| Equation | Meaning |
|---|---|
| \$pV = NkT\$ | Ideal gas law in terms of number of molecules. |
| \$pV = 31\,\text{N m}^2\$ | A specific numerical value for a sample gas. |
Start from the ideal gas law: \$pV = NkT\$.
For a monatomic ideal gas, the kinetic theory gives the pressure as
\$p = \frac{2}{3}\frac{N\,\langle E_{\text{kin}}\rangle}{V}\$
where \$\langle E_{\text{kin}}\rangle\$ is the average translational kinetic energy per molecule.
Equate the two expressions for \$p\$:
\$\frac{2}{3}\frac{N\,\langle E_{\text{kin}}\rangle}{V} = \frac{NkT}{V}\$
Cancel \$N\$ and \$V\$ and solve for \$\langle E_{\text{kin}}\rangle\$:
\$\langle E_{\text{kin}}\rangle = \frac{3}{2}kT\$
Result: The average translational kinetic energy of a molecule is \$\boxed{\frac{3}{2}kT}\$.
When you see \$pV = NkT\$ in a question, remember you can always rewrite it as \$p = \dfrac{NkT}{V}\$ and compare it with the kinetic theory form \$p = \dfrac{2}{3}\dfrac{N\langle E{\text{kin}}\rangle}{V}\$. This gives a quick route to \$\langle E{\text{kin}}\rangle = \dfrac{3}{2}kT\$.
Imagine a pool table where each ball represents a gas molecule. The table’s floor is the container walls. When the balls (molecules) collide with the walls, they push back – that’s the pressure we measure. The faster the balls move, the more force they exert on the walls. In the same way, the average kinetic energy of the molecules (\$\frac{3}{2}kT\$) tells us how “energetic” the gas is at a given temperature.