# Deriving the Ideal Gas Equation

Im specifically looking at how the IGE can be derived from the kinetic theory of gasses.

Most textbooks and webpages I've viewed start off with a wall that the particle is hitting and then writes an expression for the force $$F = \frac{dP}{dt} = \frac{2mv_x}{\Delta t}$$ where $\Delta t$ is the amount of time it takes for a particle to traverse the box and back, so $$\Delta t = \frac{2L}{v_x}$$

This gives us an expression for $$P = F/A = \frac{2mv^2_x}{2LA} = \frac{mv^2_x}{V}$$

$$\implies PV = mv_x^2 = 2KE$$

Considering the adverage over a number of particles gives us

$$PV = m \bar v^2_x$$

Assuming that we are dealing with one kind of particle in the box $\bar m = m$, and the pressure can be considered to be continuous so $\bar P = P$.

This shows how $PV$ relates to energy, but how do we show that $NkT$ also relates to energy and thus to $PV$ to finish the derivation?

Please feel free to correct me on the derivation above.

Just a sketch.

The kinetic theory shows that $$U=\frac{3}{2}PV$$ Hence $$dU=\frac{3}{2}PdV+\frac{3}{2}VdP$$ By 1st law, $$\delta Q=dU+PdV=\frac{5}{2}PdV+\frac{3}{2}VdP$$

Consider a small rectangle $R$ on the PV diagram, it can be shown that

$$\oint_R \delta Q \ne 0$$

But

$$\oint_R \frac{\delta Q}{PV}=0$$

In other words, $\frac{1}{PV}$ is the integrating factor.

Therefore,

$$T \propto PV$$

First of all, the term $PV=mv_x^2$ you got is due only to one molecule. You have to sum over all molecules, $$PV=m\sum v_x^2=N\frac{\sum v_x^2}{N}=Nm\bar v_x^2.$$ Assuming space isotropy you can write $$\bar v_x^2=\bar v_y^2=\bar v_z^2=\frac 13 \bar v^2.$$ Hence $$PV=\frac N3m\bar v^2.$$ Now you have to invoke the Equipartition Theorem It basically says every quadratic term in the energy would contribute with $kT/2$ in the mean value of the energy. Therefore, $$\frac 12 m\bar v^2=\frac 12 m(\bar v_x^2+\bar v_y^2+\bar v_z^2)=\frac 32 kT,$$ and $$PV=NkT.$$ Notice that in first equation of Kyson's answer he is using the Equipartition Theorem for a motoatomic molecule. For a molecule with $q$ degrees of freedom appearing quadratically in the energy we get $$U=N\frac q2 kT=\frac q2 PV.$$

• $$PV = \frac{1}{3} m \bar v^2$$ and $$m \bar v^2 = 3kT$$ would imply that $$PV = kT$$ Where has N gone? – jamesmartini Apr 11 '16 at 13:08
• Sorry, my mistake. I ate $N$ in my first and third equations: – Diracology Apr 11 '16 at 13:15