# Average Kinetic Energy of a Molecule ## Deriving Average Kinetic Energy - 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 combined with the equation for kinetic energy ### Starting with the Ideal Gas Equation - The ideal gas equation is: $$pV = NkT$$ - Where: - $p$ = pressure (Pa) - $V$ = volume (m³) - $N$ = number of molecules - $k$ = Boltzmann constant, $1.38 \times 10^{-23}$ (J K⁻¹) - $T$ = Temperature (K) ### Linking Pressure and Mean Square Speed - The equation linking pressure and mean square speed of the molecules is: $$pV = \frac{1}{3}Nm\overline{c^2}$$ - Where: - $p$ = pressure (Pa) - $V$ = volume (m³) - $N$ = number of molecules - $m$ = mass of one molecule of gas (kg) - $\overline{c^2}$ = mean square speed of the molecules (m² s⁻²) ### Combining the Equations - Since both equations are expressions for $pV$, we can equate them: $$\frac{1}{3}Nm\overline{c^2} = NkT$$ - $N$ can be removed from both sides of the equation, giving: $$\frac{1}{3}m\overline{c^2} = kT$$ - This can be rearranged to give: $$m\overline{c^2} = 3kT$$ ### Substituting Kinetic Energy - The equation for kinetic energy is: $$E = \frac{1}{2}mv^2$$ - Therefore, this can be substituted in to give an equation for average molecular kinetic energy: $$2E = m\overline{c^2} = 3kT$$

E = ½m c̅² = ³⁄₂kT

- Where: - $E$ = Kinetic energy of a molecule (J) - $m$ = mass of one molecule (kg) - $\overline{c^2}$ = mean square speed of the molecules (m² s⁻²) - $k$ = Boltzmann constant, $1.38 \times 10^{-23}$ (J K⁻¹) - $T$ = Temperature (K)

E is the average kinetic energy for only one molecule of the gas

## Key Relationship: Energy and Temperature - A key feature of this equation is that the mean kinetic energy of an ideal gas molecule is **proportional** to its thermodynamic temperature $$E \propto T$$ - This means: - Doubling the absolute temperature doubles the average kinetic energy - Halving the absolute temperature halves the average kinetic energy - At absolute zero (0 K), molecules have minimum kinetic energy - The graph below shows this linear relationship between average kinetic energy and temperature:

- The straight line through the origin confirms the direct proportionality - The gradient of this line equals $\frac{3}{2}k$ ## Alternative Form of the Equation - The Boltzmann constant $k$ can be replaced with $$k = \frac{R}{N_A}$$ - Where: - $R$ = Molar gas constant, 8.31 (J mol⁻¹ K⁻¹) - $N_A$ = Avogadro constant, $6.02 \times 10^{23}$ (mol⁻¹) - Substituting this into the average molecular kinetic energy equation means it can also be written as: $$E = \frac{1}{2}m\overline{c^2} = \frac{3}{2}kT = \frac{3RT}{2N_A}$$ ## Worked Example: Temperature and r.m.s. Speed **Question:** Nitrogen can be treated as an ideal gas. Nitrogen molecules have a root-mean-square (r.m.s.) speed of 520 m s⁻¹ at a temperature of 20 °C. Calculate the r.m.s. speed of the molecules at a temperature of 150 °C. **Answer:** **Step 1: Write down the known quantities** - Initial $c_{r.m.s.}$ = 520 m s⁻¹ - Initial temperature = 20 °C = 293 K - Final temperature = 150 °C = 423 K - Boltzmann constant, $k = 1.38 \times 10^{-23}$ J K⁻¹ **Step 2: Write down the equation for average kinetic energy** $$E = \frac{1}{2}m\overline{c^2} = \frac{3}{2}kT$$ **Step 3: State the relationship between $\overline{c^2}$ and $T$** $$\overline{c^2} \propto T$$ **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 $a$** $$a = \frac{520}{\sqrt{293}}$$ **Step 8: Substitute the expression for '$a$' and the final temperature into the equation and calculate $c_{r.m.s.}$ at a temperature of 150 °C** $$c_{r.m.s.} = a\sqrt{T} = \frac{520}{\sqrt{293}} \times \sqrt{423} = 625.1 \text{ m s}^{-1} = 625 \text{ m s}^{-1} \text{ (3 s.f.)}$$ ## Key Points to Remember

This kinetic energy equation is for one molecule – multiply by N for all molecules

- This particular equation for kinetic energy applies to **only 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 ## Summary Table | Quantity | Symbol | Unit | |----------|--------|------| | Average kinetic energy of one molecule | $E$ | J | | Mass of one molecule | $m$ | kg | | Mean square speed | $\overline{c^2}$ | m² s⁻² | | Root mean square speed | $c_{r.m.s.}$ or $\sqrt{\overline{c^2}}$ | m s⁻¹ | | Boltzmann constant | $k$ | $1.38 \times 10^{-23}$ J K⁻¹ | | Temperature | $T$ | K | | Number of molecules | $N$ | - |