CIE A Level Physics (9702) exams from 2022

Revision Notes

15.2.3 Average Kinetic Energy of a Molecule

Average Kinetic Energy of a Molecule

  • An important property of molecules in a gas is their average kinetic energy
  • This can be deduced from the ideal gas equations relating pressure, volume, temperature and speed
  • Recall the ideal gas equation:

pV = NkT

  • Also recall the equation linking pressure and mean square speed of the molecules:

Average Kinetic Energy of a Molecule equation 1

  • The left hand side of both equations are equal (pV)
  • This means the right hand sides are also equal:

Average Kinetic Energy of a Molecule equation 2

  • N will cancel out on both sides and multiplying by 3 obtains the equation:

m<c2> = 3kT

  • Recall the familiar kinetic energy equation from mechanics:

Average Kinetic Energy of a Molecule equation 3

  • Instead of v2 for the velocity of one particle, <c2> is the average speed of all molecules
  • Multiplying both sides of the equation by ½ obtains the average translational kinetic energy of the molecules of an ideal gas:

Average Kinetic Energy of a Molecule equation 4

  • Where:
    • EK = kinetic energy of a molecule (J)
    • m = mass of one molecule (kg)
    • <c2> = mean square speed of a molecule (m2 s-2)
    • k = Boltzmann constant
    • T = temperature of the gas (K)
  • Note: this is the average kinetic energy for only one molecule of the gas
  • A key feature of this equation is that the mean kinetic energy of an ideal gas molecule is proportional to its thermodynamic temperature

EK ∝ T

  • Translational kinetic energy is defined as:

The energy a molecule has as it moves from one point to another

  • A monatomic (one atom) molecule only has translational energy, whilst a diatomic (two-atom) molecule has both translational and rotational energy

Worked example: Average kinetic energy of a molecule

Average_Kinetic_Energy_of_a_Molecule_Worked_example_-_Average_Kinetic_Energy_of_a_Molecule_Question, downloadable AS & A Level Physics revision notes

Step 1:            Write down the equation for the average translational kinetic energy:

Average Kinetic Energy of a Molecule Worked Example equation 1

Step 2:            Find the relation between cr.m.s and temperature T

Since m and k are constant, <c2> is directly proportional to T

<c2> ∝ T

Therefore, the relation between cr.m.s and T is:

Average Kinetic Energy of a Molecule Worked Example equation 2

Step 3:            Write the equation in full

Average Kinetic Energy of a Molecule Worked Example equation 3a

where a is the constant of proportionality

Step 4:            Calculate the constant of proportionality from values given by rearranging for a:

T = 45 oC + 273.15 = 318.15 K

Average Kinetic Energy of a Molecule Worked Example equation 3

Step 5:            Calculate cr.m.s  at 80 oC by substituting the value of a and new value of T

T = 80 oC + 273.15 = 353.15 K

Average Kinetic Energy of a Molecule Worked Example equation 4

Exam Tip

Keep in mind this particular equation for kinetic energy is only for one molecule in the gas. If you want to find the kinetic energy for all the molecules, remember to multiply by N, the total number of molecules.

You can remember the equation through the rhyme ‘Average K.E is three-halves kT’.

Author: Katie

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.
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