Real Gases and the van der Waals Equation

In short

The ideal gas law PV = nRT treats gas molecules as point particles that never feel each other. Real molecules have finite size and attract each other at short range. When gas density rises — high pressure, low temperature, or close to condensation — those two effects push the measured (P, V, T) away from the ideal prediction by per cents, then tens of per cents, then more.

The van der Waals equation corrects the ideal gas law with two physical additions, for one mole of gas:

\boxed{\;\left(P + \frac{a}{V^2}\right)(V - b) = RT\;}

b is the co-volume — the effective volume excluded by the finite size of the molecules themselves. For a gas of hard spheres of diameter d, b \approx 4 \times N_A \cdot \tfrac{4}{3}\pi (d/2)^3 per mole (four times the actual molecular volume, because of the geometry of excluded pair-volumes).
a/V^2 is the internal pressure — an effective extra pressure the gas exerts on itself because molecules attract each other at moderate separations. The 1/V^2 dependence is geometric: one factor of 1/V comes from the density of molecules feeling the attraction, another from the density of molecules doing the attracting.

For carbon dioxide, a = 0.364\,\text{Pa m}^6\,\text{mol}^{-2} and b = 4.27 \times 10^{-5}\,\text{m}^3\,\text{mol}^{-1}. For nitrogen, a = 0.137\,\text{Pa m}^6\,\text{mol}^{-2}, b = 3.87 \times 10^{-5}\,\text{m}^3\,\text{mol}^{-1}. For hydrogen — weakly attracting and small — both constants are tiny. For helium — the least sticky of all gases — both are the smallest in the periodic table.

Solve the equation for P and plot P versus V at fixed T: below a critical temperature T_c = 8a/(27 R b), the isotherm develops a hump and a dip — a signature of a gas that wants to split into a dense liquid phase and a dilute vapour phase. Above T_c, the hump vanishes and the gas can no longer be liquefied by compression. Below T_c, real gases do liquefy — which is exactly why an LPG cylinder at room temperature (well below propane's T_c \approx 370 K) contains liquid, not gas.

Pick up an LPG cylinder in an Indian kitchen. Shake it gently — you can hear liquid sloshing inside. The cylinder is labelled "LPG" (liquefied petroleum gas), and the word liquefied is doing a lot of work: at room temperature, the propane and butane molecules inside are not a gas at all. They are a liquid in equilibrium with a small amount of vapour above it, at a pressure of roughly 2–8 atmospheres. Open the valve; the vapour rushes into the regulator, the pressure drops, and some more liquid boils to replace what left. That is why a 14.2 kg cylinder cooks your meals for weeks — most of its contents are packed into the liquid, not the gas.

The ideal gas law PV = nRT cannot explain any of this. Plug in 14.2 kg of propane (molar mass 44 g/mol, so n \approx 323 mol) and the internal volume of a domestic cylinder (about 33 litres) at 300 K and you get a pressure of 244 atm — huge, ridiculous, and also wrong. The reality is that the gas condenses long before those densities are reached; the cylinder contains a liquid, and the vapour above it sits at a modest 2–8 atm. The ideal gas law did not merely misquote the number — it failed to predict the phase of matter.

This is the failure mode you should carry around in your head. The ideal gas law is excellent when molecules are far apart compared to their size and when the temperature is high enough that they cannot form a liquid. Air in your room at one atmosphere — superb; an error below one per cent. A SCUBA tank at 200 atm — noticeably off. An LPG cylinder at the density of liquid propane — the equation does not even apply. The job of a real-gas equation of state is to push accuracy into the regime where molecules are close enough to feel each other, and ultimately to predict the gas–liquid transition itself.

The van der Waals equation, proposed in 1873, is the simplest real-gas equation of state that captures both of the ways real molecules deviate from ideality — and the one you need to know for JEE. It is not the final word (engineers use Peng–Robinson or Redlich–Kwong for serious work), but it is the equation that teaches you which physics matters: molecules take up space, and molecules attract each other. Everything else is quantitative refinement.

This article builds the equation from first principles. You should have read the ideal gas law (where PV = nRT comes from), kinetic theory — pressure and temperature (where pressure was derived from wall collisions of non-interacting point particles), and mean free path (where molecular diameter d entered the story through the collision cross-section). Familiarity with degrees of freedom and equipartition helps but is not essential here.

Where the ideal gas law quietly breaks

The kinetic-theory derivation of PV = nRT relied on two assumptions, both stated without apology in the ideal-gas-laws article:

Molecules are point particles — they occupy zero volume. The only volume a gas has is the container volume V.
Molecules do not interact except during instantaneous collisions — there are no attractive or repulsive forces at finite range. A molecule flies in a straight line between wall impacts.

Both assumptions are unambiguously false for real molecules. A nitrogen molecule has a diameter of roughly 0.37 nm; a water molecule, 0.28 nm. Two molecules at a separation of three or four diameters attract each other with a force called the van der Waals force — an electromagnetic dipole-dipole attraction that arises because even a neutral molecule has a briefly fluctuating dipole moment, and that fluctuation polarises its neighbour into an attractive partner. At much shorter range — one diameter or less — the electron clouds of the two molecules overlap and the force reverses sharply: the molecules repel each other hard. This is the origin of the effective "hard-sphere" size.

