Ideal gas law derivation

P = \frac{F}{A} = \frac{Δ p}{Δ t \cdot A} = \frac{2 m v_{x}}{Δ t \cdot A}

Given a small increment in time $Δ t$ , the particles need to be $Δ x \leq v_{x} Δ t$ close to the wall to hit it.
The probability that a small particle traveling at $v_{x}$ is within the small volume defined by the area we focus on is: $\frac{A v_{x} Δ t}{V}$
Pressure:

P = \frac{1}{2} \int \frac{2 m v_{x}}{Δ t \cdot A} \cdot \frac{A v_{x} Δ t}{V} \cdot N p (v_{x}) d v_{x} = \frac{m N}{V} \int v_{x}^{2} p (v_{x}) d v_{x}

1/2 comes from the fact that only half the particles are heading towards the wall
$p (v_{x})$ is the probability that a particle is traveling at $v_{x}$
there are $N$ particles we're observing
multiply the pressure by the probability of being in that region and the probability of having that velocity
integrate over all possible velocities
That integral gives the expectation value (definition: probability * the velocity)

\int v_{x}^{2} p (v_{x}) d v_{x} = ⟨ v_{x}^{2} ⟩

Isotropy of the gas: $$E = \frac{1}{2}m\langle v^2 \rangle = \frac{1}{2}m\langle v^2_x + v^2_y + v^2_z \rangle = \frac{3}{2}m\langle v_x^2\rangle$$

m ⟨ v_{x}^{2} ⟩ = \frac{2}{3} E

Pressure: $$P = \frac{2NE}{3V}$$

P V = \frac{2 N E}{3} = N k_{B} T

E = \frac{3}{2} k_{B} T