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Average Kinetic Energy of a Molecule (OCR A Level Physics)

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Katie M

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Katie M

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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 and the equation for kinetic energy

  • The ideal gas equation is:

pV = NkT

  • Where 

    • p = pressure (Pa)

    • V = volume (m3)

    • N = number of molecules

    • k = Boltzmann constant, 1.38 × 10−23 (J K−1)

    • T = Temperature (K)

  • The equation linking pressure and mean square speed of the molecules is:

p V equals 1 third N m stack c squared with bar on top

  • Where 

    • p = pressure (Pa)

    • V = volume (m3)

    • N = number of molecules

    • m = mass of one molecule of gas (kg)

    • stack c squared with bar on top = mean square speed of the molecules (m2 s−2)

  • Since both equations are expressions for pV, we can equate them:

1 third N m stack c squared with bar on top equals N k T

  • N can be removed from both sides of the equation, giving:

1 third m stack c squared with bar on top equals k T

  • This can be rearranged to give:

m stack c squared with bar on top equals 3 k T

  • The equation for kinetic energy is:

E equals 1 half m v squared

  • Therefore, this can be substituted in to give an equation for average molecular kinetic energy:

2 E space equals m stack c squared with bar on top equals 3 k T

E equals 1 half m stack c squared with bar on top equals 3 over 2 k T

  • Where:

    • E = Kinetic energy of a molecule (J)

    • m= mass of one molecule (kg)

    • stack c squared with bar on top= mean square speed of the molecules (m2 s−2)

    • k = Boltzmann constant, 1.38 × 10−23 (J K−1)

    • T = Temperature (K)

  • E 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

    • E proportional to T

  • The Boltzmann constant k can be replaced with

k equals R over N subscript A

  • Where:

    • R = Molar gas constant, 8.31 (J mol−1 K−1)

    • NA = Avogadro constant, 6.02 × 1023 (mol−1)

  • Substituting this into the average molecular kinetic energy equation means it can also be written as:

E equals 1 half m stack c squared with bar on top equals 3 over 2 k T equals fraction numerator 3 R T over denominator 2 N subscript A end fraction

Worked Example

Helium can be treated as an ideal gas.

Helium molecules have a root-mean-square (r.m.s.) speed of 730 m s−1 at a temperature of 45 °C.

Calculate the r.m.s. speed of the molecules at a temperature of 80 °C.

Answer:

Step 1: Write down the known quantities

  • Initial cr.m.s.= 730 m s−1

  • Initial temperature = 45 °C = 318 K

  • Final temperature = 80 °C = 353 K

  • Boltzmann constant, k = 1.38 × 10−23 J K−1

Step 2: Write down the equation for average kinetic energy

E equals 1 half m stack c squared with bar on top equals 3 over 2 k T

Step 3: State the relationship between stack bold c to the power of bold 2 with bold bar on top and T

stack c squared with bar on top proportional to T

Step 4: Determine the relationship between cr.m.s. and T

c subscript r. m. s. end subscript proportional to square root of T

Step 5: Change this into an equation

  • Use the letter a as the constant of proportionality to avoid confusion with k, the Boltzmann constant

c subscript r. m. s. end subscript equals a square root of T

Step 6: Rearrange the equation to make 'a' the subject

a equals fraction numerator c subscript r. m. s. end subscript over denominator square root of T end fraction

Step 7: Substitute in values for initial cr.m.s. and initial T to find an expression for a

a equals fraction numerator 730 over denominator square root of 318 end fraction

Step 8: Substitute the expression for 'a' and the final temperature into the equation and calculate cr.m.s. at a temperature of 80 °C

c subscript r. m. s. end subscript equals a square root of T equals fraction numerator 730 over denominator square root of 318 end fraction cross times square root of 353 equals 769.12 space straight m space straight s to the power of negative 1 end exponent space equals space 770 space straight m space straight s to the power of negative 1 end exponent space left parenthesis 2 space straight s. straight f. right parenthesis

Examiner Tips and Tricks

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’.

Remember that temperatures must be in Kelvin, which can be obtained by adding 273 to the temperature in °C.

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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.