Real Gases and van der Waals Equation

01 / UNDERSTAND & APPLY

1. Measure deviation before trying to explain it

The ideal gas equation assumes negligible particle volume and no intermolecular contribution to the energy between collisions. Real particles occupy space and interact, so a measured P, V, n and T need not satisfy PV=nRT exactly. The question is not simply whether a gas is “real”; every physical gas is real. The useful question is whether ideality is accurate enough at the specified state.

Define the compressibility factor Z=PV/(nRT)=PVₘ/(RT), where Vₘ=V/n. It is dimensionless. At the same T and P, Z equals the actual molar volume divided by the ideal molar volume. At the same T and molar volume, it equals actual pressure divided by ideal pressure. These are two valid comparisons, but their held-fixed variables differ.

If Z is below one, the measured pressure at fixed T and Vₘ is lower than the ideal prediction. If Z is above one, it is higher. A common qualitative interpretation associates negative deviation with net attractive effects and positive deviation with excluded-volume or repulsive effects. This is a statement about the net pressure correction, not proof that one kind of interaction is entirely absent.

For a normal dilute gas at fixed nonzero temperature, Z approaches one as pressure approaches zero. Particles become widely separated, so interaction and finite-size corrections become small relative to the ideal term. High temperature can also reduce the relative importance of attraction, but high temperature alone does not eliminate crowding at very high density.

A single state with Z=1 does not prove ideal behaviour over a whole range. Attractive and excluded-volume corrections can cancel at one state. To test a model, inspect several states and trends, rather than treating one fortunate numerical equality as evidence that molecules have zero size or no attraction.

02 / UNDERSTAND & APPLY

2. Two physical corrections act in opposite directions

Imagine fixing temperature and total volume while comparing an ideal model with a gas whose molecules attract. Cohesion draws neighbouring molecules toward each other and, in a mean-field picture, reduces the pressure transmitted to the walls. To reconstruct the ideal-like pressure term, add a positive attraction correction to the measured pressure.

Now imagine particles with finite size but no attraction. Their centres cannot explore every point in the container because two particles cannot overlap. The accessible volume for translational motion is less than the geometric container volume. A simple model replaces V with V−nb, so the same amount at the same T produces a larger kinetic pressure than nRT/V.

The constant b is a molar excluded-volume parameter, not literally the measured liquid molar volume or the exact volume occupied by a mole of molecules. In the elementary hard-sphere interpretation, b is four times the total geometric particle volume per mole. The factor follows from counting a sphere of excluded centre positions for a pair and sharing that exclusion between the two particles.

The constant a measures the attraction correction within this model. Larger a means a larger attraction contribution when the amount and volume are held fixed. Comparing a alone does not settle every question about boiling or liquefaction, because molecular size and other features also matter; the model critical temperature depends on a/b.

The van der Waals model packages complicated intermolecular behaviour into two substance-dependent parameters. It is conceptually powerful but quantitatively limited, particularly for strongly associating fluids, dense liquids and the immediate neighbourhood of a real critical point. Parameters obtained by fitting one region may not give high accuracy everywhere.

03 / UNDERSTAND & APPLY

3. Keep total volume and molar volume distinct

For n moles occupying total volume V, the equation is (P+an²/V²)(V−nb)=nRT. Solving for measured pressure gives P=nRT/(V−nb)−an²/V². The sign of the final attraction contribution is negative even though the correction is added inside the bracket on the left. This is a frequent source of sign errors.

Divide all extensive volumes by n to obtain the molar form: (P+a/Vₘ²)(Vₘ−b)=RT, or P=RT/(Vₘ−b)−a/Vₘ². Here b already has volume per mole units; inserting nb into the molar form double-counts the amount. Conversely, dropping n from the total-volume form silently assumes a one-mole sample.

Units provide a fast consistency check. If pressure is bar and Vₘ is L mol⁻¹, then a has units L² bar mol⁻² and b has units L mol⁻¹. The term a/Vₘ² then has pressure units, and Vₘ−b is a valid subtraction. Use R=0.08314 L bar mol⁻¹ K⁻¹ for those quantities.

The basic physical domain of this equation is Vₘ>b. At Vₘ=b the repulsive term diverges; below it the accessible-volume interpretation fails. A formal algebraic root outside the domain is not an admissible gas state. Even roots inside the domain need further stability and phase checks at subcritical temperature.

If P, T and amount are given but V is unknown, the equation becomes cubic in volume. A numerical solver can find mathematical roots, but it cannot replace the physical interpretation. Label which root corresponds to vapour, liquid-like or unstable behaviour, and do not automatically accept the first number returned.

