As we all know, the compressibility factor $Z$ of hydrogen and helium is always greater than $1$ at a constant moderate temperature. But, if we look at the virial equation of compressibility factor

$$Z = 1 + \frac{1}{V_\mathrm{m}}\left(a - \frac{b}{RT}\right) + \frac{b^2}{V_\mathrm{m}^2} + \cdots$$

At extremely low pressure $V_\mathrm{m}$ would be extremely large. So, the equation simplifies to

$$Z = 1 + \frac{1}{V_\mathrm{m}}\left(a - \frac{b}{RT}\right)$$

Now, the values of $a$ for hydrogen and helium are extremely small. So, we can take them as zero. The equation further simplifies to

$$Z = 1 - \frac{1}{V_\mathrm{m}}\left(\frac{b}{RT}\right)$$

Now, the value of $b$, $R$, $T$ and $V_\mathrm{m}$ are all positive. Though the value of $b$ for hydrogen is extremely small, still it is positive. So, shouldn't the value of $Z$ at some small pressure be less than $1$?

Any help would be appreciated.

  • $\begingroup$ your value of $b/RT$ is also very small $\approx 0.0014$ so is smaller than $a$ so you should ignore this term and not $a$. $\endgroup$
    – porphyrin
    Mar 3, 2019 at 11:20

1 Answer 1


In my textbook the opposite meaning is given to $a$ (an attractive parameter) and $b$ (a size parameter), but this may be a matter of differences in labelling, not in meaning. I will for the sake of consistency stick to your convention. For $\ce{He}$ the van der Waals parameters have the following values:

  • $b = \pu{3.4598 J mol-1 M-1}$
  • $a = \pu{0.023733 M-1}$

As you can see from the following figure at $RT$ ($\pu{298K}$), while your first approximation is quite accurate for $\ce{He}$ already at $V_\mathrm m \approx \pu{1 L/mol}$, your second approximation fails until higher $V_\mathrm m$. It is negative, but negliglibly so, and at that point is not particularly useful as an approximation as already $Z \approx 1$.

enter image description here

The approximation fails (at RT) because the condition $b/RT \gt a$ is never observed: $$\frac{b}{RT} = \pu{0.0014 M-1}$$ whereas $$a = \pu{0.0237 M-1}$$

You can solve for the temperature at which these two become equal:

$$T_b = \frac{b}{aR}$$

For $\ce{He}$, $T_b = 17.53\ \mathrm K$. Below $T_b$ you may observe $Z<1$.

Sure enough, below that $T$ your second approximation starts to hold, as shown here for $T = 10\ \mathrm K$:

enter image description here

or in terms of $P$:

enter image description here

Note: it is not clear that the virial equation and van der Waals parameters should capture the behavior of the gases over a wide range of $P$ or $V_m$ at such low $T$.

  • 1
    $\begingroup$ (1) Out of curiosity what temperature did you use? (2) for OP's assumption to hold it isn't sufficient that just $b/Rt \gt a$ but rather it must be that $b/Rt \gg a$. $\endgroup$
    – MaxW
    Feb 24, 2019 at 0:14

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