In the general equation of the compressibility factor $Z$, we define $Z$ as $$Z=\frac{pV}{nRT}$$ Here, what is $p$? Is it $p_\text{real}$ or $p_\text{ideal}$? Also, what is $V$? $V_\text{real}$ or $V_\text{container}$?

  • 1
    $\begingroup$ What difference do you see $V_{real}$ versus $V_{container}$ ? $\endgroup$
    – Poutnik
    Apr 12, 2019 at 6:16

2 Answers 2


Both quantities are of the real gas.

Note that for the ideal gas behaviour, the compressibility factor $$Z=\frac{V_\text{real}}{V_\text{ideal}}=\frac{pV_\text{real}}{nRT}=1$$
for any combination of $p, V, T$.

But real gases, in contrary to the ideal gas, have nonzero volume of molecules and there are intermolecular interactions.

The volume of molecules increases $Z$, as it makes the pressure higher, because the collision frequency is higher.

The cohesive forces between molecules decreases $Z$. As molecule attraction, like hydrogen bonds or dipole interactions, decreases the effective number of molecules and therefore pressure.

This is reflected in the van der Waals equation - probably the simplest state equation of real gases, for not too high pressure: $$\begin{align} \left(p + \frac{an^2}{V^2}\right)\left(V - nb\right)&=nRT \\ \left(p + \frac{a}{V_\mathrm m^2}\right)\left(V_\mathrm m - b\right)&=RT \end{align}$$

$$\begin{align} Z &= \frac{p\cdot V_\text{real}}{nRT} \\ Z&= \frac{p\cdot V_\text{real}}{\left(p + \frac{an^2}{V_\text{real}^2}\right)(V_\text{real} - nb)} \\ Z&= \frac{1}{\left(1 + \frac{an^2}{p \cdot V_\text{real}^2}\right)\left(1 - \frac{nb}{V_\text{real}}\right)} \\ Z&= \frac{1}{\left(1 + \frac{a}{pV_{\mathrm m,\text{real}}^2}\right)\left(1 - \frac{b}{V_{\mathrm m,\text{real}}}\right)} \end{align}$$

For $|x|\ll1, 1/(1+x)=1-x$

For not too far from ideal behaviour, we can apply the above approximation.

$$\begin{align} Z&=\left(1 - \frac{an^2}{pV_{\mathrm{real}}^2}\right)\left(1 + \frac{bn}{V_{\mathrm{real}}}\right) \\ Z&=\left(1 - \frac{a}{pV_{\mathrm m,\text{real}}^2}\right)\left(1 + \frac{b}{V_{\mathrm m,\text{real}}}\right) \\ \end{align}$$

We can therefore also afford to neglect minor terms.

$$\begin{align} Z&=1 - \frac{an^2}{pV_{\mathrm{real}}^2}+ \frac{bn}{V_{\mathrm{real}}}\\ Z&=1 - \frac{a}{pV_{\mathrm m,\text{real}}^2}+ \frac{b}{V_{\mathrm m,\text{real}}}\\ \end{align}$$

To address clarified scenario, as a the same amounts of the ideal and real gas are under the same temperature and pressure:

$$V_\text{ideal gas}=\frac{nRT}{p}$$ $$V_\text{real gas}=Z \frac{nRT}{p}$$

There is need to perform iteration for probably the easiest way to get the result. With the $V_{\rm{ideal gas}}\ $ as the first approximation. The other option is to solve the root of the cubic equation what we probably do not want to.

$$\begin{align} V_\text{real}&=Z(p, V_\text{real})\frac{nRT}{p}\\ V_\text{real}&=\frac{1}{\left(1 + \frac{an^2}{p \cdot V_\text{real}^2}\right)\left(1 - \frac{nb}{V_\text{real}}\right)} \frac{nRT}{p} \\ V_\text{real}&=\left( 1 - \frac{an^2}{pV_{\rm{real}}^2}+ \frac{bn}{V_{\rm{real}}}\right) \frac{nRT}{p} \\ \end{align}$$

The van der Waals equation can also be expressed in terms of reduced properties

$$\left( P_r + \frac{3}{V_\rm{r}^2}\right) \left( V_{\rm r}-\frac{1}{3}\right) = \frac{8}{3}T_\mathrm r$$

This yields a critical compressibility factor $3/8$.

The values of pressure, temperature and volume are divided by the respective critical values of the given gas.

  • $\begingroup$ We have p, V, n, T. Now we compare 1 mol of real gas and 1 mol of perfect gas. We set p and T same for both gases, then we inspect their V. eg we set for both gases p=1atm T=298K. p for perfect gas is that p(ideal), p for real gas is p(real), both set to have same value and that 'same value' plugged into that equation. I don't see how p is real or ideal in the equation. It's just a common value that we set experimentally. So, that p is both p(ideal) and p(real), both have the same value. Correct me if I am wrong. I am still a learner. $\endgroup$
    – TheLearner
    Apr 13, 2019 at 7:32
  • 1
    $\begingroup$ This all should be in your question elaboration. Your scenario was not clear, neither what quantities are given and what are dependent. $\endgroup$
    – Poutnik
    Apr 13, 2019 at 7:48
  • $\begingroup$ @Poutnik Well, shouldn't that be obvious when we talk about compression factor, which is defined as ratio of molar volume of real gas to that of perfect gas under the same conditions of pressure and temperature? The V is definitely V(real), but I doubt the statement 'Both quantities are taken as real ones.' is correct. $\endgroup$
    – TheLearner
    Apr 13, 2019 at 8:50
  • $\begingroup$ It should be obvious from the question wording, which was rather underelaborated. :-) $\endgroup$
    – Poutnik
    Apr 13, 2019 at 9:24
  • $\begingroup$ Isn't P ideal = P real + an²/V² ? $\endgroup$
    – user226375
    Apr 16, 2019 at 17:10

Compressibility factor for a gas is defined as the ratio of the volume of real gas to the volume of ideal gas . Thus $$ Z = \frac{V_\text{real}}{V_\text{ideal}} $$ Using the ideal gas equation $$ PV_\text{ideal} = nRT $$ $$ V_\text{ideal} = \frac{nRT}{P} $$ So $$ Z = \frac{V_\text{real}}{\frac{nRT}{P} } $$ $$ Z = \frac{PV_\text{real}}{nRT} $$ Hence except $V_\text{real}$ all other quantities are ideal.

  • $\begingroup$ The answer apart from this answer says that even P is real, I don't know which to trust. can you provide a source to your answer? $\endgroup$
    – user226375
    Apr 13, 2019 at 2:31
  • $\begingroup$ P is pressure on container wall. And V is container volume. $\endgroup$
    – Jay
    Apr 24, 2021 at 14:59

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