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 at 11:20

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 at 0:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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