So a real molecule is neither a point nor indifferent. It has size, and it has a short-range attraction. Both effects are negligible when the gas is dilute (few molecules per unit volume, so pairs are rarely close together) and when the temperature is high (molecules are so energetic that the tiny attractive energy barely slows them). Both effects become unavoidable when the gas is compressed toward liquid densities or cooled toward liquid temperatures — that is, exactly in the regime where liquefaction is about to happen.

Potential energy between two molecules as a function of their separation $r$. At $r < d$ (one molecular diameter), the electron clouds overlap and the potential spikes upward — this is the hard-core repulsion that underlies the van der Waals parameter $b$. At $r \approx 1.1\,d$ the curve reaches a shallow minimum of depth $\varepsilon$ — this weak attraction is what $a$ is measuring. For $r \gg d$ the molecules barely feel each other, which is why the ideal-gas limit works at low density.

The ideal gas law corresponds to pretending this entire curve is flat at zero: no repulsion, no well, no force. Van der Waals' 1873 insight was that you can capture the leading effect of each feature with one simple substitution in the equation.

Correction 1 — finite molecular volume: the V - b term

Start with the first failure: molecules are not points. If a gas occupies a container of volume V, the molecules themselves are already taking up some of that space, and each molecule has less room to roam than V would suggest.

Treat each molecule as a hard sphere of diameter d (radius d/2). Ask the following concrete question: in a container holding N molecules, what is the volume available to any one molecule — the volume in which its centre can actually wander without overlapping any other molecule?

The naive answer is "V minus the volume occupied by N spheres," i.e. V - N \cdot \tfrac{4}{3}\pi (d/2)^3. That answer is almost right but misses a factor-of-two geometric subtlety. Here is the careful version.

Step 1. Consider two molecules, 1 and 2. From molecule 1's point of view, molecule 2 is forbidden to come closer than a centre-to-centre distance of d.

Why: each molecule has radius d/2. When their surfaces just touch, the centre-to-centre distance is d/2 + d/2 = d. Any closer is overlap, which hard spheres cannot do.

Step 2. So the "forbidden zone" that molecule 2 creates around molecule 1 is a sphere of radius d around molecule 1 — a sphere of volume

v_\text{excl, pair} = \tfrac{4}{3}\pi d^3.

Why: the forbidden region for the centre of molecule 2 is all points within distance d of the centre of molecule 1 — i.e. the interior of a sphere of radius d.

Step 3. Now note that this excluded volume was computed looking at a pair of molecules (one "host", one "guest"). In a real gas you have not two molecules but N_A per mole, and every pair contributes one such excluded volume. But each pair is shared between two molecules (molecule A excludes B, and equivalently B excludes A — the same forbidden-centre event), so when you divide the total pair-excluded volume across all N_A molecules, each molecule gets half of one pair's excluded volume:

v_\text{excl, per molecule} = \tfrac{1}{2}\cdot\tfrac{4}{3}\pi d^3 = \tfrac{2}{3}\pi d^3 = 4 \times \underbrace{\tfrac{4}{3}\pi(d/2)^3}_{\text{actual volume}} .

Why: expand \tfrac{2}{3}\pi d^3 and compare to the actual molecular volume \tfrac{4}{3}\pi (d/2)^3 = \tfrac{1}{6}\pi d^3. The ratio \tfrac{2}{3}\pi d^3 \,/\, \tfrac{1}{6}\pi d^3 = 4. The excluded volume per molecule is 4\times the actual volume of a molecule — a factor often remembered as "excluded volume is four times molecular volume" in the Indian JEE textbooks.

Step 4. For one mole, multiply by Avogadro's number:

b = N_A \cdot \tfrac{2}{3}\pi d^3 = 4 N_A \cdot \tfrac{4}{3}\pi (d/2)^3. \tag{1}

Why: b is the total excluded volume per mole. The formula gives a quick estimate: for nitrogen (d \approx 0.37 nm), b \approx 4 \times 6.02\times10^{23}\times \tfrac{4}{3}\pi(0.185\times10^{-9})^3 \approx 6\times 10^{-5}\,\text{m}^3/\text{mol}. The measured value is 3.87\times 10^{-5}\,\text{m}^3/\text{mol}; the hard-sphere estimate is correct to a factor of order unity.

Step 5. Replace the ideal-gas volume V in the equation of state by the effective available volume V - b:

P(V - b) = RT \qquad \text{(first correction only)}. \tag{2}

Why: in the ideal gas law, V was the volume available for molecular centres to wander. Now that volume is reduced by the co-volume b. At low density (V \gg b), equation (2) reduces back to the ideal gas law. At high density (V \to b), the left-hand side diverges — meaning you cannot compress the gas below the co-volume; the molecules are already touching.

That single substitution accounts for the repulsive side of the molecular potential curve. It is a hard effect: it says "you cannot compress a mole of gas into less than b of volume, no matter how hard you push." For nitrogen, b \approx 39\,\text{cm}^3/\text{mol} — which is already close to the molar volume of liquid nitrogen (about 34\,\text{cm}^3/\text{mol}). The co-volume is what prevents the ideal gas law from predicting zero volume in the limit of infinite pressure.

Correction 2 — molecular attraction: the a/V^2 term

Now the attractive side of the potential curve. At moderate separations (roughly one to a few diameters) molecules attract each other. How does that attraction modify the pressure the gas exerts on its container?

Here is the physical picture. Consider a molecule deep inside the gas, surrounded symmetrically by other molecules. The attractive forces on it from all directions cancel by symmetry — it feels no net tug. Now consider a molecule just about to hit the wall. As it approaches the wall, all its attractive neighbours are behind it (there are no molecules in the wall to attract it forward). The gas pulls on the molecule with a net backward force, slowing it slightly before impact. The molecule hits the wall, but with a slightly smaller velocity — and therefore with a slightly smaller momentum change — than it would have had in the absence of attractions.

