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Charles' Law

Key Concepts

  • At constant pressure, the volume of a given quantity of gas is directly proportional to the absolute temperature : V ∝ T (in Kelvin)

    So at constant pressure, if the temperature (K) is doubled, the volume of gas is also doubled.

    OR

    At constant pressure for a given quantity of gas, the ratio of its volume and the absolute temperature is a constant : V/T = constant, V/T = k

  • At constant pressure for a given quantity of gas : Vi/Ti = Vf/Tf

    where Vi is the initial (original) volume, Ti is its initial (original) temperature (in Kelvin), Vf is its final volme, Tf is its final tempeature (in Kelvin)

    Vi and Vf must be in the same units of measurement (eg, both in litres), Ti and Tf must be in Kelvin NOT celsius.

    temperature in kelvin = temperature in celsius + 273 (approximately)

  • All gases approximate Charles' Law at high temperatures and low pressures.

    A hypothetical gas which obeys Charles' Law at all temperatures and pressures is called an Ideal Gas.

    A Real Gas is one which approaches Charles' Law as the temperature is raised or the pressure lowered.

    As a Real Gas is cooled at constant pressure from a point well above its condensation point, its volume begins to increase linearly.
    As the temperature approaches the gases condensation point, the line begins to curve (usually downward) so there is a marked deviation from Ideal Gas behaviour close to the condensation point.
    Once the gas condenses to a liquid it is no longer a gas and so does not obey Charles' Law at all.
    Absolute zero (0K, -273oC approximately) is the temperature at which the volume of a gas would become zero if it did not condense and if it behaved ideally down to that temperature.

Graphical Representations

Expansion of Hydrogen gas at constant pressure

Volume
(mL)
Temperature
(oC)
Temperature
(K)
V/T (K) Graph
25 -23 250 0.1
30 27 300 0.1
35 77 350 0.1
40 127.5 400.5 0.1
45 177 450 0.1

Calculations : Vi/Ti = Vf/Tf

  1. A sample of gas at 101.3 kPa had a volume of 1.2 L at 100oC. What would its volume be at 0oC at the same pressure?

    Vi = 1.2L                                   Vf = ?

    Ti = 100oC = 100 + 273 = 373 K     Tf = 0oC = 0 + 273 =273 K

    1.2/373 = Vf/273

    3.22 x 10-3 = Vf/273

    Vf = 3.22 x 10-3 x 273 = 0.88 L (880 mL)

  2. A balloon had a volume of 75 L at 25oC. To what does the temperature need to raised in order for the balloon to have a volume of 100 L at the same pressure?

    Vi = 75 L                                 Vf = 100 L

    Ti = 25oC = 25 + 273 = 298 K     Tf = ? (K)

    Vi/Ti = Vf/Tf

    75/298 = 100/Tf

    0.2517 = 100/Tf

    Tf = 100/0.2517 = 397 K (397-273 = 124oC)


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