Explain the physical significance of van der Waals parameters.

Exercises · 5.23

Q. Explain the physical significance of van der Waals parameters.

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Concept understanding — Van der Waals Equation

Why the Ideal Gas Law Fails

The ideal gas equation $PV = nRT$ assumes molecules are point particles with no interactions — they bounce off walls perfectly and never attract or repel each other. That works fine at low pressures and high temperatures, where molecules are far apart and moving fast. But crank up the pressure or lower the temperature, and reality bites.

Two things break the ideal picture:

Molecules have finite size. In the ideal model, the entire container volume $V$ is available for motion. But real molecules occupy some space themselves. The volume actually free for them to move in is less than $V$ .
Molecules attract each other. At close range, intermolecular forces (van der Waals forces) pull molecules together. This reduces the force with which they hit the walls — the measured pressure is lower than what an ideal gas would exert.

The van der Waals equation corrects both problems in a simple, physically motivated way.

The Corrections, Step by Step

Volume correction — the $b$ term

If each mole of gas molecules occupies a volume $b$ that is excluded (the molecules themselves take up space), then for $n$ moles the free volume is not $V$ but $V - nb$ . So we replace $V$ with $V - nb$ :

P_{\text{ideal}} (V - nb) = nRT

Here $b$ is called the co-volume or excluded volume per mole. It is roughly four times the actual volume of one mole of molecules.

Pressure correction — the $a$ term

Attractive forces between molecules reduce the pressure. A molecule about to hit the wall is pulled back by neighbours. The reduction in pressure is proportional to:

the number of molecules near the wall (density $n/V$ ), and
the number of molecules pulling them back (also density $n/V$ ).

So the pressure drop is proportional to $(n/V)^2$ . We add a correction term $a n^2 / V^2$ to the measured pressure $P$ to recover the pressure an ideal gas would exert:

P_{\text{ideal}} = P + \frac{a n^2}{V^2}

The constant $a$ measures the strength of intermolecular attraction — larger $a$ means stronger attraction.

The Full Equation

Putting both corrections together:

\left( P + \frac{a n^2}{V^2} \right) (V - nb) = nRT

This is the van der Waals equation for $n$ moles of a real gas.

For one mole ( $n = 1$ ), it simplifies to:

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

What $a$ and $b$ Mean Physically

Constant	What it corrects	Typical values	Units
$a$	Intermolecular attraction	Larger for polar molecules (e.g., water: 5.46 L²·atm/mol²)	Pressure × (volume/mol)²
$b$	Finite molecular volume	Larger for bigger molecules (e.g., hexane: 0.174 L/mol)	Volume/mol

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Step 1 – Recall the van der Waals equation

\left(p + \frac{an^2}{V^2}\right)(V - nb) = nRT

This equation modifies the ideal gas law with two correction terms, one for pressure and one for volume, to account for real-gas (non-ideal) behaviour.

Step 2 – Significance of the parameter $a$ (pressure correction)

Real gas molecules attract each other (intermolecular attractive forces), so a molecule about to strike the container wall is pulled slightly backward by neighbouring molecules. This makes the OBSERVED (measured) pressure LOWER than the pressure an ideal, non-interacting gas would exert. The term $an^2/V^2$ is added back to the observed pressure to correct for this:

p_{\text{ideal}} = p_{\text{observed}} + \frac{an^2}{V^2}

$a$ is therefore a measure of the MAGNITUDE OF INTERMOLECULAR ATTRACTIVE FORCES — a LARGER value of $a$ means stronger attractions between the gas's molecules (and correlates with a higher critical temperature and easier liquefaction).

Step 3 – Significance of the parameter $b$ (volume correction)

Real gas molecules have a finite size and occupy actual physical space (they are not point masses, as the ideal gas model assumes), so the volume truly available for the molecules to move around in is LESS than the container's total volume $V$ . The term $nb$ subtracts this excluded/unavailable volume: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQ
$\left( P + \frac{a}{4V^2} \right) = \frac{RT}{(V-b)}$
›Reveal solutionSolution
The van der Waals equation must be written for a specific number of moles $n$ . Substituting $n = 0.5$ into the general form $\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$ and simplifying leads to option (B).

The van der Waals equation corrects the ideal gas law for two real-gas effects:

Attractive forces between molecules reduce the pressure, so we add a term $\frac{an^2}{V^2}$ to the measured pressure $P$ .

Finite molecular volume reduces the available volume, so we subtract $nb$ from the total volume $V$ .

The general form for $n$ moles is:
$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$
Here, $a$ and $b$ are constants specific to the gas. The key is to plug in $n = 0.5$ and simplify carefully — many students forget to square $n$ in the pressure correction term.

Start with the general van der Waals equation for $n$ moles:

$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$

Substitute $n = 0.5$ into both the pressure correction and the volume correction:

For the pressure term: $n^2 = (0.5)^2 = 0.25 = \frac{1}{4}$ , so

$\frac{an^2}{V^2} = \frac{a \cdot \frac{1}{4}}{V^2} = \frac{a}{4V^2}$

For the volume term: $nb = 0.5\,b$ , so

$V - nb = V - 0.5b$

On the right-hand side: $nRT = 0.5\,RT$

Write the equation with these substitutions: $\left(P + \frac{a}{4V^2}\right)(V - 0.5b) = 0.5\,RT$ …
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TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQ
$N_2$
›Reveal solutionSolution
The van der Waals constant a (measure of intermolecular attraction) is smallest for the smallest, least polarisable molecule (N₂), while b (measure of molecular size) is largest for the bulkiest molecule (CCl₄). Hence the correct pair is N₂ and CCl₄ — option (C).

The van der Waals equation corrects the ideal gas law for real gases using two constants:

a accounts for intermolecular attractive forces (larger a → stronger attraction).

b accounts for the volume occupied by the molecules themselves (larger b → bigger molecules).

So the question reduces to: Which gas has the weakest attractions (smallest a) and which has the largest molecular size (largest b)?

Identify the molecules and their key properties

N₂: small, nonpolar diatomic; only weak London dispersion forces.

Cl₂: diatomic, larger than N₂, more electrons → stronger dispersion forces.

CH₂Cl₂: polar molecule (dichloromethane); dipole-dipole forces add to dispersion.

CCl₄: large, tetrahedral, nonpolar but with many electrons → very strong dispersion forces.

Rank by intermolecular attraction (constant a)

Attraction strength generally increases with molecular size and polarity.

N₂ is the smallest and nonpolar → weakest attraction → smallest a.

Cl₂ is larger than N₂ → larger a.

CH₂Cl₂ is polar → a larger still.

CCl₄ is large and highly polarisable → largest a (but we only need the smallest a here).

So smallest a = N₂.

Rank by molecular size (constant b)

Constant b is roughly proportional to the volume of one mole of molecules (their “excluded volume”).

N₂ is tiny → smallest b.

Cl₂ is larger → bigger b. …
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Q.Explain the physical significance of van der Waals parameters.

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