Van der Waals equation

If a mixture of ${\displaystyle n}$ gases is being considered, and each gas has its own ${\displaystyle a}$ (attraction between molecules) and ${\displaystyle b}$ (volume occupied by molecules) values, then ${\displaystyle a}$ and ${\displaystyle b}$ for the mixture can be calculated as

${\displaystyle m}$ = total number of moles of gas present,

for each ${\displaystyle i}$, ${\displaystyle m_{i}}$ = number of moles of gas ${\displaystyle i}$ present, and ${\displaystyle x_{i}={\frac {m_{i}}{m}}}$

${\displaystyle a=\sum _{i=1}^{i=n}\sum _{j=1}^{i=n}(x_{i}x_{j}{\sqrt {a_{i}a_{j}}})}$[10][11]

${\displaystyle b=\sum _{i=1}^{i=n}\sum _{j=1}^{i=n}(x_{i}x_{j}{\sqrt {b_{i}b_{j}}})}$[10][11]

and the rule of adding partial pressures becomes invalid if the numerical result of the equation ${\displaystyle \left(p+({n^{2}a}/{V^{2}})\right)\left(V-nb\right)=nRT}$ is significantly different from the ideal gas equation ${\displaystyle pV=nRT}$ .

The Van der Waals equation can also be expressed in terms of reduced properties:

${\displaystyle \left(P_{r}+{\frac {3}{V_{r}^{2}}}\right)\left(V_{r}-{\frac {1}{3}}\right)={\frac {8}{3}}T_{r}}$

The equation in reduced form is exactly the same for every gas, this is consistent with the Theorem of corresponding states.

This yields a critical compressibility factor of 3/8.

The compressibility factor for the Van der Waals equation is:

${\displaystyle Z={\frac {PV_{m}}{RT}}={\frac {V_{m}}{V_{m}-b}}-{\frac {a}{V_{m}RT}}}$

Or in reduced form by substitution of ${\displaystyle P_{r}=P/P_{c},T_{r}=T/T_{c},V_{r}=V_{m}/V_{c}}$:

${\displaystyle Z={\frac {V_{r}}{V_{r}-{\frac {1}{3}}}}-{\frac {9}{8V_{r}T_{r}}}}$

At the critical point:

${\displaystyle Z_{c}={\frac {P_{c}V_{c}}{RT_{c}}}={\frac {3}{8}}=0.375}$

Consider one mole of gas composed of non-interacting point particles that satisfy the ideal gas law:(see any standard Physical Chemistry text, op. cit.)

${\displaystyle p={\frac {RT}{V_{\mathrm {m} }}}.}$

Next, assume that all particles are hard spheres of the same finite radius r (the Van der Waals radius). The effect of the finite volume of the particles is to decrease the available void space in which the particles are free to move. We must replace V by V − b, where b is called the excluded volume (per mole) or "co-volume". The corrected equation becomes

${\displaystyle p={\frac {RT}{V_{\mathrm {m} }-b}}.}$

The excluded volume ${\displaystyle b}$ is not just equal to the volume occupied by the solid, finite-sized particles, but actually four times the total molecular volume for one mole of a Van der waals' gas. To see this, we must realize that a particle is surrounded by a sphere of radius 2r (two times the original radius) that is forbidden for the centers of the other particles. If the distance between two particle centers were to be smaller than 2r, it would mean that the two particles penetrate each other, which, by definition, hard spheres are unable to do.

The excluded volume for the two particles (of average diameter d or radius r) is

${\displaystyle b'_{2}=4\pi d^{3}/3=8\times (4\pi r^{3}/3)}$,

which, divided by two (the number of colliding particles), gives the excluded volume per particle:

${\displaystyle b'=b'_{2}/2\quad \rightarrow \quad b'=4\times (4\pi r^{3}/3)}$,

So b′ is four times the proper volume of the particle. It was a point of concern to Van der Waals that the factor four yields an upper bound; empirical values for b′ are usually lower. Of course, molecules are not infinitely hard, as Van der Waals thought, and are often fairly soft. To obtain the excluded volume per mole we just need to multiply by the number of molecules in a mole, i.e. by the avogadro number:

${\displaystyle b=N_{A}b'}$.

