There are a number of conditions which must be satisfied for a gas to be considered ideal

- There must be a large number of molecules and they should move in random directions with a range of different speeds.
- The spacing between molecules should be much greater than the size of the molecules.
- Molecules are assumed to interact only through collisions.
- The collisions are assumed to be elastic.

We can work out the pressure exerted by an ideal gas on it's container by starting from the change in momentum of a molecule when it strikes the container wall

The average force due to one molecule is then

$F_{molecule}=\frac{\Delta(mv)}{\Delta t}=\frac{2mv_{x}}{2l/v_{x}}=\frac{mv_{x}^{2}}{l}$

The net force on the wall will be the sum of the forces from all $N$ molecules

$F_{net}=\frac{m}{l}\Sigma_{i=1..N} v_{xi}^{2}$

$\frac{\Sigma_{i=1..N} v_{xi}^{2}}{N}=\bar{v_{x}^{2}}$ → $F_{net}=\frac{m}{l}N\bar{v_{x}^{2}}$

$v^{2}=v_{x}^{2}+v_{y}^{2}+v_{z}^{2}$ → $\bar{v^{2}}=\bar{v_{x}^{2}}+\bar{v_{y}^{2}}+\bar{v_{z}^{2}}=\bar{3v_{x}^{2}}$

$F_{net}=\frac{m}{l}N\frac{\bar{v^{2}}}{3}$

$P=\frac{F}{A}=\frac{1}{3}\frac{Nm\bar{v^{2}}}{Al}=\frac{1}{3}\frac{Nm\bar{v^{2}}}{V}$

$PV=\frac{2}{3}N(\frac{1}{2}m\bar{v^{2}})=NkT$

$\bar{KE}=\frac{1}{2}m\bar{v^{2}}=\frac{3}{2}kT$

If we look at a simulation of particles moving according to the kinetic theory we can get some idea of the distribution of the speeds of the particles.

The Maxwell-Boltzmann Distribution gives the probability $f(v)$ that a particular particle in an ideal gas has a given speed $v$.

$f(v)=4\pi (\frac{m}{2\pi k T})^{\frac{3}{2}}v^{2}e^{-\frac{1}{2}\frac{mv^{2}}{kT}}$

It was originally derived by James Clerk Maxwell based on symmetry arguments, later Ludwig Boltzmann derived it on a more general basis.

Now that we have this we can find various useful quantities, such as the average velocity

$\bar{v}=\int_{0}^{\infty}vf(v)dv=2\sqrt{\frac{2}{\pi}}\sqrt{\frac{kT}{m}}\approx 1.6 \sqrt{\frac{kT}{m}}$

or the most probable speed of a particle, from the condition $\frac{df}{dv}=0$ which gives

$v=\sqrt{\frac{2kT}{m}}\approx 1.41 \sqrt{\frac{kT}{m}}$

A third measure of velocity that we could already obtain before deriving the Maxwell Boltzmann distribution, is the root-mean-square, or rms velocity. Here we take the square root of the average of the squared velocity

$\frac{1}{2}m\bar{v^{2}}=\frac{3}{2}kT$ → $\sqrt{\bar{v^{2}}}=\sqrt{\frac{3kT}{m}}\approx 1.73 \sqrt{\frac{kT}{m}}$

The distribution of speeds is highly temperature dependent, as can be seen in simulations and is predicted by the Maxwell distribution.

The plot below is for He atoms at various temperatures.

Another parameter that describes a gas is the mean free path, the average distance a molecule moves without a collision.

To determine this we should consider the distance a molecule traveling at the mean speed of the gas moves in a time interval $\Delta t$, $\bar{v}\Delta t$ divided by the number of collisions in that time.

The number of collisions that would occur if the other molecules in the gas were stationary is $\frac{N}{V}\pi (2r)^{2}\bar{v}\Delta t$

$l_{m}=\frac{\bar{v}\Delta t}{(N/V)\pi (2r)^{2}\bar{v}\Delta t}=\frac{1}{4\pi r^{2}(N/V)}$

However, as the other molecules are **not** stationary the number of collisions will be due to the relative velocity

$N/v\pi (2r)^{2}\bar{v}\Delta t$ → $N/v\pi (2r)^{2}v_{rel}\Delta t$

It can be shown that $v_{rel}=\sqrt{2}\bar{v}$, so

$l_{m}=\frac{\bar{v}\Delta t}{(N/V)\pi \sqrt{2} (2r)^{2}\bar{v}\Delta t}=\frac{1}{4\pi \sqrt{2} r^{2}(N/V)}$

Some corrections can be made to the ideal gas law which help to make it a better representation of a real gas. These corrections were developed by Johannes Diderik Van Der Waals.

