5.3.4 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 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:

$p V = \frac{1}{3} N m \bar{c^{2}}$

Where
- p = pressure (Pa)
- V = volume (m³)
- N = number of molecules
- m = mass of one molecule of gas (kg)
- $\bar{c^{2}}$ = mean square speed of the molecules (m² s⁻²)

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

$\frac{1}{3} N m \bar{c^{2}} = N k T$

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

$\frac{1}{3} m \bar{c^{2}} = k T$

This can be rearranged to give:

$m \bar{c^{2}} = 3 k T$

The equation for kinetic energy is:

$E = \frac{1}{2} m v^{2}$

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

$2 E = m \bar{c^{2}} = 3 k T$

$E = \frac{1}{2} m \bar{c^{2}} = \frac{3}{2} k T$

Where:
- E = Kinetic energy of a molecule (J)
- m= mass of one molecule (kg)
- $\bar{c^{2}}$ = 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
- $E \propto T$

The Boltzmann constant k can be replaced with

$k = \frac{R}{N_{A}}$

Where:
- R = Molar gas constant, 8.31 (J mol⁻¹K⁻¹)
- N_{A =}Avogadro constant, 6.02 × 10²³ (mol⁻¹)

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

$E = \frac{1}{2} m \bar{c^{2}} = \frac{3}{2} k T = \frac{3 R T}{2 N_{A}}$

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.

Step 1: Write down the known quantities

- Initial c_r.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

$E = \frac{1}{2} m \bar{c^{2}} = \frac{3}{2} k T$

Step 3: State the relationship between $\bar{c^{2}}$ and T

$\bar{c^{2}} \propto T$

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

$c_{r . m . s .} \propto \sqrt{T}$

Step 5: Change this into an equation

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

$c_{r . m . s .} = a \sqrt{T}$

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

$a = \frac{c_{r . m . s .}}{\sqrt{T}}$

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

$a = \frac{730}{\sqrt{318}}$

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

$c_{r . m . s .} = a \sqrt{T} = \frac{730}{\sqrt{318}} \times \sqrt{353} = 769.12 m s^{- 1} = 770 m s^{- 1} (2 s . f .)$

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’.

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

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OCR A Level Physics

Revision Notes

5.3.4 Average Kinetic Energy of a Molecule

Average Kinetic Energy of a Molecule

Worked Example

Exam Tip

Author: Deva

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