Next, we introduce a (not necessarily pairwise) attractive force between the particles. Van der Waals assumed that, notwithstanding the existence of this force, the density of the fluid is homogeneous; furthermore, he assumed that the range of the attractive force is so small that the great majority of the particles do not feel that the container is of finite size. Given the homogeneity of the fluid, the bulk of the particles do not experience a net force pulling them to the right or to the left. This is different for the particles in surface layers directly adjacent to the walls. They feel a net force from the bulk particles pulling them into the container, because this force is not compensated by particles on the side where the wall is (another assumption here is that there is no interaction between walls and particles, which is not true, as can be seen from the phenomenon of droplet formation; most types of liquid show adhesion). This net force decreases the force exerted onto the wall by the particles in the surface layer. The net force on a surface particle, pulling it into the container, is proportional to the number density. On considering one mole of gas, the number of particles will be N_A

${\displaystyle C=N_{\mathrm {A} }/V_{\mathrm {m} }}$.

The number of particles in the surface layers is, again by assuming homogeneity, also proportional to the density. In total, the force on the walls is decreased by a factor proportional to the square of the density, and the pressure (force per unit surface) is decreased by

${\displaystyle a'C^{2}=a'\left({\frac {N_{\mathrm {A} }}{V_{\mathrm {m} }}}\right)^{2}={\frac {a}{V_{\mathrm {m} }^{2}}}}$,

so that

${\displaystyle p={\frac {RT}{V_{\mathrm {m} }-b}}-{\frac {a}{V_{\mathrm {m} }^{2}}}\Rightarrow \left(p+{\frac {a}{V_{\mathrm {m} }^{2}}}\right)(V_{\mathrm {m} }-b)=RT.}$

Upon writing n for the number of moles and nV_m = V, the equation obtains the second form given above,

${\displaystyle \left(p+{\frac {n^{2}a}{V^{2}}}\right)(V-nb)=nRT.}$

It is of some historical interest to point out that Van der Waals, in his Nobel prize lecture, gave credit to Laplace for the argument that pressure is reduced proportional to the square of the density.

The canonical partition function Z of an ideal gas consisting of N = nN_A identical (non-interacting) particles, is:[17][18]

${\displaystyle Z={\frac {z^{N}}{N!}}\quad {\hbox{with}}\quad z={\frac {V}{\Lambda ^{3}}}}$

where ${\displaystyle \Lambda }$ is the thermal de Broglie wavelength,

${\displaystyle \Lambda ={\sqrt {\frac {h^{2}}{2\pi mkT}}}}$

with the usual definitions: h is the Planck constant, m the mass of a particle, k the Boltzmann constant and T the absolute temperature. In an ideal gas z is the partition function of a single particle in a container of volume V. In order to derive the Van der Waals equation we assume now that each particle moves independently in an average potential field offered by the other particles. The averaging over the particles is easy because we will assume that the particle density of the Van der Waals fluid is homogeneous. The interaction between a pair of particles, which are hard spheres, is taken to be

${\displaystyle u(r)={\begin{cases}\infty &{\hbox{when}}\quad r<d,\\-\epsilon \left({\frac {d}{r}}\right)^{6}&{\hbox{when}}\quad r\geq d,\end{cases}}}$

r is the distance between the centers of the spheres and d is the distance where the hard spheres touch each other (twice the Van der Waals radius). The depth of the Van der Waals well is ${\displaystyle \epsilon }$.

Because the particles are not coupled under the mean field Hamiltonian, the mean field approximation of the total partition function still factorizes,

${\displaystyle Z=z^{N}/N!}$,

but the intermolecular potential necessitates two modifications to z. First, because of the finite size of the particles, not all of V is available, but only V − Nb', where (just as in the conventional derivation above)

${\displaystyle b'=2\pi d^{3}/3}$.

