Charles Law

Charles Law in General Gas Equation

In a popular laboratory experiment, a fully inflated balloon is placed inside a container with liquid nitrogen (which is at a temperature of -196 ˚C), and we can observe that the balloon shrinks to about 1/1000th of its former size. Similarly, if we keep an inflated balloon in the fridge it too shrinks considerably. After this, if the balloon is carefully removed from the container or the fridge is allowed to slowly warm itself to room temperature, it will regain the size that it was as before; it was shrunk down due to the cooling of liquid nitrogen. This is one of the simplest experiments which can be performed to demonstrate the effects of temperature on gases.

The law that is used to explain the effects of temperature on gases under constant pressure is called as Charles Law. When a gas is heated, the molecules gain more energy and move about, thus expanding the space that they need to occupy. It is named in recognition of the French scientist and balloon flight pioneer Jacques Alexandre César Charles. The statement of Charles law is "The volume of any gas is directly proportional to its temperature on the Kelvin scale under the same amount of pressure”

Mathematically, the directly proportional relationship between temperature and volume in Charles law can be expressed as

V T
Or
V/T =k

Where:

     V is the volume of the given gas
     T is the temperature of the given gas measured in kelvins
     k is a constant, obtained by dividing V by T

Charles also stated that the volume of a fixed mass of gas decreases or increases by 1/273 times the volume of that gas at 0 °C for every fall or rise in the temperature of the liquid by 1 °C.
Therefore,
          VT = V0 + (1/273 × V0) × T
          VT = V0 + (1 + T/273)

Where:    


 VT is the volume of the given gas at the temperature T
     V0 is the volume of the given gas at 0 °C

Discovery and naming of the law:



The law was named after Jacques Alexandre César Charles, who was a French mathematician, scientist, inventor, and balloonist. He is popularly known for being the first to launch a manned hydrogen-filled gas balloon. Only a few days after the first manned launch of the hot air balloon, natural philosopher Joseph Louis Gay-Lussac in 1802 published a paper describing how gases tend to expand when heated. He credited it to unpublished work of Jacques Charles in the 1780s. Gay-Lussac also named this law as ‘Charles Law' in his honour.

Charles did an experiment at 1787 in which he filled five containers to the same volume with different gases. He then raised the temperature of these containers to 80 °C. He then noticed that all the gases increased by a similar amount. This experiment was referenced by Gay-Lussac in his paper in 1802. From this, Charles observed that under constant pressure, the volume of a gas increases linearly with the absolute temperature of that gas
The formula created by Charles was V1/T1 = V2/T2.
We can also further consider this to give V1/T1 = V2/T2 = V3/T3 = V4/T4.

Charles Law in General Gas Equation:

The combined gas laws describing the properties of gases relating to pressure, temperature and volume give the General Gas Equation. General Gas Equation is obtained by combining Charles’s Law, Boyle’s Law and Gay-Lussac’s Law. Boyle’s Law explains the relationship between volume and pressure at constant pressure. Gay-Lussac’s law explains the relationship between pressure and temperature under constant volume. Along with Charles’s law that explains the relationship between temperature and volume, they are the three foundations that constitute the General Gas Equation.

The combined gas law states that the volume of any gas is proportional to the temperature of the gas divided by its pressure. It can be expressed as

V T

Or

PV/T =k

Where:

     V is the volume of the given gas
     T is the temperature of the given gas measured in kelvins

P is the pressure which the gives gas is subjected to
     k is a constant

Problems with Charles’s Law:

Charles’s Law says that volume directly increases or decreases with an increase or decrease in volume. Hence as the temperature decreases, the volume also decreases proportionally. As the temperature becomes very low, theoretically the volume of the gas should decrease until it starts approaching zero. This seems to imply that at absolute zero (a theoretical temperature at which a gas possesses zero energy and hence restrict motion), the gas should have no volume. Of course, during the time of Charles's or Gay-Lussac's research, they had no experience with liquid air, which was not prepared until 1877 even though he believed that gases like hydrogen and oxygen could be liquified. Gay-Lussac worked with vapours of volatile liquids (liquids with low boiling points) and was aware that Charles's law does not apply to liquids just above the boiling point of that liquid. Later after liquefaction of gases like hydrogen and oxygen became possible, it was found that Charles's law cannot be applied to gases like these just above their boiling point. Hence Charles's law does not hold true at extremely low temperature or just above the boiling point of liquids.

Examples for problems requiring the application of Charles’s Laws:


  • 1) A balloon is filled to a volume of 2.40 L at a temperature of 20 °C. The balloon is then heated to a temperature of 70 °C. Assuming the pressure remains constant throughout, find the new volume of the balloon.

  • Solution: Given:
    V1 = 2.40 L
    T1 = 20 °C = 293 K
    T2 = 70 °C = 343 K
    V2 =?
    We know V1/T1 = V2/T2
                   V2 = V1 ×T2 / T1
                   V2 = 2.40 × 343 / 293
                   V2 = 2.80 L

  • 2) A sample of a gas has an initial volume of 30.8 L and an initial temperature of −67°C. What must be the temperature of the gas for its volume to be 21.0 L?

  • Solution: Given:
    V1 = 30.80 L
    T1 = -67 °C = 206 K
    T2 =?
    V2 = 21.0 L
    We know V1/T1 = V2/T2
                   T2 = V2×T1 / V1
                   T2 = 30.80 × 206 / 21.0
                   T2 = 302 K =29 °C

  • 3) If V1 = 3.60 L and T1 = 255 K, what is V2 if T2 = 102 K?

  • Solution: Given:
    V1 = 3.60 L
    T1 = 255 K
    T2 = 102 K
    V2 =?
    We know V1/T1 = V2/T2
                   V2 = V1 ×T1 / T2
                   V2 = 3.60 × 255 / 102
                   V2 = 9 L