The wall therefore experiences a slightly smaller pressure than a non-interacting gas at the same density and temperature would produce. The attractions have reduced the measured pressure P below its kinetic-theory value. If you wanted to predict the pressure from microscopic motion, you would compute the ideal kinetic pressure and then subtract a correction proportional to the strength of the cohesive pull.

The size of that correction depends on two factors:

How many molecules are there to be pulled back from the wall? This is proportional to the density n/V of molecules approaching the wall — i.e. proportional to 1/V for one mole.
How many molecules are there behind them doing the pulling? The strength of the cohesive tug on each approaching molecule is proportional to the density of its attracting neighbours — again, proportional to 1/V.

The pressure correction is therefore the product of two density factors, one for the target and one for the source:

\Delta P_\text{attr} \propto \frac{1}{V^2}, \qquad \text{write } \Delta P_\text{attr} = \frac{a}{V^2}. \tag{3}

Why: you can think of this as a mean-field treatment. Each molecule near the wall is retarded by a mean attraction that depends on the density of all the others. Doubling the density doubles both the number of candidates being slowed and the strength of the average slowing force on each — so the total pressure correction scales as density squared, i.e. as 1/V^2. The constant a absorbs the depth of the attractive well and the range of the attraction; it is positive, with units of Pa·m⁶·mol⁻².

Step 6. If P_\text{kinetic} is the pressure a gas of the same density would exert in the absence of attraction, and P is the pressure the wall actually measures, then

P_\text{kinetic} = P + \frac{a}{V^2}.

Why: the wall measurement is smaller than the kinetic value by a/V^2; to recover the kinetic value you must add the attraction correction back.

Step 7. Apply this to equation (2). Equation (2) says the kinetic pressure obeys P_\text{kinetic}(V - b) = RT. Substitute P_\text{kinetic} = P + a/V^2:

\left(P + \frac{a}{V^2}\right)(V - b) = RT. \tag{4}

Why: this is the van der Waals equation for one mole of gas. Both corrections are in: V - b accounts for the excluded volume from finite molecular size; a/V^2 accounts for the attractive pull that reduces wall pressure below the kinetic value. For n moles in a volume V, replace V \to V/n and b \to b per mole, giving \left(P + n^2 a / V^2\right)(V - n b) = nRT.

That is the equation you derive, use, and memorise. Two corrections — one per flaw in the ideal gas assumptions.

Interpreting the two constants

Both a and b are positive — never negative, for any gas. Their relative sizes tell you what kind of gas you are dealing with.

Gas	a (Pa·m⁶/mol²)	b (×10⁻⁵ m³/mol)	Boiling point at 1 atm	T_c (K)	Comment
He	0.0035	2.37	4.2 K	5.2	Smallest a of any gas — almost ideal everywhere above 10 K
H₂	0.0247	2.65	20.3 K	33.2	Small, weakly attracting
N₂	0.137	3.87	77 K	126.2	Air's main component; nearly ideal at 300 K
O₂	0.138	3.18	90 K	154.6	Similar to N₂
CO₂	0.364	4.27	sublimes at 195 K	304.1	Strong attractions; close to ideal only at low density
H₂O (vapour)	0.554	3.05	373 K	647.1	Very strong attractions (hydrogen bonds)
Propane (C₃H₈)	0.939	9.05	231 K	369.8	Large a and b; LPG gas

The pattern is simple. Bigger molecules have bigger b. Propane, with three carbon atoms, has a co-volume about four times that of hydrogen. More-strongly-attracting molecules have bigger a. Water's a is about 20 times helium's because water molecules hydrogen-bond to each other, while helium atoms barely attract anything. Hydrogen bonds, polar dipoles, and dispersion forces all end up packed into the single empirical number a.

The table also hints at a relationship between a, b, and the temperature at which a gas will liquefy. Notice that gases with large a have high boiling points and high critical temperatures. That is not an accident. The same attractive force that reduces wall pressure is what ultimately binds molecules together into a liquid. The van der Waals equation can be used to predict the critical temperature from a and b.

The critical point — where a gas stops being compressible into a liquid

Here is the most beautiful consequence of the van der Waals equation. Rearrange (4) to solve for P as a function of V at fixed T (for one mole):

P(V;T) = \frac{RT}{V - b} - \frac{a}{V^2}. \tag{5}

For each T this defines an isotherm — a curve on the P–V plane. At high T, the curve is a smooth hyperbola-like decrease of P with increasing V, not very different from Boyle's law. As you lower T, something remarkable happens: at some critical temperature T_c, the isotherm develops an inflection point — a horizontal tangent point with zero curvature. Below T_c, the isotherm grows a hump: as V increases from the small side, P rises, reaches a local maximum, falls through a local minimum, then rises to approach the high-V limit. The hump means the equation P(V) = P_0 can have three solutions for a single pressure P_0.

Van der Waals isotherms in units where $a = 0.5$, $b = 0.5$, $R = 1$. From top to bottom: $T = 1.3, 1.0, 0.83, 0.7, 0.6$. The middle (dark) curve is at the critical temperature $T_c = 8a/(27 R b) \approx 0.83$ — it has a single horizontal inflection at the critical point $(V_c, P_c) = (1.5, 0.67)$. Above $T_c$ the isotherms descend monotonically. Below $T_c$ the isotherms develop a wiggle (a maximum and a minimum) — the unstable wiggle is replaced in reality by a horizontal plateau where liquid and vapour coexist.

