Problem — Derivation of PV = nRT

Problem

Starting from the expression for the average translational kinetic energy of a molecule $\langle E_\text{kin,transl}\rangle = \tfrac{3}{2}k_B T$ and the microscopic relation $P = \tfrac{2}{3}\,E_\text{kin}/V$, derive the ideal gas equation of state $PV = nRT$. Clearly show the step from $N$ to $n$ (number of moles).

Solution

In the microscopic relation $E_\text{kin}$ is the total translational kinetic energy of the gas: the sum of the contributions of all $N$ molecules. If each molecule has on average $\langle E_\text{kin,transl}\rangle = \tfrac{3}{2}k_B T$, then: $E_\text{kin} = N\,\langle E_\text{kin,transl}\rangle = N\cdot \tfrac{3}{2}k_B T$ Substituting into the microscopic pressure relation: $P = \frac{2}{3}\frac{E_\text{kin}}{V} = \frac{2}{3}\cdot \frac{1}{V}\cdot N\cdot \tfrac{3}{2}k_B T$ The factor $\tfrac{2}{3}\cdot\tfrac{3}{2}=1$ simplifies: $PV = N k_B T$ I now move from $N$ (number of molecules) to $n$ (number of moles) using $N = n N_A$: $PV = n N_A k_B T$ Defining the universal gas constant $R = N_A k_B$ (the product of Avogadro’s number and the Boltzmann constant): $PV = nRT$ $\ev{PV = nRT\quad \text{with}\ R = N_A k_B}$

Connections

Topics: Teoria cinetica dei gas Concepts: Legge dei gas perfetti · Energia cinetica Skills: Impostazione simbolica