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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?

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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:

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

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:

$$ T_B = \frac{a}{bR} $$

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  • $\begingroup$ This is a very good explanation of the derivation for the Boyle temperature, but it still does not answer my question. Above the Boyle temperature, gases deviate from ideal gas behavior. Why? A higher temperature should mean less intermolecular interactions and so a higher resemblance to ideal behavior. $\endgroup$ – Gerard Feb 11 '14 at 3:25
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    $\begingroup$ @Gerard: in the limit of very high temperature all gases become ideal, 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. $\endgroup$ – John Rennie Feb 11 '14 at 6:49
  • $\begingroup$ You should add your comment into the answer, because what you just said was what I really was looking for, not a derivation of the Boyle temperature. $\endgroup$ – Gerard Feb 11 '14 at 6:52

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