Second, we insert a Boltzmann factor exp[ - ϕ/2kT] to take care of the average intermolecular potential. We divide here the potential by two because this interaction energy is shared between two particles. Thus

${\displaystyle z={\frac {(V-Nb')\,e^{-\phi /(2kT)}}{\Lambda ^{3}}}.}$

The total attraction felt by a single particle is

${\displaystyle \phi =\int _{d}^{\infty }u(r){\frac {N}{V}}4\pi r^{2}dr,}$

where we assumed that in a shell of thickness dr there are N/V 4π r²dr particles. This is a mean field approximation; the position of the particles is averaged. In reality the density close to the particle is different than far away as can be described by a pair correlation function. Furthermore, it is neglected that the fluid is enclosed between walls. Performing the integral we get

${\displaystyle \phi =-2a'{\frac {N}{V}}\quad {\hbox{with}}\quad a'=\epsilon {\frac {2\pi d^{3}}{3}}=\epsilon b'.}$

Hence, we obtain,

${\displaystyle \ln Z=N\ln {(V-Nb')}+{\frac {N^{2}a'}{VkT}}-N\ln {(\Lambda ^{3})}-\ln {N!}}$

From statistical thermodynamics we know that

${\displaystyle p=kT{\frac {\partial \ln Z}{\partial V}}}$,

so that we only have to differentiate the terms containing ${\displaystyle V}$. We get

${\displaystyle p={\frac {NkT}{V-Nb'}}-{\frac {N^{2}a'}{V^{2}}}\Rightarrow \left(p+{\frac {N^{2}a'}{V^{2}}}\right)(V-Nb')=NkT\Rightarrow \left(p+{\frac {n^{2}a}{V^{2}}}\right)(V-nb)=nRT.}$

The Maxwell equal area rule can also be derived from an assumption of equal chemical potential μ of coexisting liquid and vapour phases.[23] On the isotherm shown in the above plot, points a and c are the only pair of points which fulfill the equilibrium condition of having equal pressure, temperature and chemical potential. It follows that systems with volumes intermediate between these two points will consist of a mixture of the pure liquid and gas with specific volumes equal to the pure liquid and gas phases at points a and c.

The Van der Waals equation may be solved for V_G and V_L as functions of the temperature and the vapor pressure p_V. Since:

${\displaystyle p=-\left({\frac {\partial A}{\partial V}}\right)_{T,N}}$

where A is the Helmholtz free energy, it follows that the equal area rule can be expressed as:

${\displaystyle p_{V}={\frac {A(V_{L},T,N)-A(V_{G},T,N)}{V_{G}-V_{L}}}}$is

Since the gas and liquid volumes are functions of p_V and T only, this equation is then solved numerically to obtain p_V as a function of temperature (and number of particles N), which may then be used to determine the gas and liquid volumes.

Locus of coexistence for two phases of Van der Waals fluid

A pseudo-3D plot of the locus of liquid and vapor volumes versus temperature and pressure is shown in the accompanying figure. One sees that the two locii meet at the critical point (1,1,1) smoothly. An isotherm of the Van der Waals fluid taken at T _r = 0.90 is also shown where the intersections of the isotherm with the loci illustrate the construct's requirement that the two areas (red and blue, shown) are equal.

We reiterate that the extensive volume V  is related to the volume per particle v=V/N  where N = nN_A  is the number of particles in the system. The equation of state does not give us all the thermodynamic parameters of the system. We can take the equation for the Helmholtz energy A [24]

${\displaystyle A=-kT\ln Q.\,}$

From the equation derived above for lnQ, we find

${\displaystyle A(T,V,N)=-NkT\left[1+\ln \left({\frac {(V-Nb')T^{3/2}}{N\Phi }}\right)\right]-{\frac {a'N^{2}}{V}}.}$

