In my chemistry textbook, there is a table of the molar volumes for different gases. Most of them are below 22.42 l/mol, the ideal gas molar volume, but I noticed that hydrogen and helium do not. Hydrogen is 22.433 l/mol and helium is 22.434 l/mol. Why is this? What makes these particles, with mass, have higher molar volumes than the basically mass-less particles of an ideal gas? And why does the molar volume increase when going from Hydrogen to Helium?

  • 2
    $\begingroup$ Mass is hardly relevant here. The deviation from ideal molar volume is an interplay of two factors, both of which are absent in ideal and present in real gases: (1) volume of molecules themselves and (2) attractive interactions between molecules. $\endgroup$ Aug 16, 2016 at 20:24
  • 5
    $\begingroup$ And of those two factors, the nonzero volume may outweigh the attractive force in the case of hydrogen and (to a greater extent) helium, giving a positive deviation from the ideal gas volume--whereas attractive forces dominate in the other gases, causing a negative deviation from ideal volume. $\endgroup$
    – iad22agp
    Aug 16, 2016 at 21:06
  • 1
    $\begingroup$ Are these tiny differences between hydrogen and helium really significant? $\endgroup$ Aug 17, 2016 at 13:20
  • 1
    $\begingroup$ @ChesterMiller The major differences that come about from the tiny differences in IMF's are in the boiling points (H2: -252.9 degC, He: -268.9) [values from wikipedia] $\endgroup$
    – Dan Burden
    Aug 18, 2016 at 20:17
  • 1
    $\begingroup$ The ideal gas approximation has nothing to do with mass. Ideal gases just have neither molecular volume nor intermolecular interactions! $\endgroup$ Sep 19, 2021 at 15:53

2 Answers 2


The ideal gas law gives a very good approximation at STP, so at this point the question is mostly trifling over decimal places. For a nonideal gas, increasing the pressure and decreasing the temperature will tend to increase negative deviations from ideality: the molecules are interacting and want to stick together. Increasing the temperature while increasing the pressure, however, can introduce positive deviations from ideality, where the molecules repel each other on average, compared to a non-interacting ideal gas. If you give them enough energy, the molecules slam into each other so hard that they never even feel the underlying attractive potential.

A very useful principle here is that of corresponding states. If we take the temperature and pressure and divide them by the critical temperature and pressure of the gas we're using, we get the "reduced" temperature and pressure. The principle of corresponding states says that all gases should behave the same when they're at the same reduced temperature and pressure. If we do this for hydrogen, helium, and nitrogen at STP, we get:

$$\begin{array}{cr} \text{Gas} & T_\mathrm{crit}/\pu{K} & P_\mathrm{crit}/\pu{bar} & T_\mathrm{r} & P_\mathrm{r}\\ \hline \text{H}_{\text{2}} & 32.9 & 12.9 & 8.3 & 0.08 \\ \text{He} & 5.2 & 2.3 & 52.5 & 0.44 \\ \text{N}_\text{2} & 126 & 33.9 & 2.2 & 0.03 \end{array}$$

From this you can see that from hydrogen and helium's perspective STP conditions are quite hot indeed! Since we've introduced corresponding states, we can look at a chart of the "generalized compressibility factor" and estimate the deviation from ideality. This link gives such a chart and from it we can see that at such high reduced temperatures, we expect a positive deviation from ideality (Z>1) and thus a larger molar volume than the ideal gas.


As hydrogen and helium's molecular weight are too less, the intermolecular attractions are also too less. So the a/V^2 is negligible. So the v.d.o equation becomes P (V - nb) =RT . By which, it can be said as the pressure of these elements are too less and from Boyle's law , the volume is higher.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.