# Derive the kinetic gas equation [closed]

I need a proof of high school level I have confusion in taking probabilities in all directions not yet concluded this, and I need a proof using momentum i.e I need a high school level proof

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• I have a short paper I used in algebra/trig intro mechanics (i.e. exactly the level of a typical "high school" course in the USA), that covers the topic reasonably well, but "short" is a relative thing ... it's six pages and more than two thousand words. For that matter "covers" is relative as well, as I had to pull several facts out of thin air (bolstered with rather thin plausibility arguments) to get through it all that quickly. I would never characterize such a treatment as "a proof", though I do think that it is useful. – dmckee May 4 at 5:32
• OK. I got ambitious and adapted a large subset of that document for Stack Exchange. – dmckee May 4 at 5:54

This treatment adapted from one I wrote for class and may eventually compile with others in the form of a book.

## Introduction

The kinetic theory of ideal gasses'' provides an accessible connection between the macroscopic variables of a diffuse, single-component, atomic gas (pressure, volume, and temperature) and a simple model of the average behavior of the atoms that make up that gas. The ideal gas law can be recovered from the kinematics of the gas molecules with few assumption beyond Boltzmann's identification of the temperature with the average kinetic energy of the atoms.

By way of an aside, I would like to note that this is one of the most beautiful results in basic physics and almost alone among other results of similar elegance in being decidedly accessible to all comers.

Here we will see the marriage of the microscopic world---which we shall assume follows the same rules for discrete particles we state for classroom demos---with the macroscopic behavior of smooth fluids through a simple statistical assumption.

## Ideal Gas

For our purposes, an ideal gas is one made of identical, spherical atoms so small that their volume and rotational behavior can be neglected, and whose interactions are always elastic and at once negligible over short time scales and frequent enough to share the internal energy of the gas without preference between all the atoms. Physical gasses that approximate this definition are observed to obey a macroscopic relation between their pressure $$P$$, volume $$V$$, and temperature $$T$$ called the "ideal gas law" and written \begin{align} PV &= N k T \tag{1}\\ &= n R T \,, \end{align} where $$k \approx 1.381 \times 10^{-23} \,\mathrm{J/K}$$ is "Boltzmann's constant", $$N$$ is the number of atoms present, $$n = N/N_A$$ is the number of moles present, $$R = N_A k \approx 8.314\,\mathrm{J/(mol \cdot K)}$$ is the "universal gas constant", and $$N_A \equiv 6.02214 \times 10^{23}\,\mathrm{objects/mol}$$ is Avogadro's Constant.

We'll write the mass of an individual atom of the gas $$m$$, which makes the molar mass $$m N_A$$.

## Kinetic Theory

In this section we will begin by considering the interaction of a single atom in a cubical box of side $$L$$ and volume $$V = L^3$$ having mass much larger than the gas contained within it and from the wall of which the atoms bounces elastically.1 We will then treat the behavior of many such atoms to find a relationship between the pressure of the gas, the volume of the box and the average kinetic energies of the atoms.

From there we'll show that we can recover the ideal gas law by giving a definition of temperature in terms of average atomic kinetic energy.

### One Atom in a Box

In a gas we expect to find atoms moving in all possible directions with equal probability, and further to have a distribution of speeds with many going at some typical speed, some going faster or slower than that, and just a few going much slower or much faster.

For the moment we'll concern ourselves, with one particular atom with speed $$v$$. Moreover, we'll only ask about it's motion in the $$x$$-direction, so we really only care about the $$x$$-component of it's velocity $$v_x$$. The atom will typically have non-zero components of velocity in both $$y$$ and $$z$$ directions as well, but the geometry of reflection insures that while all three component change sign from time to time (when they hit the wall) they keep their magnitude.2

At one particular moment we find the atom approaching the right-hand wall, where it reflects elastically, picking up a change in momentum of \begin{align*} \Delta p_x &= J_x \\ &= (-mv_x) - (mv_x) \\ &= -2 m v_x \,. \end{align*} Of course the box feels the opposite force and opposite impulse due to Newton's Third Law. Before it touches the right hand wall again, this atom must first coast to the left-hand wall, reflect and coast back. The total distance is $$2L$$ which means this takes time \begin{align*} \Delta t &= \frac{2L}{v_x} \,. \end{align*}

Impulse is average force times average time, so we can compute the average force'' on the right-hand wall due to this one particle which is \begin{align*} \bar{F}_{OP} &= \frac{J_x}{\Delta t} \\ &= \frac{2 m v_x}{1} \frac{v_x}{2L} \\ &= \frac{m\,v_x^2}{L} \,. \end{align*} The "OP" here means "one particle", and we should keep in mind that it represents an average, because most of the time the particle is far from the wall and it only collides briefly on each back and forth trip.

