂

Gay-Lussac’s Law

• Statement: At constant volume, the pressure of a fixed mass of gas is directly proportional to its absolute temperature (Kelvin).
• P T
• Equation:
  P₁/T₁ = P₂/T₂

Student Exam Tips when Charles' Law and Gay-Lussac' Law

• Always convert °C to K before using gas law equations.
• State assumptions: fixed mass of gas, constant pressure/volume depending on the law.
• Rearrange equations carefully:
  • V₁/V₂ = T₁/T₂  and  P₁/T₁ = P₂/T₂
• Graphs:
  • Charles’ Law: Volume versus Temperature (K) → straight line through origin.
  • Gay-Lussac’s Law: Pressure versus Temperature (K) → straight line through origin.
• In multiple-choice, check if the question asks for absolute temperature not Celsius.

Typical Misconceptions about using Charles' Law and Gay-Lussac' Law

•  Forgetting to convert °C to K — laws only work with Kelvin.
•  Thinking volume or pressure is proportional to Celsius temperature — it’s proportional to Kelvin.
•  Assuming particles stop moving at 0 °C — they only reach minimum kinetic energy at 0 K (absolute zero).
•  Mixing up which variable is constant:
  • Charles’ Law → pressure constant.
  • Gay-Lussac’s Law → volume constant.
•  Misinterpreting graphs: lines must pass through the origin (0 K), not 0 °C.

Quick Recap on Charles' Law and Gay-Lussac' Law

• Charles’ Law: ( V proportional to T ) (at constant pressure).
• Gay-Lussac’s Law: ( p proportional to T ) (at constant volume).
• Both explained by kinetic particle theory: higher temperature → faster particles → stronger/more frequent collisions.
• Exam boards require calculations, graphs, and theory links.
• Tips: convert to Kelvin, state assumptions, interpret graphs correctly.
• Misconceptions: Celsius versus Kelvin, wrong constant variable, misunderstanding particle motion at absolute zero.

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ANSWERS to Charles' Law and Gay Lussac's Law calculations

• Q1

  • The pressure exerted by a gas in sealed container is 100kPa at 17^oC. It was found that the container might leak if the internal pressure exceeds 120kPa. Assuming constant volume, at what temperature in ^oC will the container start to leak?

  • 17^oC + 273 = 290K

  • p₁/T₁ =  p₂/T₂

  • rearranging to scale up to the higher temperature

  • T₂ = T₁ x p₂/p₁

  • T₂ = 290 x 120/100 = 348 K or 348 – 273 = 75^oC when the container might leak

• Q2

  • A cylinder of propane gas at 20^oC exerted a pressure of 8.5 atmospheres. When exposed to sunlight it warmed up to 28^oC. What pressure does the container side now experience?

  • 20^oC = 273 + 20 = 293K, 28^oC = 273 + 28 = 301K

  • p₂ = p₁ x T₂/T₁

  • p₂ = 8.5 x 301/293 = 8.73 atm

• Q3

  • 12.0 dm³ of gas in a cylinder and piston system is heated from 290 K to 340 K. If the pressure remains constant, calculate the final volume of gas in the cylinder.

  • V/T = constant

  • V₁/V₂ = T₁/T₂

  • V₁ x T₂ = V₂ x T₁

  • V₂ = V₁ x T₂/T₁

  • V₂ = 12 x 340/290 = 14.1 dm³

• Q4

  • The fuel and air gases in the cylinders of a 1200 cm³ car engine go from 25^oC before combustion and rise to a peak temperature of 2100^oC after combustion. If normal atmospheric pressure is 101 kPa, calculate the peak pressure reached after combustion assuming the volume is constant at both ends of the cycle.

  • T₁ = 25 + 273 = 298 K, T₂ = 2100 + 273 = 2373 K, P1 = 101 KPa

  • p/T = constant

  • p₁/p₂ = T₁/T2

  • p₂ = p₁ x T₂/T₁

  • p₂ = 101 x 2373/298 = 804 kPa

If you think there are any errors, email details to chem55555@hotmail.com