04 / UNDERSTAND & APPLY

4. Z corrects state estimates, while compressibility is a derivative

For a gas with known Z at the actual state, PV=ZnRT. Thus n=PV/(ZRT) and V=ZnRT/P. The ideal amount PV/(RT) equals Z times the true amount. When Z<1, an ideal calculation from measured P, V and T underestimates the amount; this direction is easy to reverse unless the algebra is written explicitly.

The density relation becomes ρ=PM/(ZRT), or M=ρZRT/P. If one infers a molar mass with the ideal equation while the real Z is below one, the apparent ideal molar mass is M/Z and is too large. A nonideal density anomaly must not automatically be interpreted as chemical association without additional evidence.

Despite its name, the compressibility factor is not the isothermal compressibility κ_T=−(1/V)(∂V/∂P)_T. The latter measures the fractional volume response to a pressure change and has inverse-pressure units. Z is a state ratio and has no units. Knowing one isolated Z value does not provide the slope needed to determine κ_T.

For a mechanically stable homogeneous phase, κ_T is positive, so its isotherm has a negative pressure–volume slope. A region of a model isotherm with pressure increasing as volume increases would imply negative κ_T and mechanical instability. This derivative criterion becomes important when inspecting the oscillating subcritical van der Waals curve.

In a problem that supplies Z as a function of pressure or temperature, evaluate it at the requested state and retain the dependence when differentiating. Treating a strongly varying Z as a universal constant can give a reasonable state estimate but a wrong response coefficient.

05 / UNDERSTAND & APPLY

5. Expand the model to expose the low-density behaviour

At large molar volume, write 1/(Vₘ−b)=(1/Vₘ)/(1−b/Vₘ). Expanding the geometric series gives 1/Vₘ+b/Vₘ²+b²/Vₘ³+…. Substituting this into the pressure equation and multiplying by Vₘ/(RT) gives Z=1+(b−a/(RT))/Vₘ+b²/Vₘ²+….

The coefficient of 1/Vₘ is the second density virial coefficient B(T). For the van der Waals model, B(T)=b−a/(RT). Its units are volume per mole. It describes the leading dilute-gas correction and can be positive or negative depending on temperature. Higher virial terms remain when that leading coefficient vanishes.

The Boyle temperature is defined by B(T_B)=0, giving T_B=a/(Rb) for this model. At that temperature, the low-pressure Z versus P curve has zero initial slope. It does not follow that Z stays exactly one at every pressure, nor that the gas has become a collection of noninteracting point particles.

The leading pressure expansion follows from 1/Vₘ≈P/(RT) at sufficiently low pressure: Z≈1+[B(T)/(RT)]P. The slope coefficient in this pressure expansion has inverse-pressure units, whereas B(T) in the density expansion has volume-per-mole units. Authors sometimes use different symbols for these coefficients; read the defining equation rather than relying on a letter alone.

Below the model Boyle temperature, B is negative and attraction dominates the first dilute correction. Above it, B is positive. This statement concerns the initial low-density trend. At high density the excluded-volume term can dominate even when B is negative, causing Z eventually to rise above one.

06 / UNDERSTAND & APPLY

6. Derive the critical constants from a horizontal inflection

A subcritical fluid can show distinct liquid and vapour phases in equilibrium. Their density difference diminishes as temperature approaches the critical temperature from below. At the critical point the distinction disappears. Above the critical temperature, compression does not produce the ordinary liquid–vapour coexistence transition, although the fluid can become very dense.

In the van der Waals equation, the critical isotherm has a horizontal inflection at Vₘ,c. Therefore both (∂P/∂Vₘ)_T and (∂²P/∂Vₘ²)_T vanish. Differentiating gives −RT/(Vₘ−b)²+2a/Vₘ³=0 and 2RT/(Vₘ−b)³−6a/Vₘ⁴=0 at the critical point.

Rearrange the first equation to RT=2a(Vₘ−b)²/Vₘ³ and the second to RT=3a(Vₘ−b)³/Vₘ⁴. Equating them and cancelling positive common factors yields 2Vₘ=3(Vₘ−b), hence Vₘ,c=3b. Substituting back gives RT_c=8a/(27b), and substituting both into the original pressure equation gives P_c=a/(27b²).

These formulas are predictions of the two-parameter equation, not exact universal identities for all real fluids. They imply Z_c=P_cVₘ,c/(RT_c)=3/8. Real substances do not all have this critical compressibility factor. Fitting a and b to two measured critical quantities does not guarantee that the third is reproduced.

For a sample of n moles, the total critical volume is V_c=3nb, while the critical molar volume is 3b. Pressure and temperature are intensive and do not scale with sample amount. Confusing these two meanings of critical volume can introduce a hidden factor of n into an otherwise correct calculation.