If we know the average force we can compute the average pressure on the right-hand wall, which comes to \begin{align*} \bar{P}_{OP} &= \frac{\bar{F}_{OP}}{A} \\ &= \frac{\bar{F}_{OP}}{L^2} \\ &= \frac{m\, v_x^2}{L^3} \\ &= \frac{m\,v_x^2}{V} \,. \end{align*}

So far we've only worried about the pressure on one side of the box, but we can also ask about the pressure on the others. By symmetry, the left hand side will be just like the right hand side. The top and bottom will be the same as each other and similar to the left and right, but using $$v_y$$. The front and back will also be the same as each other and will use $$v_z$$.

Which leaves us with a little problem: it doesn't make sense for some sides of the box to have different pressure than the others. This problem doesn't actually bother us as long as it is a single atom, because it's be silly to treat one atoms as a fluid, but we should keep it in mind as we begin to look at a system that can be considered a fluid.

### Many Atoms in a Box

The total pressure on the walls of our box due to many atoms should just be the sum of the pressures due to a single atoms.3 But our atoms are not all going in the same direction, nor are they all going at the same speed regardless of their headings.

However, if we could find the right average speed, we should be able to get the right answer anyway. In that case we could say the total pressure on the wall was \begin{align*} P = N P_{avg} \,, \end{align*} for $$N$$ particles where $$P_{avg}$$ is the correctly chosen average4 all the one-particle pressures, and the average will take care of not only the different speeds, but also the different directions (which means different ratios of $$v_x$$ to $$v_y$$ to $$v_z$$).

By assumption all the atoms have the same mass $$m$$, and they're all in the same box so they all use the same volume $$V$$, so the average is going to concern itself only with the speed $$v_{avg}$$. Which depends on the speeds in all possible directions; but notice we've assumed all direction are equally likely, so we expect the average $$v_x$$ and the average $$v_y$$ and the average $$v_z$$ to all be the same:5 \begin{align*} v_{avg}^2 &= v_{avg,x}^2 + v_{avg,y}^2 + v_{avg,z}^2\\ &= 3 v_{avg,x}^2 \,. \end{align*} Now we proceed by writing \begin{align} P &= N P_{avg} \\ &= N \frac{m v_{avg,x}^2}{V} \\ &= N \frac{m v_{avg}^2}{3V} \\ &= N \frac{2 \, K_{avg}}{3V} \tag{2} \,. \end{align} Suddenly the pressure depends on the average kinetic energy $$K_{avg}$$ of the atoms. I'm just going to tell you (without justifying it) that we can use the arithmetic mean of the kinetic energies $$\bar{K}$$, which means that we use the root-mean-square6 the velocities if we want to express it that way.

We should admit to a hidden assumption here: that the distribution of the speeds and directions of the atoms stays the same over time. In a thin enough gas this can be true simply because the atoms don't collide with each other during the time we examine the system, but for generality we would have to prove that the averages we're taking are constant as the atoms collide with one another over and over again. This is a difficult computation that was first done by Boltzmann.

## Temperature

If we rearrange EQN~(2), we get \begin{align} PV = \frac{2}{3} N \overline{K} \,, \end{align} which is very similar to the ideal gas law (EQN~(1)).7

In fact, EQN~(2) would be exactly the ideal gas law if we had $$\frac{2}{3} \overline{K} = kT$$, and that would agree well with our hand-waving model of temperature being related to the total amount of energy being used to jiggle atoms and molecules about inside a substance.

This is also very interesting because we can recognize $$U = N \overline{K}$$ as the internal energy of our ideal gas, which is a fairly special thing because most types of matter have a very complicated internal energy and formulas for them are hard to find or even not known with any precision.

Writing that in it's conventional form we have \begin{align} \overline{K} = \frac{3}{2} k T \tag{3} \,, \end{align} as a definition of the temperature of an ideal gas in terms of the kinetic energy of the constituent atoms. Similarly \begin{align*} U = \frac{3}{2} N k T \,. \end{align*}

### Generalization

At this point our mathematical expressions have no explicit reference to the original assumption that the box was rectangular, and so we drop the assumption that the box even exists and treat the results here as simply describing gasses that meet the assumptions we made for an ideal gas.