The three solutions for a given pressure are interpreted as follows. The smallest V is a dense liquid-like state; the largest is a dilute gas-like state; the middle root (on the descending part of the wiggle) is mechanically unstable (squeezing it would raise P, which would want to compress further — unstable). So the physical gas, below T_c, does not smoothly trace the wiggly curve. What happens instead is a phase separation: at some intermediate pressure, a portion of the container fills with the dense liquid root and the rest with the dilute gas root. The true equation-of-state plateau at that pressure is a horizontal line (the Maxwell construction) stretching from the liquid volume to the gas volume.

Above T_c, no such plateau exists. The gas cannot be liquefied by compression alone — no matter how hard you squeeze, the molecules never settle into two distinct phases, and you get a single dense "supercritical" fluid instead.

Deriving T_c, P_c, and V_c from a and b

The critical isotherm is the one where the hump has just collapsed to a single inflection point. At an inflection, both \partial P/\partial V and \partial^2 P/\partial V^2 vanish. Apply those two conditions to equation (5).

Step 1. Compute \partial P/\partial V from equation (5).

\frac{\partial P}{\partial V} = -\frac{RT}{(V-b)^2} + \frac{2a}{V^3}.

Why: differentiate (5) term by term. The first term, RT/(V-b), differentiates to -RT/(V-b)^2. The second term, -a/V^2, differentiates to +2a/V^3.

Step 2. Compute \partial^2 P/\partial V^2.

\frac{\partial^2 P}{\partial V^2} = \frac{2RT}{(V-b)^3} - \frac{6a}{V^4}.

Why: differentiate again. The first term becomes +2RT/(V-b)^3; the second becomes -6a/V^4.

Step 3. At the critical point, set both derivatives to zero:

\frac{RT_c}{(V_c-b)^2} = \frac{2a}{V_c^3} \tag{i}

\frac{2RT_c}{(V_c-b)^3} = \frac{6a}{V_c^4} \tag{ii}

Why: both first and second derivatives must vanish at the critical inflection. Equation (i) is \partial P/\partial V = 0 rearranged; (ii) is \partial^2 P/\partial V^2 = 0 rearranged.

Step 4. Divide equation (ii) by equation (i) to eliminate RT_c:

\frac{(ii)}{(i)}:\quad \frac{2RT_c/(V_c-b)^3}{RT_c/(V_c-b)^2} = \frac{6a/V_c^4}{2a/V_c^3}

\frac{2}{V_c - b} = \frac{3}{V_c}.

Why: cancelling RT_c and a on each side leaves pure geometric factors. The equation is now a single constraint on V_c and b alone.

Step 5. Solve for V_c.

2 V_c = 3(V_c - b) = 3 V_c - 3b

V_c = 3b. \tag{6}

Why: a beautifully simple result. The critical volume is three times the co-volume — i.e. the molecules at the critical point are packed to one-third of their maximum possible density. The numerical ratio "3" is universal: every gas obeying the van der Waals equation has V_c/b = 3.

Step 6. Substitute V_c = 3b into equation (i) and solve for T_c.

\frac{RT_c}{(3b - b)^2} = \frac{2a}{(3b)^3}

\frac{RT_c}{4 b^2} = \frac{2a}{27 b^3}

RT_c = \frac{4 b^2 \cdot 2a}{27 b^3} = \frac{8a}{27 b}

\boxed{\;T_c = \frac{8a}{27 R b}\;}. \tag{7}

Why: substitute V_c = 3b into \partial P/\partial V = 0 and solve algebraically. The result expresses T_c in terms of the two van der Waals constants alone. For propane: T_c = 8 \times 0.939 / (27 \times 8.314 \times 9.05\times 10^{-5}) \approx 370 K — matching the measured propane T_c = 369.8 K to three significant figures. The van der Waals prediction is quantitatively correct for small molecules.

Step 7. Plug V_c and T_c back into (5) for P_c.

P_c = \frac{R T_c}{V_c - b} - \frac{a}{V_c^2} = \frac{(8a/27b)}{2b} - \frac{a}{9 b^2} = \frac{4a}{27 b^2} - \frac{3a}{27 b^2} = \frac{a}{27 b^2}

\boxed{\;P_c = \frac{a}{27 b^2}\;}. \tag{8}

Why: substitute V_c - b = 2b into the first term and V_c^2 = 9 b^2 into the second. The arithmetic gives \tfrac{4a}{27b^2} - \tfrac{a}{9b^2} = \tfrac{4a - 3a}{27 b^2} = \tfrac{a}{27b^2}, which is the critical pressure.

The three critical parameters are therefore:

V_c = 3b, \qquad T_c = \frac{8a}{27 R b}, \qquad P_c = \frac{a}{27 b^2}.

A fourth, dimensionless combination is especially useful — the critical compressibility:

Z_c \equiv \frac{P_c V_c}{R T_c} = \frac{(a/27b^2)(3b)}{R \cdot (8a/27Rb)} = \frac{3}{8} = 0.375.

Why: multiply and cancel. The van der Waals equation predicts that every real gas has Z_c = 0.375 at its critical point — a universal number. Measured values are mostly in the range 0.27–0.31, so the prediction is about 20–30% high. Better equations of state (Peng–Robinson gets Z_c \approx 0.307) do better, but 3/8 is a useful first estimate and a famous result.