07 / UNDERSTAND & APPLY

7. Reduced variables remove a and b from this model

Define reduced pressure P_r=P/P_c, reduced temperature T_r=T/T_c and reduced molar volume V_r=Vₘ/Vₘ,c. Substitute the model critical constants into the equation and simplify to (P_r+3/V_r²)(3V_r−1)=8T_r, or P_r=8T_r/(3V_r−1)−3/V_r².

The same reduced equation applies to every van der Waals gas, independent of its separate a and b values. This is the model version of corresponding states: gases at equal reduced conditions have the same reduced behaviour within the model. It motivates comparisons across substances but is not an exact law for every molecular fluid.

The physically allowed volume domain becomes V_r>1/3. The critical point is (T_r,P_r,V_r)=(1,1,1). A plot made with these dimensionless variables lets the qualitative critical structure be displayed without suggesting that a particular real substance has exactly the model curve.

Combining Z=P_rV_r Z_c/T_r with Z_c=3/8 yields Z=3P_rV_r/(8T_r). This is useful for checking reduced-state calculations. Also T_B/T_c=27/8 for the van der Waals model. The ratio is model-specific, just as its fixed critical compressibility is.

08 / UNDERSTAND & APPLY

8. An algebraic loop is not the observed equilibrium path

Below T_c, a van der Waals isotherm can develop a local minimum and maximum, giving an S-shaped loop. Part of that loop has positive slope and negative isothermal compressibility. A uniform phase in that region is mechanically unstable: a density fluctuation tends to grow rather than being restored.

An actual equilibrium isothermal compression through liquid–vapour coexistence changes the proportions of the two phases at a common saturation pressure. In a P–V plot for a fixed amount, the coexistence interval is a horizontal segment. The gas volume decreases as more material enters the denser liquid phase; pressure need not rise during that interval.

For the van der Waals model, the coexistence pressure is selected by the Maxwell equal-area construction. The horizontal line intersects the low-volume liquid-like and high-volume vapour-like branches, and the signed integral of P_model−P_coexistence between those endpoint molar volumes is zero. This condition expresses equality of the two phases’ chemical potentials.

The equal-area construction does not mean that every point on the original loop is an accessible stable state. Portions outside the spinodal interval can be metastable under suitable constraints, such as delayed condensation, but the equilibrium two-phase state has lower free energy. Distinguish equilibrium, metastability and mechanical instability instead of labelling the entire curve simply “real”.

At T_c the coexistence endpoints merge and the plateau contracts to the critical point. Above T_c there is no ordinary liquid–vapour coexistence plateau. This does not mean that the substance must remain dilute or that it cannot have liquid-like density. The terms supercritical and ideal describe different ideas.

09 / UNDERSTAND & APPLY

9. Use limits to catch incorrect signs and powers

Let a and b tend to zero: the equation must reduce to P=RT/Vₘ. At fixed T and Vₘ, increasing a must lower the model pressure. At fixed T and Vₘ>b, increasing b must raise it. These checks catch a misplaced plus sign without requiring a full recalculation.

At sufficiently large Vₘ, both correction terms are small relative to the ideal term. At Vₘ approaching b from above, the positive excluded-volume contribution diverges while the attraction term remains finite for fixed positive b. Extrapolating that simple divergence into an exact description of real matter is unjustified; it only marks the boundary of the model’s available-volume interpretation.

If measured T_c and P_c are used to fit the model, rearrangement gives b=RT_c/(8P_c) and a=27R²T_c²/(64P_c). With those fitted parameters, Vₘ,c is forced to equal 3RT_c/(8P_c). A measured critical molar volume that differs is evidence of model limitation, not an arithmetic contradiction.

Two measured pressures at different molar volumes and the same temperature can also constrain parameters, but solving nonlinear equations may produce inadmissible or poorly conditioned values. Verify positivity where the simple attraction/excluded-volume interpretation is intended, check Vₘ>b, and test another state if one is available.

A negative pressure returned by a formal model expression in some low-temperature region should not be accepted as an ordinary stable gas pressure. It can signal extrapolation into an unstable or tensile branch outside the intended gas calculation. Interpret phase behaviour before using a root merely because it satisfies the polynomial.

10 / UNDERSTAND & APPLY

10. Solved practice: calculation plus interpretation

Always say whether a result is a measured state property, a dilute-limit approximation or a van der Waals prediction. Those labels matter when a neat algebraic result such as Z_c=3/8 is compared with experimental data. The following original problems use the supplied parameters and do not claim a particular fluid has exactly those properties.