= Van der Waals molar equation
{c}
{title2=$p=RT/(v_m-b_m)-a_m/v_m^2$}
The <molar volume> $v_m=V/n_{\rm mol}$ is the <volume (thermodynamics)> per <amount of substance>, not the total container <volume (thermodynamics)> or the <volume per particle>. In the molar convention of the <Van der Waals equation>, take constant positive parameters
$$
[a_m]=\mathrm{Pa\,m^6\,mol^{-2}},\qquad
[b_m]=\mathrm{m^3\,mol^{-1}},\qquad
[R]=\mathrm{Pa\,m^3\,mol^{-1}\,K^{-1}}.
$$
Then
$$
p=\frac{RT}{v_m-b_m}-\frac{a_m}{v_m^2},\qquad v_m>b_m,
$$
or, for any fixed <amount of substance>,
$$
\left(p+\frac{a_m n_{\rm mol}^2}{V^2}\right)
(V-n_{\rm mol}b_m)=n_{\rm mol}RT.
$$
Both terms in the first equation are <pressures>. The total excluded <volume (thermodynamics)> is $n_{\rm mol}b_m$: using $V-b_m$ would subtract quantities with different units. Changing the sample size at fixed <molar volume> and <temperature> changes $V$ and $n_{\rm mol}$ 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>:
$$
\left(\frac{\partial p}{\partial v_m}\right)_T
=-\frac{RT}{(v_m-b_m)^2}+\frac{2a_m}{v_m^3},\qquad
\left(\frac{\partial^2 p}{\partial v_m^2}\right)_T
=\frac{2RT}{(v_m-b_m)^3}-\frac{6a_m}{v_m^4}.
$$
Setting both to zero gives $RT_c=2a_m(v_{m,c}-b_m)^2/v_{m,c}^3$ and $RT_c=3a_m(v_{m,c}-b_m)^3/v_{m,c}^4$. Equating them and dividing by the nonzero factors yields $2v_{m,c}=3(v_{m,c}-b_m)$, so
$$
v_{m,c}=3b_m,\qquad
T_c=\frac{8a_m}{27Rb_m},\qquad
p_c=\frac{a_m}{27b_m^2}.
$$
Here $v_{m,c}$ has units $\mathrm{m^3\,mol^{-1}}$, $T_c$ has units $\mathrm K$, and $p_c$ has units $\mathrm{Pa}$. This is the molar version of the particle derivation below: $v_{m,c}=N_Av_c$, 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 gives
$$
q=\frac{8t}{3x-1}-\frac{3}{x^2}.
$$
Because $v_m/v_{m,c}=v/v_c$, 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 $p_cv_{m,c}/(RT_c)=3/8$, providing a third check on any proposed parameter fit.
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