Explore the isotherms yourself

The interactive figure below is the van der Waals equation of state with a = 0.5, b = 0.5, and R = 1 in dimensionless units (so T_c = 8/(27) \times 1 / 0.5 \approx 0.59 and things are easy to see). Drag the temperature slider to watch the isotherm change shape. Above T_c it is a smooth monotonic curve. Below T_c you see the wiggle that signals the liquid–vapour transition.

The van der Waals isotherm at reduced temperature $T$. Drag the red dot vertically along the right edge to change $T$ between 0.3 and 1.2. For $T > T_c \approx 0.593$, the curve falls monotonically — a normal gas. For $T < T_c$, the curve grows a wiggle — the signature of an impending liquid–gas transition. In reality the wiggle is replaced by a horizontal plateau (Maxwell construction) at the coexistence pressure.

The critical point is exactly where the hump has just disappeared — the single temperature at which liquid and vapour become indistinguishable. Above it, no amount of squeezing will condense the gas.

Worked examples

Example 1: LPG in a domestic cylinder

A 14.2 kg LPG cylinder sold for domestic use in India contains roughly pure propane (C₃H₈, molar mass 44.1 g/mol) for estimation purposes. The cylinder's internal volume is 33.3 L. Using the van der Waals constants for propane (a = 0.939\,\text{Pa m}^6\,\text{mol}^{-2}, b = 9.05\times 10^{-5}\,\text{m}^3\,\text{mol}^{-1}), first estimate the pressure the cylinder would register if the propane were all in the gas phase at 300 K. Then explain, using T_c, why the cylinder actually contains a liquid.

Schematic LPG cylinder. The bulk of the 14.2 kg charge sits as a liquid at the bottom; a small vapour space above it delivers gas to the regulator as the liquid boils to replace it.

Step 1. Compute the number of moles.

n = \frac{14.2\,\text{kg}}{44.1\times 10^{-3}\,\text{kg/mol}} \approx 322\,\text{mol}.

Why: moles is mass divided by molar mass. 14.2 kg is 322 moles of C₃H₈.

Step 2. Ideal-gas prediction at T = 300 K, V = 33.3 \times 10^{-3}\,\text{m}^3.

P_\text{ideal} = \frac{nRT}{V} = \frac{322 \times 8.314 \times 300}{33.3\times 10^{-3}} \approx 2.41\times 10^{7}\,\text{Pa} \approx 241\,\text{atm}.

Why: plug into PV = nRT. The ideal-gas answer is absurdly large — an LPG cylinder rated for a working pressure of about 17 atm would explode at 241 atm.

Step 3. Check the critical temperature for propane.

T_c = \frac{8a}{27 R b} = \frac{8 \times 0.939}{27 \times 8.314 \times 9.05\times 10^{-5}} = \frac{7.512}{0.02031} \approx 370\,\text{K}.

Why: apply equation (7) directly. The result matches the measured value of 369.8 K, confirming that the van der Waals constants for propane are doing their job.

Step 4. Compare T with T_c.

At T = 300\,\text{K}, you are at T / T_c = 300/370 \approx 0.81. This is well below the critical temperature — so propane can exist simultaneously as a liquid and its vapour at 300 K. In fact, at 300 K the saturated-vapour pressure of propane is about 9.8 atm; the cylinder will sit at that pressure, with most of the propane as liquid at the bottom.

Why: T < T_c is the necessary and sufficient condition for the gas to be liquefiable by compression. Below T_c, the van der Waals isotherm has the liquid–vapour plateau; above it, no liquid phase exists.

Step 5. Estimate the volume of the liquid.

Liquid propane at 300 K has a density of about 493 kg/m³. So 14.2 kg of liquid occupies about 14.2 / 493 \approx 0.0288\,\text{m}^3 = 28.8 L. The vapour space is 33.3 - 28.8 \approx 4.5 L. The vapour pressure in that space at 300 K is about 9.8 atm.

Why: the cylinder is designed so that after filling, roughly 85 per cent of the volume is liquid and 15 per cent is vapour. This vapour headspace is what is meant by "the gas in an LPG cylinder"; when you open the valve, the vapour leaves, the pressure drops, and the liquid boils to replace it.

Result: The ideal-gas prediction is off by nearly 25×. The real cylinder holds the propane as a liquid at about 10 atm because T < T_c. The van der Waals equation is what tells you — using only a and b — that liquefaction is possible at 300 K.

What this shows: the ideal gas law has no concept of liquefaction. The van der Waals equation, through T_c = 8a/(27Rb), predicts exactly which gases can be liquefied at room temperature and which cannot. Propane, butane, ammonia, CO₂ — all have T_c > room temperature and all can be shipped in liquid form. Nitrogen, oxygen, argon, methane — all have T_c < room temperature and all require cryogenic cooling to liquefy.

Example 2: Carbon dioxide at Diwali dry-ice conditions

For Diwali celebrations, some party organisers scatter blocks of "dry ice" — solid CO₂ at 195 K — around a stage. The CO₂ sublimes directly to gas, producing the white fog that hovers at ground level (cold CO₂ is denser than air). Estimate the pressure a container would need to hold one mole of CO₂ at 250 K in a volume of 100\,\text{cm}^3 = 10^{-4}\,\text{m}^3, using (a) the ideal gas law and (b) the van der Waals equation. CO₂ has a = 0.364\,\text{Pa m}^6\,\text{mol}^{-2} and b = 4.27\times 10^{-5}\,\text{m}^3\,\text{mol}^{-1}.

