27 May 2025
Real Gases, Virial Equation, and Critical Constants
Compressibility, virial corrections, Boyle temperature, and the van der Waals critical point.
An ideal gas neglects molecular size and intermolecular forces. A real gas approaches $pV=nRT$ only in the low-density limit. For one mole, define the compressibility factor
\[Z=\frac{pV_m}{RT},\]where $V_m$ is molar volume. An ideal gas has $Z=1$ at every state. Attraction commonly gives $Z<1$ at moderate density, while short-range repulsion and finite molecular size give $Z>1$ at high density.
Virial equation
At sufficiently low density, the equation of state can be expanded as
\[\boxed{Z=1+\frac{B(T)}{V_m}+\frac{C(T)}{V_m^2}+\cdots}.\]$B(T)$, $C(T)$, and higher virial coefficients describe two-body, three-body, and higher correlations. Since $Z$ is dimensionless, $B$ has units $\mathrm{m^3\,mol^{-1}}$ and $C$ has units $\mathrm{m^6\,mol^{-2}}$.
van der Waals equation
Let $b$ be the excluded molar volume. Molecular centres then move in $V_m-b$, not $V_m$. Attractions reduce the measured pressure below the kinetic pressure by $a/V_m^2$. The van der Waals equation is
\[\boxed{\left(p+\frac{a}{V_m^2}\right)(V_m-b)=RT},\]or
\[p=\frac{RT}{V_m-b}-\frac{a}{V_m^2}.\]Here $b$ has units $\mathrm{m^3\,mol^{-1}}$ and $a$ has units $\mathrm{Pa\,m^6\,mol^{-2}}$. For $V_m\gg b$,
\[\frac{1}{V_m-b}=\frac1{V_m} \left(1+\frac b{V_m}+\cdots\right).\]Multiplying the equation of state by $V_m/(RT)$ and retaining the leading correction gives
\[Z=1+\frac1{V_m}\left(b-\frac{a}{RT}\right)+\cdots.\]Thus the van der Waals second virial coefficient is
\[B(T)=b-\frac{a}{RT}.\]The Boyle temperature is the temperature at which this leading correction vanishes:
\[B(T_B)=0 \quad\Longrightarrow\quad \boxed{T_B=\frac{a}{Rb}}.\]Near $T_B$ and at low pressure, a real gas therefore obeys Boyle’s law to first order in density.
Critical constants
At the critical point, the liquid and vapour become indistinguishable. The critical isotherm has a horizontal inflection:
\[\left(\frac{\partial p}{\partial V_m}\right)_{T_c}=0, \qquad \left(\frac{\partial^2p}{\partial V_m^2}\right)_{T_c}=0.\]For the van der Waals equation,
\[\left(\frac{\partial p}{\partial V_m}\right)_T =-\frac{RT}{(V_m-b)^2}+\frac{2a}{V_m^3},\] \[\left(\frac{\partial^2p}{\partial V_m^2}\right)_T =\frac{2RT}{(V_m-b)^3}-\frac{6a}{V_m^4}.\]Solving the two critical-point conditions and substituting into the equation of state gives
\[\boxed{V_c=3b},\qquad \boxed{T_c=\frac{8a}{27Rb}},\qquad \boxed{p_c=\frac{a}{27b^2}}.\]Consequently,
\[Z_c=\frac{p_cV_c}{RT_c}=\frac38, \qquad T_B=\frac{27}{8}T_c.\]With reduced variables $p_r=p/p_c$, $v_r=V_m/V_c$, and $T_r=T/T_c$, all van der Waals gases obey the reduced equation
\[\boxed{p_r=\frac{8T_r}{3v_r-1}-\frac{3}{v_r^2}}.\]
For $T_r<1$, the analytic isotherm contains mechanically unstable portions. A real sample instead separates into liquid and vapour over the coexistence interval. At $T_r=1$, the inflection is $(v_r,p_r)=(1,1)$; for $T_r>1$, compression alone does not produce a discontinuous gas-liquid transition.
The Unit I Maxima worksheet substitutes the critical and Boyle values directly. Its displayed residuals are
\[p(V_c,T_c)-p_c=0,\qquad \left.\frac{\partial p}{\partial V_m}\right\rvert_c=0,\] \[\left.\frac{\partial^2p}{\partial V_m^2}\right\rvert_c=0, \qquad B(T_B)=0.\]
Discussion