The first correction is to consider the molecules as having a finite volume, this reduces the volume in which molecules can move around without collisions. We reduce the amount of volume by a parameter b which represents the amount of unavailable volume in a mole of gas.

$PV=nRT$ → $P(V-nb)=nRT$

or $P(\frac{V}{n}-b)=RT$

The second correction is to consider attractive forces between molecules. These reduce the pressure by a factor which is proportional to the square of the density. It is proportional to the square of the density because we consider both the number of molecules subject to the force, and the number of molecules exerting the force.

$P=\frac{RT}{\frac{V}{n}-b}-a(\frac{n}{V})^{2}$

$P=\frac{RT}{\frac{V}{n}-b}-\frac{a}{(\frac{V}{n})^{2}}$

or $(P+\frac{a}{(\frac{V}{n})^{2}})(\frac{V}{n}-b)=RT$

The values of $a$ and $b$ are empirical parameters which depend on the gas and can be found here.

The Van Der Waals equation works pretty well under certain conditions. We should be careful that below the critical point it predicts oscillations that are not observed. Maxwell suggested replacing this path with a straight line so that areas below and above the line are equal, which is described here.

The phase diagram of water is fairly complicated, particularly at high pressures, but we are not going to pay attention to all the different kinds of ice that can exist!

The lines represent conditions where 2 phases co-exist, referred to as a phase boundary. Where two lines intersect we have a triple-point. A critical point is a point where a phase boundary disappears. At temperatures above the critical point the liquid and vapor phases are indistinguishable, we call this a supercritical fluid, or more simply a gas. The dotted line shows the behavior of water, which is anomalous (ie. it expands upon freezing).

A phase region on a phase diagram represents the equilibrium state of the material under the stated conditions. Materials can, however, exist in non-equilibrium states and we will try to produce one by the end of the lecture.

We know that if we add salt to ice it will melt it. Does it do this by raising or lowering the temperature?

In fact the salt lowers the melting point of the ice and thus also it's temperature (I'll explain why when we discuss entropy). We'll use this fact to try to cool some bottles of water below their freezing point.

We can see from the phase diagram of water that increased pressure also leads to a decreased melting point. Does this explain why the ice under an iceskate blade is slippery (the commonly accepted theory before 1950)?

We can do a simple calculation to estimate the pressure exerted by the blades of a skate.

If the area is small the pressure is large. In the case of the blades, the surface in contact with the ice is tiny because the width of the blade is less than 0.5 mm.

$P=\frac{F}{A}=\frac{mg}{lw}$

For an 80 Kg person, assuming a blade of 30 cm length and 0.5 mm wide, the pressure on each foot will be ~ 27 atm. Not so large, especially considering that the slope P Vs T for melting ice has such a large slope. This means that this pressure will change the melting temperature by less than 1 $^o$C!

In fact the reason that iceskates work is because there is always layer of water on the surface of ice (as long as the temperature is above -35^{0}C). More on this here.

Within a liquid there will be a distribution of velocities for the molecules and the most energetic ones can leave the liquid and join the vapor phase, which is evaporation. The reverse process can also happen where lower energy molecules in the vapor phase join the liquid through condensation. In a closed system at equilibrium these processes occur at the same rate. The amount of vapor in the air under this condition is the saturated vapor pressure, which will depend on the temperature and the overall pressure.

Boiling occurs when the saturated vapor pressure equals the external pressure. Normally we boil water by increasing the temperature…but it can also be done by decreasing the pressure. See here for the vapor pressure of water.

If we do it this way will the temperature of the water increase or decrease?

Air can contain a varying amount of water, with the total pressure of the air being due to many different kinds of molecule. The partial pressure of a particular gas is the pressure due to a particular kind of gas molecule.

We define humidity in terms of the partial pressure of H_{2}O compared to the saturated vapor pressure of H_{2}O. The saturated vapor pressure is a function of temperature. The dew point is the temperature at which, when the temperature is dropping, the partial pressure of water in the atmosphere is equal to the saturated water pressure and dew starts to form on surfaces.

$\mathrm{Humidity=\frac{\textrm{partial pressure of } \mathrm{H_{2}O}}{\textrm{saturated vapor pressure of }\mathrm{H_{2}O}}\times 100\%}$.