The same state of CO₂ predicted by the two equations of state. The ideal-gas estimate is nearly three times too high.

Step 1. Ideal-gas prediction.

P_\text{ideal} = \frac{RT}{V} = \frac{8.314 \times 250}{10^{-4}} = 2.08\times 10^{7}\,\text{Pa} \approx 205\,\text{atm}.

Why: plug the numbers in. RT/V with V = 10^{-4}\,\text{m}^3 gives a large number — about 205 atm — because the density is very high.

Step 2. Van der Waals prediction. Solve equation (4) for P:

P = \frac{RT}{V - b} - \frac{a}{V^2}.

With V - b = 10^{-4} - 4.27\times 10^{-5} = 5.73\times 10^{-5}\,\text{m}^3:

\frac{RT}{V - b} = \frac{8.314 \times 250}{5.73\times 10^{-5}} = \frac{2078.5}{5.73\times 10^{-5}} \approx 3.63\times 10^{7}\,\text{Pa}.

With V^2 = 10^{-8}\,\text{m}^6:

\frac{a}{V^2} = \frac{0.364}{10^{-8}} = 3.64\times 10^{7}\,\text{Pa}.

Subtract:

P = 3.63\times 10^{7} - 3.64\times 10^{7} \approx -0.1 \times 10^7\,\text{Pa}.

Why: uh oh. With these specific numbers the two terms nearly cancel and the equation is giving a tiny or even negative pressure — a sign that we are in the liquid–gas coexistence region, where the van der Waals equation is unphysical and should be replaced by the Maxwell construction. Let us retry with a slightly larger molar volume where the answer is well-defined.

Step 2′. Retry with V = 2\times 10^{-4}\,\text{m}^3 (200 cm³ per mole), still well below ideality but now unambiguously in the gas branch.

\frac{RT}{V - b} = \frac{8.314 \times 250}{2\times 10^{-4} - 4.27\times 10^{-5}} = \frac{2078.5}{1.573\times 10^{-4}} \approx 1.32\times 10^{7}\,\text{Pa}.

\frac{a}{V^2} = \frac{0.364}{4\times 10^{-8}} = 9.1\times 10^{6}\,\text{Pa}.

P_\text{vdW} = 1.32\times 10^{7} - 9.1\times 10^{6} = 4.1\times 10^{6}\,\text{Pa} \approx 41\,\text{atm}.

Ideal-gas prediction at the same state:

P_\text{ideal} = \frac{8.314 \times 250}{2\times 10^{-4}} \approx 1.04\times 10^{7}\,\text{Pa} \approx 103\,\text{atm}.

Why: at V = 2\times 10^{-4}\,\text{m}^3 the ideal-gas prediction (103 atm) is about two and a half times the van der Waals prediction (41 atm). The attractive a/V^2 term dominates over the excluded-volume correction at this density — the molecules are pulling each other back so hard that the actual pressure is less than half what the ideal-gas law would predict.

Step 3. Interpret.

At 250 K, T/T_c = 250/304 \approx 0.82 — we are well below the CO₂ critical temperature. Compressing the gas into 100\,\text{cm}^3/\text{mol} would actually liquefy a portion of it; the van der Waals answer in step 2 (close to zero or negative) is the mathematical signature that we had entered the coexistence plateau. At 200\,\text{cm}^3/\text{mol} the gas is dense but still gaseous; the van der Waals equation gives a sensible answer, and it is badly different from the ideal prediction because the molecules are strongly attracting each other.

Result: At modestly dense conditions (V = 200\,\text{cm}^3/\text{mol}, T = 250 K), CO₂ exerts about 41 atm, less than half of the 103 atm the ideal gas law predicts. At still higher density, the gas liquefies, a regime the ideal gas law cannot represent at all.

What this shows: the attractive term is often the more important correction for cold, polar-ish gases like CO₂. Ignoring it can easily lead to a factor-of-two or worse error. The van der Waals equation captures the physics qualitatively and quantitatively across this regime — and it warns you (through negative pressures in the naive solution) when you have crossed into the two-phase region.

Example 3: Estimating the size of a molecule from $b$

For nitrogen, the measured van der Waals constant is b = 3.87\times 10^{-5}\,\text{m}^3/\text{mol}. Estimate the molecular diameter d of N₂ and compare with the value inferred from mean-free-path measurements (d \approx 0.37 nm).

The co-volume $b$ encodes the molecular diameter $d$. Solving $b = 4 N_A \cdot \tfrac{4}{3}\pi (d/2)^3$ for $d$ gives an estimate within 20 per cent of the value from mean-free-path data.

Step 1. Start from equation (1):

b = 4 N_A \cdot \tfrac{4}{3}\pi \left(\frac{d}{2}\right)^3 = \frac{2\pi N_A d^3}{3}.

Why: substitute the sphere-volume formula into equation (1) and simplify. The combined constant on the right is 2\pi N_A / 3.

Step 2. Solve for d:

d^3 = \frac{3 b}{2\pi N_A} \quad \Rightarrow \quad d = \left(\frac{3b}{2\pi N_A}\right)^{1/3}.

Why: invert the algebra. The cube root isolates d from the cubic relation.

Step 3. Plug numbers in.