Where Φ is an undetermined constant, which may be taken from the Sackur–Tetrode equation for an ideal gas to be:

${\displaystyle \Phi =T^{3/2}\Lambda ^{3}=\left({\frac {h}{\sqrt {2\pi mk}}}\right)^{3}}$

This equation expresses A  in terms of its natural variables V  and T , and therefore gives us all thermodynamic information about the system. The mechanical equation of state was already derived above

${\displaystyle p=-\left({\frac {\partial A}{\partial V}}\right)_{N,T}={\frac {NkT}{V-Nb'}}-{\frac {a'N^{2}}{V^{2}}}.}$

The entropy equation of state yields the entropy (S )

${\displaystyle S=-\left({\frac {\partial A}{\partial T}}\right)_{N,V}=Nk\left[\ln \left({\frac {(V-Nb')T^{3/2}}{N\Phi }}\right)+{\frac {5}{2}}\right]}$

from which we can calculate the internal energy

${\displaystyle U=A+TS={\frac {3}{2}}\,NkT-{\frac {a'N^{2}}{V}}.}$

Similar equations can be written for the other thermodynamic potential and the chemical potential, but expressing any potential as a function of pressure p  will require the solution of a third-order polynomial, which yields a complicated expression. Therefore, expressing the enthalpy and the Gibbs energy as functions of their natural variables will be complicated.

Although the material constant a and b in the usual form of the Van der Waals equation differs for every single fluid considered, the equation can be recast into an invariant form applicable to all fluids.

Defining the following reduced variables (f_R, f_C are the reduced and critical variable versions of f, respectively),

${\displaystyle p_{R}={\frac {p}{p_{C}}},\qquad v_{R}={\frac {v}{v_{C}}},\quad {\hbox{and}}\quad T_{R}={\frac {T}{T_{C}}}}$,

where

${\displaystyle p_{C}={\frac {a'}{27b'^{2}}},\qquad \displaystyle {v_{C}=3b'},\quad {\hbox{and}}\quad kT_{C}={\frac {8a'}{27b'}}}$

as shown by Salzman.[25]

The first form of the Van der Waals equation of state given above can be recast in the following reduced form:

${\displaystyle \left(p_{R}+{\frac {3}{v_{R}^{2}}}\right)(3v_{R}-1)=(8T_{R}}$)

This equation is invariant for all fluids; that is, the same reduced form equation of state applies, no matter what a and b may be for the particular fluid.

This invariance may also be understood in terms of the principle of corresponding states. If two fluids have the same reduced pressure, reduced volume, and reduced temperature, we say that their states are corresponding. The states of two fluids may be corresponding even if their measured pressure, volume, and temperature are very different. If the two fluids' states are corresponding, they exist in the same regime of the reduced form equation of state. Therefore, they will respond to changes in roughly the same way, even though their measurable physical characteristics may differ significantly.

The Van der Waals equation is a cubic equation of state; in the reduced formulation the cubic equation is:

${\displaystyle {v_{R}^{3}}-{\frac {1}{3}}\left({1+{\frac {8T_{R}}{p_{R}}}}\right){v_{R}^{2}}+{\frac {3}{p_{R}}}v_{R}-{\frac {1}{p_{R}}}=0}$

At the critical temperature, where ${\displaystyle T_{R}=p_{R}=1}$ we get as expected

${\displaystyle {v_{R}^{3}}-3v_{R}^{2}+3v_{R}-1=\left(v_{R}-1\right)^{3}=0\quad \Longleftrightarrow \quad v_{R}=1}$

For T_R < 1, there are 3 values for v_R. For T_R > 1, there is 1 real value for v_R.

The solution of this equation for the case where there are three separate roots may be found at Maxwell construction.

The equation is also usable as a PVT equation for compressible fluids (e.g. polymers). In this case specific volume changes are small and it can be written in a simplified form:

${\displaystyle (p+A)(V-B)=CT,\,}$

where p is the pressure, V is specific volume, T is the temperature and A, B, C are parameters.