Gas Laws (AQA A Level Physics)

Revision Note

Katie M

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Katie M

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Ideal Gas Laws

  • The ideal gas laws are the experimental relationships between pressure (P), volume (V) and temperature (T) of an ideal gas

  • You need to be able to explain the following three ideal gas laws:

    • Boyles Law

    • Charles's Law

    • Pressure Law

  • The mass and the number of molecules of the gas is assumed to be constant for all of these

Boyle’s Law

  • If the temperature T of an ideal gas is constant, then Boyle’s Law is given by:

Boyle's Law_2
  • This means the pressure is inversely proportional to the volume of a gas

  • The relationship between the pressure and volume for a fixed mass of gas at constant temperature can also be written as:

P1V1 = P2V2

  • Where:

    • P1 = initial pressure (Pa)

    • P2 = final pressure (Pa)

    • V1 = initial volume (m3)

    • V2 = final volume (m3)

2-3-3-gas-laws-volume-pressure-graph

Boyle's Law graph representing pressure inversely proportional to volume

  • If the temperature is higher but still constant, then the graph has the same shape but is shifted further above the origin

Charles's Law

  • If the pressure P of an ideal gas is constant, then Charles’s law is given by:

V ∝ T

  • This means the volume is directly proportional to the temperature of a gas

  • The relationship between the volume and thermodynamic temperature for a fixed mass of gas at constant pressure can also be written as:

Charles's Law
  • Where:

    • V1 = initial volume (m3)

    • V2 = final volume (m3)

    • T1 = initial temperature (K)

    • T2 = final temperature (K)

2-3-3-gas-laws-volume-temperature-graph

Charles's Law graph representing temperature (in K) directly proportional to the volume

 

  • A straight line through the origin shows direct proportionality

Pressure Law

  • If the volume V of an ideal gas is constant, the Pressure law is given by:

PT

  • This means the pressure is proportional to the temperature

  • The relationship between the pressure and thermodynamic temperature for a fixed mass of gas at constant volume can also be written as:

Pressure Law
  • Where:

    • P1 = initial pressure (Pa)

    • P2 = final pressure (Pa)

    • T1 = initial temperature (K)

    • T2 = final temperature (K)

pressure-temp-kelvin

Pressure Law graph representing temperature (in K) directly proportional to the volume

  • A straight line through the origin shows direct proportionality between pressure and temperature in kelvin

Worked Example

The pressure inside a bicycle tyre is 5.10 × 105 Pa when the temperature is 279 K.After the bicycle has been ridden, the temperature of the air in the tyre is 299 K. Calculate the new pressure in the tyre, assuming the volume is unchanged.

Answer:

Pressure Law Worked Example _2

Examiner Tips and Tricks

Remember when using any ideal gas law, including the ideal gas equation, the temperature must always be in kelvin (K).

For Charles' Law, for example, if the x axis is temperature in degrees Celsius, not kelvin, the straight line no longer passes through the origin. 0 K is −273 °C so the line passes through the x axis at this value:

Charles Law

Relationships Between Pressure, Volume & Temperature

  • An ideal gas is one that obeys the relation:

pVT

  • Where:

    • p = pressure of the gas (Pa)

    • V = volume of the gas (m3)

    • T = thermodynamic temperature (K)

  • The molecules in a gas move around randomly at high speeds, colliding with surfaces and exerting pressure upon them

Gas molecules move about randomly at high speeds

  • Imagine molecules of gas free to move around in a box

  • The temperature of a gas is related to the average speed of the molecules:

    • The hotter the gas, the faster the molecules move

    • Hence the molecules collide with the surface of the walls more frequently

  • Since force is the rate of change of momentum:

    • Each collision applies a force across the surface area of the walls

    • The faster the molecules hit the walls, the greater the force on them

  • Since pressure is the force per unit area

    • Higher temperature leads to higher pressure

  • If the volume V of the box decreases, and the temperature T stays constant:

    • There will be a smaller surface area of the walls and hence more collisions

    • This also creates more pressure

  • Since this equates to a greater force per unit area, pressure in an ideal gas is therefore defined by:

    The frequency of collisions of the gas molecules per unit area of a container

Gas Laws Molecular Model (1), downloadable AS & A Level Physics revision notes
Gas Laws Molecular Model (2), downloadable AS & A Level Physics revision notes
Gas Laws Molecular Model (3), downloadable AS & A Level Physics revision notes

Molecular model of the three ideal gas laws

Worked Example

An ideal gas is in a container of volume 4.5 × 10-3 m3.The gas is at a temperature of 30 °C and a pressure of 6.2 × 105 Pa. Calculate the pressure of the ideal gas in the same container when it is heated to 40 °C.

Answer:

Step 1: Ideal gas relation between pressure, volume and temperature

pVT

Step 2: Write the equation in full

pV = kT

  • Where k = the constant of proportionality

Step 3: Rearrange for the constant of proportionality

k italic equals fraction numerator p V over denominator T end fraction

Step 4: Convert temperature T into Kelvin

θ °C + 273.15 = T K

30 °C + 273.15 = 303.15 K

Step 5: Substitute in known value into constant of proportionality equation

k equals fraction numerator 6.2 cross times 10 to the power of 5 cross times 4.5 cross times 10 to the power of negative 3 end exponent over denominator 303.15 end fraction equals 9.203...

Step 6: Rearrange ideal gas relation equation for pressure

p italic equals fraction numerator k T over denominator V end fraction

Step 7: Substitute in new values

k = 9.203...

V stays the same = 4.5 × 10-3 m3

T = 40 °C = 40 + 273.15 = 313.15 K

p equals fraction numerator open parentheses 9.203... close parentheses cross times 313.15 over denominator 4.5 cross times 10 to the power of negative 3 end exponent end fraction equals 640.45... cross times 10 cubed space Pa equals 640 space kPa

Examiner Tips and Tricks

Don’t round too early in your working out! In the worked example, the unrounded value of k is represented by “…” to show its full value is to be carried over to the next step of the calculation. On your calculator, this can be done by using the “ans” button instead of typing in the whole number.

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Katie M

Author: Katie M

Expertise: Physics

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.