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

19. Charles Law and Gay-Lussac's Law Calculations

involving V1/V2 = T1/T2  and  P1T1 =P2T2

Doc Brown's chemistry revision notes: basic school chemistry science GCSE chemistry, IGCSE  chemistry, O level and ~US grades 8, 9 and 10 school science courses or equivalent for ~14-16 year old science students for national examinations in chemistry and also helpful for UK advanced level chemistry students aged ~16-18 and US grades 11-12 K12 honors.


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.

  • Charles's Law, V versus T graph(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

    • 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 -273oC (Celsius scale).

  • Charles's Law, P versus T graph(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

    • 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 -273oC (Celsius scale).

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


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

  • Q1

    • The pressure exerted by a gas in sealed container is 100kPa at 17oC. 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 20oC exerted a pressure of 8.5 atmospheres. When exposed to sunlight it warmed up to 28oC. 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 25oC before combustion and rise to a peak temperature of 2100oC 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


Learning objectives for calculations based on Charles's Law and Gay-Lussac's Law calculations

Know that for a fixed mass of gas the volume is directly proportional to the absolute temperature on the Kelvin scale (Charles's Law).

Know that for a fixed mass of gas at constant volume the pressure is directly proportional to the absolute temperature (K) of the gas (Gay Lussac's Law).

Be able explain Charles's Law and Gay Lussac's Law in terms of the kinetic particle theory i.e. the frequency, and more importantly, the kinetic energy of the particles and their force of impact.

Be able to do calculations based on Charles's Law i.e. V = constant x T and V1 x T2 = V2 x T1  for a fixed mass of gas at constant pressure:

Be able to do calculations based on Gay Lussac's Law i.e. p = constant x T and p1/p2 = T1/T2  for a fixed mass of gas at constant volume:


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

  • Q1

    • The pressure exerted by a gas in sealed container is 100kPa at 17oC. 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?

    • 17oC + 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 = 75oC when the container might leak

  • Q2

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

    • 20oC = 273 + 20 = 293K, 28oC = 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 25oC before combustion and rise to a peak temperature of 2100oC 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

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