## Energy and the Speed of Air

What is the typical kinetic energy and speed of an air molecule? Air, being mostly nitrogen (and most of the rest oxygen) we'll take the molar mass to by $$28\,\mathrm{g}$$ which makes the molecular mass $$m_{\mathrm{N}_2} = 4.6 \times 10^{-26}\,\mathrm{kg}$$. Using EQN~(3) and taking $$T = 290\,\mathrm{K}$$ we find \begin{align*} \overline{K}_{\mathrm{N}_2} = 6.0 \times 10^{-21}\,\mathrm{J}\;, \end{align*} and \begin{align*} \overline{v}_{\text{RMS},\mathrm{N}_2} = 510\,\mathrm{m/s} \;. \end{align*} This is slightly more than the speed of sound,8 and gives us confidence that neglecting gravitational interaction on the scale of humans is a safe move.

In addition to asking "Is it OK to neglect gravity in the lab?" we could ask "At what height is it clear that we definitely need to be taking gravity into account?" We could take that to be the height over which the average kinetic energy drops by 10% (or equivalently when the average speed drops by 5%).

Conserving energy between a molecule moving upward from the floor with a speed of $$v_i = 510\,\mathrm{m/s}$$ and seeking the height at which it has speed $$v_f = 485\,\mathrm{m/s}$$ we get \begin{align*} E_f &= E_i \\ \frac{1}{2}m v_f^2 + mgh &= \frac{1}{2} m v_i^2 \\ v_f^2 + 2gh &= v_i^2 \\ h &= \frac{v_i^2 - v_f^2}{2g} \\ &\approx 1270\,\mathrm{m} \;. \end{align*} For the sake of being conservative we might say that $$1000\,\mathrm{m}$$ elevation changed demand that we worry about gravity. For higher precision work we might want to insure that velocity changes don't exceed 1%, and begin to worry around $$200\,\mathrm{m}$$.

## Conclusion

A model that treats atoms almost as if they were tiny billiard balls (using the same rules we use in class for macroscopic objects) was employed to arrive at the ideal gas law for sufficiently thin gasses. Along the way we relied on a few additional assumptions: a choice of an averaging procedure (take the arithmetic mean of kinetic energies), an assertion that the situation described represents and equilibrium, and an identification of that mean kinetic energy of the constituent atoms with the temperature of the gas.

This is one of several results from around the turn of the twentieth century that linked atomic theory to experimentally confirmed macroscopic behaviors,9 leading to the acceptance of atoms as real physical objects in the first couple of decades of the twentieth century.

1 We do have one limitation on the value of $$L$$: it must be small enough that we can ignore the static pressure of the gas in the box so that we can treat the top and bottom of the box as the same. We'll show later that the kinetic energy of the atoms must be much larger than the potential energy change between the top and bottom $$v^2 \gg Lg$$.

2 We chose the coordinate system so that the sides of the box are aligned with the coordinate axes for simplicity.

3 Recall that we assumed the atoms rarely interact so they mostly fly back and forth unperturbed. This is not actually a great assumption for air at STP, and ought to be replaced by an argument about the statistical distribution of directions and speeds remaining the same over time. Proving this is quite a difficult task and well beyond the scope of this document.

4 There are many ways to take averages. The one you are used to ("add up all the numbers then divide by how many there are"; written $$\bar{x} = \frac{1}{N} \left( \sum_{i=1}^N x_i \right)$$ in mathematics) is called the arithmetic mean, but sometimes different averages are useful. For the moment we'll leave the question of which average to use undecided.

5 This equality also forces the pressure to be the same on all sides of the box, putting to rest a concern that cropped up earlier.

6 "Root-mean-square" is a way of averaging. You square all the numbers, add up the squares, divide by how many there are and then take the square root. $$v_\text{rms} = \sqrt{\frac{1}{N}\left( \sum_{i=1}^N v_i^2\right) }\,.$$

7 Here is where we see the importance of the condition $$v^2 \gg Lg$$: in order for the pressure to uniform the gas molecules must have the same average kinetic energy at the bottom of the box as at the top (to the level of approximation that we require).

8 The speed of individual atoms should be faster than the speed of sound because sound waves are propagated by the motion of individual molecules and can't go faster than their carriers.

9 Though it seems strange from the modern perspective, the assorted Laws of Proportions from Chemistry were not historically taken as sufficient to prove the existence of atoms. Even the theory presented here left some prominent critics unconvinced. Albert Einstein's quantitative theory of Brownian motion published in 1905 seems to have sealed the deal.