For a fixed amount of a real gas when a graph of $Z$ vs $P$ was plotted, then at a very high pressure slope was observed to be $\pu{0.01 atm^-1}.$ At the same temperature and pressure if a graph is plotted between $PV$ vs $P,$ then for $\pu{2 mol}$ of the gas the $y$ intercept is found to be $\pu{40 atm L}.$ Calculate excluded volume in litres for $\pu{20 mol}$ of the real gas.

Kindly note that Z is the compressibility factor of the gas.

I am not sure whether or not this problem involves differentiation in order to find the slope. Maybe the excluded volume is actually the product of the van der Waals constant $b$ times the amount of substance (in this case, $\pu{20 mol}).$

So, how can the value of $b$ be determined? Also, why is the slope between $PV$ and $P$ has its units in terms of atm-litre? Shouldn't it be just in terms of litres?

The answer was given to be 4. But I wish to know how it was calculated.

  • $\begingroup$ Can you define $Z$? It's not obvious from the statement of the problem what that is. $\endgroup$ – Zhe Oct 19 '19 at 16:54
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    $\begingroup$ Z = PV/nRT, is the compressibility factor of the gas under the given conditions. $\endgroup$ – user84951 Oct 19 '19 at 17:07

The van der Waals equation is $$\left(p+\frac{a}{V_m^2}\right)\left(V_m-b\right)=RT$$ which for $p \gg a/V_m^2$ $^\ast$ can be rewritten as $$pV_m=RT+bp$$ or $$pV=nRT+nbp \tag{1}$$ The problem says that the tangent of this curve evaluated at the same $T$ has an intercept (the value of $pV$ when $p$ goes to zero) of $\pu{40 Latm}$ when $n=2$, which means that $RT=\pu{20 Latm/mol}$. But equation (1) can be written as $$\frac{pV}{nRT}=\frac{pV_m}{RT}=Z=1+\frac{bp}{RT}$$ which means that $$\left(\frac{\partial Z}{\partial p}\right)_T=\frac{b}{RT}$$ The problem states that this slope is $\pu{0.01atm^{-1}}$ so that $b=\pu{0.2 L/mol}$ and so that when $n=20$, the excluded volume is $nb=\pu{4 L}$.

$^\ast$ More precisely, assuming ideal gas behavior, the inequality is satisfied when $$\frac{(RT)^2}{a}\gg p$$ Looking up typical values of a (typically $\pu{1-20 atmL^2/mol^2}$) means that for the given problem this equality holds if $p\ll\pu{20 atm}$.

  • $\begingroup$ Hmm, But for the pressure term of the van Der Waals equation, for small and medium pressures, a/Vm^2 grows faster then pressure. The simpification is applicable, when gas is not well compressible anymore and the equation is not well applicable. $\endgroup$ – Poutnik Oct 20 '19 at 8:27
  • $\begingroup$ @Poutnik good point, I will add a comment to my answer. $\endgroup$ – Buck Thorn Oct 20 '19 at 8:46

Just complementing the first answer of @Buck Thorn :

Full rewriting of the van Der Waals equation in terms of $Z=f(p,V_\mathrm{m})$:


$$pV_\mathrm{m} - pb + a/V_\mathrm{m} - ab/{V_\mathrm{m}}^2=RT$$

$$pV_\mathrm{m} \left( 1 - \frac{b}{V_\mathrm{m}} + \frac{ a}{ p{V_\mathrm{m}}^2} - \frac{ab}{p{V_\mathrm{m}}^3}\right)=RT$$

$$Z=\frac{pV_\mathrm{m}}{RT}\\ =\frac{1}{1 - \frac{b}{V_\mathrm{m}} + \frac{ a}{ p{V_\mathrm{m}}^2} - \frac{ab}{p{V_\mathrm{m}}^3}}\\=\frac{p{V_\mathrm{m}}^3}{p{V_\mathrm{m}}^3 - pb{V_\mathrm{m}}^2 + aV_\mathrm{m} - ab}\\ =\frac{p{V_\mathrm{m}}^3}{(pV_\mathrm{m}^2 + a)( V_\mathrm{m} - b)}\\ =\frac{1}{(1 + \frac{a}{pV_\mathrm{m}^2})( 1 - \frac{b}{V_\mathrm{m}})}\\ \overset{p\gg \frac{a}{ {V_\mathrm{m}}^2}} =\frac{V_\mathrm{m}}{V_\mathrm{m}-b}$$

$$\overset{b \ll V_\mathrm{m}}=1 + \frac{b}{ V_\mathrm{m}}$$

  • $\begingroup$ There is an error in equation 4, if I understand correctly what you are doing. When you correct that you get $Z=1+b/V_m$. Also, I am not sure the series in Z is useful. $\endgroup$ – Buck Thorn Oct 20 '19 at 9:41
  • $\begingroup$ @Buck Thorn Well,I guess you are right, it should be reciprocal. More eyes see more. :-) $\endgroup$ – Poutnik Oct 20 '19 at 9:47
  • $\begingroup$ I think you'd agree that in general $b<V_m$ and at lower pressures, $b \ll V_m$, meaning that $(1-b/V_m)^{-1} \approx 1+b/V_m$ $\endgroup$ – Buck Thorn Oct 20 '19 at 10:05
  • $\begingroup$ Yes, sure. :-) But the point is, for low pressures is Z often < 1, as the cohesive pressure often overtakes non ideal compressibility. $\endgroup$ – Poutnik Oct 20 '19 at 10:08
  • $\begingroup$ But your inequality is OK. $\endgroup$ – Poutnik Oct 20 '19 at 10:40

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