= Molar Van der Waals worked example
{c}
Choose a hypothetical <fluid> with molar constants $a_m=0.360\,\mathrm{Pa\,m^6\,mol^{-2}}$ and $b_m=4.00\times10^{-5}\,\mathrm{m^3\,mol^{-1}}$, using $R=8.314462618\,\mathrm{J\,mol^{-1}\,K^{-1}}$. These stipulated constants are not a fit to <carbon dioxide>. The molar critical formulas give
$$
v_{m,c}=120.0\,\mathrm{cm^3\,mol^{-1}},\qquad
T_c\simeq320.7263\,\mathrm K,\qquad
p_c=8.333333\,\mathrm{MPa}.
$$
At $T=360\,\mathrm K$ and $v_m=240\,\mathrm{cm^3\,mol^{-1}}$, first convert $v_m$ to $2.40\times10^{-4}\,\mathrm{m^3\,mol^{-1}}$. The repulsive-volume term and attractive correction are
$$
\frac{RT}{v_m-b_m}=14.966033\,\mathrm{MPa},\qquad
\frac{a_m}{v_m^2}=6.250000\,\mathrm{MPa},
$$
so $p=8.716033\,\mathrm{MPa}$. The <ideal gas law> would instead give $RT/v_m=12.471694\,\mathrm{MPa}$. The <excluded volume> raises <pressure> relative to that ideal value, but attraction more than offsets the increase here. Neither correction alone describes this example.
\Image[/van-der-waals-molar-pressure-model.png]
{title=Molar <Van der Waals equation>: competing corrections and a calculated critical point}
{description=Original calculations for the stipulated hypothetical <fluid>. Left: the <ideal gas law>, the excluded-volume term and their attraction-corrected result at $360\,\mathrm K$. Right: molar <Van der Waals isotherms>; the dotted subcritical curve is a homogeneous continuation, while the green <Maxwell construction> plateau is equilibrium <phase coexistence>. None of the curves are measured <carbon dioxide> data.}
For $n_{\rm mol}=0.500\,\mathrm{mol}$, the same state occupies $V=n_{\rm mol}v_m=120\,\mathrm{cm^3}$. The total-volume formula gives $n_{\rm mol}b_m=20\,\mathrm{cm^3}$ and the same $8.716033\,\mathrm{MPa}$. Halving the sample has halved both its physical <volume (thermodynamics)> and its excluded <volume (thermodynamics)>, while retaining its <pressure>. Conversely, if the two critical measurements $T_c,p_c$ are used to determine model constants, inversion gives
$$
b_m=\frac{RT_c}{8p_c},\qquad
a_m=\frac{27R^2T_c^2}{64p_c}.
$$
An independent critical <molar volume> then tests the prediction $v_{m,c}=3RT_c/(8p_c)$; agreement with two fitted numbers does not ensure agreement with the third.
At $t=0.9$, the <Van der Waals liquid-gas coexistence> example below becomes $T\simeq288.6537\,\mathrm K$, $p_*\simeq5.391653\,\mathrm{MPa}$, $v_{m,l}\simeq72.40823\,\mathrm{cm^3\,mol^{-1}}$ and $v_{m,g}\simeq281.8611\,\mathrm{cm^3\,mol^{-1}}$. For a mean <molar volume> of $120\,\mathrm{cm^3\,mol^{-1}}$, the <lever rule> gives gas <mole fraction> $f_g\simeq0.227219$. Its <volume fraction> is instead $f_gv_{m,g}/v_m\simeq0.533703$: less than one quarter of the molecules occupy more than half the <volume (thermodynamics)>. The flat equilibrium <pressure> corresponds to changing these proportions. It does not describe a homogeneous sample sliding along the unstable loop.
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