 Van der Waals molar equation

The molar volume $v_m=V/n_{\mathrm{mol}}$ is the volume per amount of substance, not the total container volume or the volume per particle. In the molar convention of the Van der Waals equation, take constant positive parametersThenor, for any fixed amount of substance,Both terms in the first equation are pressures. The total excluded volume is $n_{\mathrm{mol}}{b}_m$: using $V-b_m$ would subtract quantities with different units. Changing the sample size at fixed molar volume and temperature changes $V$ and $n_{\mathrm{mol}}$ proportionally, while leaving pressure unchanged. This is a useful unit and extensivity check before fitting a real gas.

For the molar Van der Waals critical point, differentiate with respect to molar volume at fixed temperature:Setting both to zero gives $R{T}_c=2a_m(v_{m,c}-b_m{)}^2/v_{m,c}^3$ and $R{T}_c=3a_m(v_{m,c}-b_m{)}^3/v_{m,c}^4$. Equating them and dividing by the nonzero factors yields $2v_{m,c}=3(v_{m,c}-b_m)$, soHere $v_{m,c}$ has units ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, $T_c$ has units $\mathrm{K}$, and $p_c$ has units $\mathrm{Pa}$. This is the molar version of the particle derivation below: $v_{m,c}=N_A{v}_c$, with the same physical $T_c$ and $p_c$. If one writes $v_c=3b$, $T_c=8a/(27Rb)$ and $p_c=a/(27b^2)$ without subscripts, then all of $v_c,b,a$ in those formulas must be molar quantities.

Now define the dimensionless reduced variables explicitly as $x=v_m/v_{m,c}$, $t=T/T_c$ and $q=p/p_c$. Substituting the three critical values cancels $a_m,b_m,R$ and givesBecause $v_m/v_{m,c}=v/v_c$, these are exactly the reduced variables in the Van der Waals isotherm discussion. The Van der Waals law of corresponding states is therefore independent of whether one starts with particle or molar units. It predicts a critical ratio $p_c{v}_{m,c}/(R{T}_c)=3/8$, providing a third check on any proposed parameter fit.