\frac{3 b}{2\pi N_A} = \frac{3 \times 3.87\times 10^{-5}}{2\pi \times 6.022\times 10^{23}} = \frac{1.161\times 10^{-4}}{3.784\times 10^{24}} \approx 3.07\times 10^{-29}\,\text{m}^3.

d = (3.07\times 10^{-29})^{1/3} \approx 3.13\times 10^{-10}\,\text{m} = 0.313\,\text{nm}.

Why: take the cube root of 3.07 \times 10^{-29}. Break it: 3.07^{1/3} \approx 1.45 and (10^{-29})^{1/3} = 10^{-29/3} = 10^{-9.67} \approx 2.15\times 10^{-10}. Product: 1.45 \times 2.15 \approx 3.12\times 10^{-10} m, or 0.31 nm.

Step 4. Compare with the mean-free-path value.

The mean-free-path analysis of viscosity and diffusion data gives d_\text{N₂} \approx 0.37 nm. The van-der-Waals estimate, d \approx 0.31 nm, is about 16 per cent smaller.

Why: the discrepancy is genuine, not arithmetic. The "hard sphere" diameter inferred from different physical effects is slightly different because real molecules are not hard spheres. The mean-free-path diameter measures the collision cross-section, which depends on how the molecules deflect at finite speed. The van-der-Waals diameter measures the excluded volume in a dense gas, which depends on how they pack at lower speeds. Both are about 0.3 nm; the ratio agreement between two independent methods is reassuring.

Result: The co-volume b = 3.87 \times 10^{-5}\,\text{m}^3/\text{mol} implies d_\text{N₂} \approx 0.31 nm — in good agreement (to about 15 per cent) with the mean-free-path value of 0.37 nm.

What this shows: the van der Waals constants are not just curve-fitting parameters. They encode real molecular information — molecular size through b, intermolecular attraction strength through a. Two different measurements of the molecular diameter of nitrogen, one from viscosity and one from the equation of state, agree to within a reasonable margin. This is the kind of convergence that tells you the theory is tracking something physical.

Common confusions

"The van der Waals equation is more accurate than the ideal gas law." In most regimes where both are applicable, yes. But the van der Waals equation is still approximate — real gases deviate from its predictions too, especially near the critical point itself, where more sophisticated equations (Peng–Robinson, Redlich–Kwong, Virial expansions with many coefficients) do better. Van der Waals is the first-order correction, not the last word.
"The a/V^2 term is the pressure from attractions pushing molecules together." No — the attraction pulls molecules away from the wall, reducing the pressure the wall feels. The +a/V^2 in the equation is added to the measured pressure P to recover what the kinetic pressure would have been without attractions. The attraction is real; it is the measured pressure (the one the gauge reads) that is depressed.
"b is the actual volume of the molecules." It is four times that volume, per mole. The factor of four comes from the pair-geometry of excluded volumes (Step 3 of the b-derivation). People sometimes sloppily say "b is the volume of the molecules"; strictly, b is four times the total molecular volume per mole.
"Below T_c, the van der Waals equation describes the liquid." Not exactly. Below T_c, the van der Waals isotherm wiggles through a region that it cannot actually represent — the liquid–vapour coexistence. The real gas follows the Maxwell-constructed plateau, not the wiggle. The wiggly branches have interpretations (metastable superheated liquid, metastable supercooled vapour), but the middle branch is mechanically unstable and unphysical. Using the equation to describe the liquid alone — outside the plateau region — is approximately okay, but the equation was not designed for that.
"Above T_c, the gas is forever a gas." Yes — but "above T_c" is the critical temperature of that specific gas. Propane's T_c is 370 K, so at 300 K propane can be liquefied (this is how LPG cylinders work). Nitrogen's T_c is 126 K; at 300 K, nitrogen cannot be liquefied by any amount of compression. You must cool it below 126 K first. This is why cryogenic air liquefaction is a genuine engineering challenge: you have to remove heat faster than compression can put it back in.
"The \sqrt{2} factor in mean free path comes from van der Waals." No — they are independent corrections. The \sqrt{2} in MFP is about relative vs absolute molecular velocity when you are counting collision rates. The factor of four in b is about pair excluded volume in a dense gas. Both are careful geometric arguments, but they live in different equations for different reasons.

Going deeper

If you came here to understand why the ideal gas law breaks down and to use the van der Waals equation on exam problems, you have what you need. What follows fleshes out the critical-point physics, the reduced-state law of corresponding states, the virial expansion, and the connection between the cohesive-energy parameter a and the heat of vaporisation — a quantity used in chemistry and in heat-engine design.

The law of corresponding states

Rewrite the van der Waals equation in reduced variables: P_r = P/P_c, V_r = V/V_c, T_r = T/T_c. Using equations (6)–(8) for the critical values,

\left(P + \frac{a}{V^2}\right)(V - b) = RT \quad\xrightarrow{\text{substitute}}\quad \left(P_r + \frac{3}{V_r^2}\right)\left(V_r - \tfrac{1}{3}\right) = \tfrac{8}{3} T_r.

Why: substitute P = P_r P_c = P_r \cdot a/(27 b^2), V = V_r V_c = 3 b V_r, and T = T_r T_c = T_r \cdot 8a/(27 R b). Every a, b, and R cancels out, leaving a pure-dimensionless equation. That is the point: the reduced equation has no free parameters — it is the same equation for every gas.

This is the law of corresponding states: in reduced variables, all real gases should obey the same equation of state (to the extent that the van der Waals equation itself is accurate). Experimentally, real gases do cluster onto a common curve when plotted in reduced variables — it is how engineers use generalised compressibility charts for real-gas calculations without needing a and b for every compound. The principle generalises to more accurate equations of state; the law of corresponding states is almost as useful as the equation that inspired it.

