compare pV = 31Nm<c2> with pV = NkT to deduce that the average translational kinetic energy of a molecule is 23 kT, and recall and use this expression

Kinetic Theory of Gases

Learning Objective

Compare the two forms of the ideal‑gas equation,

\$pV = \frac{1}{3}Nm\langle c^{2}\rangle\$

and

\$pV = NkT\$

to deduce that the average translational kinetic energy of a molecule is \$\displaystyle \frac{3}{2}kT\$, and recall and use this expression in problem solving.

Key Symbols

Derivation

1. Write the two expressions for \$pV\$:

   • From kinetic theory: \$pV = \dfrac{1}{3}Nm\langle c^{2}\rangle\$
   • From the ideal‑gas law: \$pV = NkT\$

2. Equate the right‑hand sides because they both equal \$pV\$:

   \$\frac{1}{3}Nm\langle c^{2}\rangle = NkT\$

3. Cancel the common factor \$N\$:

   \$\frac{1}{3}m\langle c^{2}\rangle = kT\$

4. Multiply both sides by \$\dfrac{3}{2}\$:

   \$\frac{1}{2}m\langle c^{2}\rangle = \frac{3}{2}kT\$

5. Recognise that the left‑hand side is the average translational kinetic energy of a single molecule, \$\displaystyle \langle E_{\text{trans}}\rangle\$:

   \$\boxed{\langle E_{\text{trans}}\rangle = \frac{3}{2}kT}\$

Physical Interpretation

The result shows that each molecule carries, on average, three degrees of freedom (one for each Cartesian direction) and each degree contributes \$\dfrac{1}{2}kT\$ to the kinetic energy. This is a direct statement of the equipartition theorem for a monatomic ideal gas.

Using the Expression

Typical applications in A‑Level questions include:

• Finding the root‑mean‑square speed \$c_{\text{rms}} = \sqrt{\langle c^{2}\rangle}\$ from a known temperature.
• Calculating the internal energy of \$n\$ moles: \$U = \dfrac{3}{2}nRT\$, where \$R = N_{\!A}k\$.
• Estimating the temperature change when a gas does work or absorbs heat, using \$\Delta U = \dfrac{3}{2}nR\Delta T\$.

Worked Example

Problem: A sample contains \$2.0\times10^{23}\$ molecules of helium at \$300\,\$K. Find the average translational kinetic energy per molecule and the total translational kinetic energy of the sample.

Solution:

1. Average energy per molecule:

   \$\langle E_{\text{trans}}\rangle = \frac{3}{2}kT = \frac{3}{2}(1.38\times10^{-23}\,\text{J K}^{-1})(300\,\text{K}) = 6.21\times10^{-21}\,\text{J}\$

2. Total energy:

   \$E{\text{total}} = N\langle E{\text{trans}}\rangle = (2.0\times10^{23})(6.21\times10^{-21}\,\text{J}) = 1.24\times10^{3}\,\text{J}\$

Summary

• The kinetic‑theory form \$pV = \frac{1}{3}Nm\langle c^{2}\rangle\$ links macroscopic pressure to microscopic molecular motion.
• Equating it with \$pV = NkT\$ yields the fundamental result \$\displaystyle \langle E_{\text{trans}}\rangle = \frac{3}{2}kT\$.
• This expression underpins calculations of internal energy, root‑mean‑square speed, and temperature changes for ideal gases.