IB Chemistry SL

Revision Notes

1.2.5 Gas Law Relationships

Gas Law Relationships

  • Gases in a container exert a pressure as the gas molecules are constantly colliding with the walls of the container

States of Matter Pressure, downloadable IB Chemistry revision notes

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

States of Matter Volume and Pressure, downloadable IB Chemistry revision notes

Decreasing the volume of a gas causes an increased collision frequency of the gas particles with the container wall

  • The pressure is therefore inversely proportional to the volume (at constant temperature)
  • This is known as Boyle’s Law
  • Mathematically, we say P ∝ 1/V or PV = a constant
  • 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

Graphs of Boyle’s Law, downloadable IB Chemistry revision notes

Three graphs that show Boyle’s Law

Changing gas temperature

  • When a gas is heated (at constant pressure) the particles gain more kinetic energy and undergo more frequent collisions with the container walls
  • To keep the pressure constant, the molecules must get further apart and therefore the volume increases
  • The volume is therefore directly proportional to the temperature in Kelvin (at constant pressure)
  • This is known as Charles’ Law
  • Mathematically, V ∝ T or V/T = a constant
  • A graph of volume against temperature in Kelvin gives a straight line

States of Matter Volume and Temperature, downloadable AS & A Level Chemistry revision notes

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
  • This means that the particles will move faster and collide with the container walls more frequently
  • The pressure of the gas increases
  • The temperature is therefore directly proportional to the pressure (at constant volume)
  • Mathematically, we say that P ∝ T or P/T = a constant 
  • A graph of temperature in Kelvin of a gas plotted against pressure gives a straight line

States of Matter Temperature and Pressure, downloadable IB Chemistry revision notes

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:
    • P/V = 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

  • For a fixed amount of gas, n and R will be constant, so if you change the conditions of a gas we can ignore n and R in the ideal gas equation
  • This leads to a very useful expression for problem solving
  • Where P1, V1 and T1 are the initial conditions of the gas and P2, V2 and T2 are the final conditions

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

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

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

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

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