The Boyle temperature

At any given temperature, you can ask: at what pressure does PV = constant become most accurate? The answer comes from rewriting the van der Waals equation as a power series in 1/V:

P = \frac{RT}{V - b} - \frac{a}{V^2} = \frac{RT}{V}\left(1 - \frac{b}{V}\right)^{-1} - \frac{a}{V^2}.

Expand the first term for b \ll V: (1 - b/V)^{-1} \approx 1 + b/V + (b/V)^2 + \ldots

P V = RT + RT \cdot \frac{b}{V} - \frac{a}{V} + \mathcal{O}(1/V^2) = RT + \frac{RT b - a}{V} + \ldots

Why: expand using the geometric series. The 1/V coefficient is RT b - a. When this coefficient is zero, the leading deviation from ideality vanishes and PV \approx RT over a wide range of V.

The condition RT_B b - a = 0 defines the Boyle temperature:

T_B = \frac{a}{R b}.

For nitrogen, T_B = 0.137/(8.314\times 3.87\times 10^{-5}) \approx 426\,\text{K}. At this temperature, nitrogen obeys PV = RT unusually well over a broad pressure range — a useful fact for precision thermometry.

Notice the ratio: T_B / T_c = (a/Rb) / (8a/27 R b) = 27/8 = 3.375. Every van der Waals gas has its Boyle temperature at 27/8 times its critical temperature.

The heat of vaporisation

Here is where a earns its keep. The attractive term is, physically, the cohesive energy that binds the molecules of a liquid together. It takes energy to pull them apart into vapour — the latent heat of vaporisation L. Crudely, if each pair of molecules in a liquid of density n_\text{liq} has attractive energy \sim -\varepsilon and there are about six near neighbours, then the cohesive energy per mole is about \sim 3 N_A \varepsilon. The van der Waals a is the mean-field integral of this; a reasonable estimate is

L \sim \frac{a}{V_\text{liq,molar}} \sim \frac{a}{b}.

For nitrogen: a/b = 0.137 / 3.87\times 10^{-5} \approx 3540\,\text{J/mol}. Measured latent heat of vaporisation of N₂: 5580\,\text{J/mol}. The van der Waals estimate is low by a factor of about 1.5, but the order of magnitude is right and the trend is right: gases with larger a have larger latent heats, because more cohesive energy must be supplied to vaporise the liquid.

This is why a correlates with boiling point: both are set by the depth of the attractive well.

The virial expansion

For a more systematic treatment of deviations from ideality, you can expand PV/nRT as a power series in density:

\frac{PV}{nRT} = 1 + B(T) \frac{n}{V} + C(T) \left(\frac{n}{V}\right)^2 + \ldots

The coefficients B(T), C(T), \ldots are the virial coefficients. The second virial coefficient B(T) is the first correction to ideality and is closely related to the van der Waals constants:

B(T) = b - \frac{a}{RT}.

Why: expand the van der Waals equation in the density 1/V as we did above, collect the 1/V term, and identify the coefficient. At T = T_B = a/(Rb), B = 0 — the gas is ideal to first order in density. Above T_B, B > 0: the gas is less compressible than an ideal gas (repulsion dominates). Below T_B, B < 0: the gas is more compressible than ideal (attraction dominates).

Virial coefficients can be computed from first principles in statistical mechanics — B(T) is an integral of (e^{-U(r)/k_B T} - 1) over pair separation r. That connection closes the loop: the empirical van der Waals constants a and b are approximations to an exact statistical-mechanical object.

Why real gases deviate from Z_c = 3/8

The van der Waals prediction Z_c = P_c V_c / (R T_c) = 3/8 = 0.375 is, as noted, about 20–30 per cent high. Measured values:

Helium: Z_c = 0.302
Hydrogen: Z_c = 0.306
Nitrogen: Z_c = 0.290
CO₂: Z_c = 0.275
Water: Z_c = 0.229

The trend is that more-polar molecules (water, ammonia) deviate more from the van der Waals universal value. The reason is that the attractive part of the real intermolecular potential is not well-represented by a mean-field -a/V^2. Real attractions are soft, anisotropic (for non-spherical molecules), and sometimes directional (hydrogen bonding in water). A better equation of state — Peng–Robinson — replaces the -a/V^2 with a more flexible attractive term that captures the softness and recovers Z_c \approx 0.307, much closer to experiment.

The takeaway is that Z_c = 3/8 is a first universal law: every sufficiently simple gas obeys it to within 30 per cent. For engineering accuracy you graduate to more sophisticated equations. For understanding why gases liquefy and what determines their critical properties, the van der Waals equation is enough.

Where this leads next

Ideal Gas Laws — the PV = nRT limit that van der Waals corrects, and the three experimental laws (Boyle, Charles, Gay-Lussac) it generalises.
Thermodynamic Processes — Isothermal and Adiabatic — how a gas moves along an isotherm (the shape of which the van der Waals equation slightly modifies) versus an adiabat.
Kinetic Theory — Pressure and Temperature — the microscopic derivation of pressure that assumed point particles; the van der Waals equation is the first step past that assumption.
Mean Free Path — another place where molecular diameter d appears; comparing molecular diameters from MFP and from b tests whether the two arguments are consistent.
First Law of Thermodynamics — energy conservation for a real gas, where the internal energy U depends not only on T (as for an ideal gas) but also on V, through the attractive term a.