My textbook says that Boyle temperature is the temperature at which a real gas shows maximum ideal gas behavior. Below the Boyle temperature, molecules come too close and intermolecular forces skew off its behavior. But what about above the Boyle temperature? Why do gases deviate from the ideal gas law above it? I mean, rise in temperature can only mean less of intermolecular attraction, right?
[Physics] Why do real gases deviate from ideal behavior above the Boyle temperature
ideal-gasinert-gasesthermodynamics
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No you cannot.
we can show that (via conservation of kinetic energy and momentum) the speed of the particle before and after collision is the same.
Only in the center of mass of two elastically colliding particles the momentum remains the same. Each pairwise collision has a different center of mass. In the laboratory frame, which is the frame one is trying to model the ideal gas, all the momentum might be taken by one of the particles, leaving the other motionless in the lab. This happens with billiard ball collisions all the time. See this analysis. So even if one made an experimental setup with all the particles of the ideal gas with the same speed, after the first scatter, speeds will change because they will not all be head on, there will be angles, and then the laboratory versus center of mass argument prevails.
The distribution functions for the ideal gas were given by Maxwell using simple and reasonable assumptions. Boltzman refined this.
It is a model, i.e. a theoretical formula, that has been validated by data over and over again.
The answer to your question is quite interesting: the ideal gas equation of state is very general; it applies whenever the particles don't interact with one another (except by very short range forces to allow collisions). The relation between momentum and energy (called the dispersion relation when you do the analysis using statistical mechanics) can be anything at all and you still get $p V = N k_B T$.
In the Boltzmann factor it is always the energy that appears, so when the relationship between momentum and energy is different (e.g. in a relativistic gas) the distribution over momentum is different. For example one gets $$ f(p) \propto p^2 e^{-E/k_B T} $$ where $p = \gamma m v$ and $E = \gamma m c^2$ and $\gamma = 1/\sqrt{1-v^2/c^2}$. This distribution is not the MB distribution but remarkably you still get $p V = N k_B T$. The easiest way to show this is using statistical mechanics via the single particle partition function.
This is an example of a macroscopic phenomenon (the way pressure in a gas relates to volume and temperature) being independent of many of the details of the micro-physics (it could be any dispersion relation, and you could treat the parts of the gas using either classical or quantum physics). This raises some very interesting points about the way micro-physics connects to macro-physics.
Best Answer
In the limit of very high temperature, all gases become ideal (assuming they don't ionise, dissociate, etc), but this regime is far above the Boyle temperature. Around the Boyle temperature the long range attractive forces are still significant and cause non-ideal behaviour. It's just that there is a sweet spot where the attractive forces are balanced by the molecular volume.
With real gases there are two effects. At short range, there is a repulsion, and at long range, there is an attraction. A common model for this is the Van der Waals equation. You're probably familiar with the ideal gas law:
$$ PV = nRT $$
The Van der Waals equation modifies this to be:
$$ \left(P + \frac{n^2a}{V^2}\right) (V - nb) = nRT $$
where $a$ and $b$ are constants. $a$ is a measure of the long range attraction and $b$ is a measure of the short range repulsion. The Boyle temperature is the temperature at which the attraction and repulsion balance out and the gas behaves approximately ideally. Below this temperature the short range repulsion dominates over the long range attraction, while above this temperature the long range attraction dominates over the short range repulsion.
The derivation of the Boyle temperature is straightforward and easily Googlable. I'll sketch it here. For an ideal gas we can rearrange the ideal gas law to get:
$$\frac{P}{RT} = \frac{n}{V} = \rho $$
where $\rho$ is the molar density $n/V$. For a non-ideal gas we can use a similar expression:
$$\frac{P}{RT} = \rho + B_2(T)\rho^2 + B_3(T)\rho^3 + ... $$
where $B_2$, $B_3$, etc are the second, third, etc virial coefficients and are functions of temperature. For most gases we expect $B_2$ to be much smaller than one, $B_3$ to be much smaller than $B_2$, and so on so they are small corrections.
Anyhow, the Boyle temperature is the temperature at which $B_2$ is zero so the expression most closely matches the ideal gas law. So we just need to write the VdW equation in this form and we can calculate $B_2$ as a function of $a$, $b$ and $T$. To do this we rerrange the VdW equation to get:
$$\begin{align} P &= \frac{nRT}{V-b} - \frac{n^2a}{V^2} \\ &= \rho RT \frac{1}{1-\rho b} - \rho^2 a \end{align}$$
Then we assume that $\rho b$ is much less than one, so we can use a binomial expansion:
$$ \frac{1}{1-\rho b} = 1 + \rho b + (\rho b)^2 + ... $$
Substitiuting this in the equation above and neglecting terms with a factor of $\rho^3$ or higher gives:
$$ \frac{P}{RT} = \rho + \rho^2 \left(b - \frac{a}{RT} \right) $$
The Boyle temperature, $T_B$, is the temperature at which the $\rho^2$ term becomes zero so:
$$ b - \frac{a}{RT_B} = 0 $$
or:
$$ T_B = \frac{a}{bR} $$