Molar Van der Waals worked example

Choose a hypothetical fluid with molar constants $a_m=0.360\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$ and $b_m=4.00\times 10^{-5}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, using $R=8.314462618\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. These stipulated constants are not a fit to carbon dioxide. The molar critical formulas giveAt $T=360\mathrm{K}$ and $v_m=240\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, first convert $v_m$ to $2.40\times 10^{-4}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. The repulsive-volume term and attractive correction areso $p=8.716033\mathrm{MPa}$. 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.

For $n_{\mathrm{mol}}=0.500\mathrm{mol}$, the same state occupies $V=n_{\mathrm{mol}}{v}_m=120\mathrm{c}{\mathrm{m}}^3$. The total-volume formula gives $n_{\mathrm{mol}}{b}_m=20\mathrm{c}{\mathrm{m}}^3$ and the same $8.716033\mathrm{MPa}$. Halving the sample has halved both its physical volume and its excluded volume, while retaining its pressure. Conversely, if the two critical measurements $T_c,p_c$ are used to determine model constants, inversion givesAn independent critical molar volume then tests the prediction $v_{m,c}=3R{T}_c/(8p_c)$; 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 $T\simeq 288.6537\mathrm{K}$, $p_{\ast }\simeq 5.391653\mathrm{MPa}$, $v_{m,l}\simeq 72.40823\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$ and $v_{m,g}\simeq 281.8611\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. For a mean molar volume of $120\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, the lever rule gives gas mole fraction $f_g\simeq 0.227219$. Its volume fraction is instead $f_g{v}_{m,g}/v_m\simeq 0.533703$: less than one quarter of the molecules occupy more than half the volume. The flat equilibrium pressure corresponds to changing these proportions. It does not describe a homogeneous sample sliding along the unstable loop.