Take the 2-minute tour ×

I know that I can use the ideal gas law with pure gases or pure liquids. But can I also use the ideal gas law at saturated gases and saturated liquids as long as they aren't two phase substances?

share|improve this question
Combative comments deleted. –  dmckee Nov 23 '11 at 17:51
@dmckee I feel that the comments that were said by Georg were very unacceptable. I thought I could come here and ask a question about something that was giving me problems and have it answered but instead my intelligence is ridiculed and my education gets insulted multiple times. –  Greg Harrington Nov 23 '11 at 17:57
I try to avoid commenting with just "great question", but I will make an exception here. I think this is a good, insightful, question. Understanding the conditions that lead to ideal gas behavior is a great thing for any physics student. I would add that, obviously, we should formally limit this conversation to processes around the saturation point that don't cause any phase transitions. Say you have a saturated gas in a constant-pressure container and begin to heat it. How much will it behave or not behave like an ideal gas and how relevant is the fact that it's close to saturation? –  Alan Rominger Nov 23 '11 at 18:16

3 Answers 3

up vote 6 down vote accepted

dmckee gives some good qualitative considerations, but we can also develop rules for when the ideal gas law is and isn't appropriate. To start:

  • The law applies perfectly in the case of a gas when $P\rightarrow 0$.
  • The law does not apply to liquids.

Between these two states is a gray area. In that case you should look at the compressibility factor, $Z=P_\text{actual}/P_\text{ideal}$. $Z$ is a function of reduced pressure $P_r$ and reduced temperature $T_r$ (more on these later), and this correlation is given in standard charts which apply for most substances (I use one from Koretsky 2004, p. 198). If you accept errors up to 10%, you may apply the ideal gas law as long as $0.9<Z<1.1$. So:

  • The law is a good approximation when $P_r<0.1$ (even for a saturated vapor).
  • The law is a good approximation when $0.1<P_r<7$ if $T_r>1.819-\dfrac{0.3546}{P_{\!r}^{\,0.6}}\,$.
  • The law is not a good approximation when $P_r>7$, no matter the temperature.

$P_r$ is defined as $P/P_c$ and $T_r$ is defined as $T/T_c$, where $P_c$ and $T_c$ are the substance's critical properties. For pure substances, these can be looked up in tables. For mixtures of vapors and gases which don't interact strongly, calculate each by multiplying the critical property of each pure component with its volume fraction and adding them together.

For example, pure water has $P_c=217~\text{atm}$ and $T_c=647~\text{K}$. Pure water vapor at 1 atm and 373 K has $P_r=1/217=0.0046$, so the ideal gas law applies to within 10% error. Pure water vapor at 25 atm and 498 K has $P_r=0.12$ and $T_r=0.77$, and $$0.77\not>1.819-\frac{0.3546}{0.12^{0.6}}$$ Thus the ideal gas law is no longer a good approximation. But if the vapor is mixed with 80% air $(P_c=37~\text{atm},\ T_c=133~\text{K})$ and kept at the same total pressure, we get $$P_c=80\%\cdot 37+20\%\cdot 217=73\Rightarrow P_r=0.34$$ $$T_c=80\%\cdot 133+20\%\cdot 647=236\Rightarrow T_r=2.1$$ $$2.1>1.819-\frac{0.3546}{0.34^{0.6}}$$ So the ideal gas law applies again.

But these rules only apply if you accept errors up to 10%. If accuracy is important, only use the ideal gas law for $P_r<0.025$ and don't use it for saturated vapors at all. When the ideal gas law doesn't apply, correct it using the compressibility factor $(P_\text{actual}=ZP_\text{ideal})$ or use a better equation of state like Soave-Redlich-Kwong or Peng-Robinson (not van der Waals; it's bad for general use).

share|improve this answer
""The law does not apply to liquids "" This could be derived from the name "ideal gas " maybe? :=) But Greg asked about saturated liquids! Maybe that is something different? I don't know saturated liquids, but Greg mentions them in two questions meanwhile. –  Georg Nov 24 '11 at 14:08
The ideal gas law is an equation of state, and some equations of state apply for liquids as well as gases. Besides, strictly speaking, "ideal gas law" wouldn't even suggest that the law could be used for vapors – but it works well for vapors when $P_r<<1$. And a saturated liquid is just a liquid at a temperature where it can boil. There is very little volume change associated with saturated liquid versus subcooled liquid, so the ideal gas law fails equally badly in either case – even though it works better for superheated vapors than for saturated vapors. –  rdhs Nov 24 '11 at 14:20
""And a saturated liquid is just a liquid at a temperature where it can boil. "" Any textbook with this definition? I'd call that a boiling liquid. –  Georg Nov 24 '11 at 14:26
Wikipedia: Saturated fluid cites Çengel and Boles. Also, just about every steam table calculator I've run across lists boiling-point properties as saturation properties: example. –  rdhs Nov 24 '11 at 14:39
EWiki isn't a source. Too "popular". And boiling point properties are among those "saturation" properties because of the saturated steam! The word saturation has a meaning! –  Georg Nov 24 '11 at 16:40

The ideal gas law is derived from a model (the ideal gas), and like every other model it applies where it's underling assumptions are good approximations to reality.

So, important assumptions for the idea gas law:

  • Point particles In the ideal gas, the particles occupy no volume. A real gas in which the atoms of molecules occupy a vanishing fraction of the volume is a good approximation. This suggests low density. Note that most liquids decidedly do not qualify.

  • Non-interacting Philosophically the non-interaction assumption is a little bit difficult when you go to treat the ideal gas in thermodynamics and especially in statistical mechanics. I will suggest that having no internal degrees of freedom that are available for excitation at the energy of the gas and no significant interactions at ranges comparable to the average inter-particle distance qualifies. Because the accessible energy levels scales as the temperature of this gas, we should perhaps require moderate temperatures. Mono-atomic gasses will be be a good approximation to higher energies than more complex molecules (which generally have rotation and vibrational modes at lower energies than the atomic electron excitations).

  • Random motion Situations where the other conditions apply and this one fails are rare, so I'm going to skip it.

So, what happens if this assumptions are violated? Well, the Van der Waals gas for the space occupied by the molecules and a bulk attractive force between molecules. This makes it applicable to higher density materials (but still ones whose internal degrees of freedom are not excited) and causes it to exhibit the gas-to-liquid first-order phase change (which is not present in the ideal gas).

share|improve this answer
This kind of helps. But my question is can I use the ideal gas law with sauturated liquids or saturated vapors? –  Greg Harrington Nov 23 '11 at 17:35
The ideal gas does not exhibit any behaviors related to saturation and condensation, which makes it very risky to apply in any situation where those behaviors might matter. Note that just replacing it with the Van der Waals equation of state won't help unless you understand the thermodynamics of the situation. This is rather the point of my answer: you have to understand the conditions for validity of the models you're thinking of applying and you are responsible for checking that they are satisfied. –  dmckee Nov 23 '11 at 17:54
So how do I know when I should use the Ideal Gas Law? I was looking at examples in the book and there were times where I thought I could use it, and they don't end up using it. –  Greg Harrington Nov 23 '11 at 17:55
@Greg: "Is it true that I can use the ideal gas equation with pure liquids or pure gases?" - pure gases, yes; pure liquids, no. All that's needed is that the ideal gas assumptions should be "applicable" to the situation at hand. Usually, this means high temperatures (keeps the particles moving randomly) and low pressures (particles can be as far apart and noninteracting as possible). A liquid is a "gas" at low temperature and/or high pressure, so one requires a different equation of state. –  user172 Nov 23 '11 at 23:58
Liquids that exhibit a surface tension (that is, essentially all of them) fail the no inter-particle forces at mean distance test (because those inter-particle fores are responsible for the surface tension). –  dmckee Nov 24 '11 at 0:52

Aside from the classical cases where the ideal gas law applies, it also applies to describe the exact entropy of a dilute solution, even if that solution is in a dense liquid. The reason is that the entropy of a dilute solution in a dense liquid is exactly the same as the entropy of a dilute gas, the number of possible positions for the solute particles is the same as the number of possibilities for the gas.

share|improve this answer

Your Answer


