Which of the following is the correct expression for the equation of state of van der Waals gas? (a) (P+(a)/(n

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Q. Which of the following is the correct expression for the equation of state of van der Waals gas?

(a)

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

(b)

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

(c)

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

(d)

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

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Concept understanding — Van der Waals Equation of State

J. D. van der Waals corrected the ideal gas equation for the two assumptions kinetic theory makes that no real gas actually satisfies -- negligible molecular volume and zero intermolecular attraction -- by adding a pressure correction and a volume correction.

Pressure correction. A molecule about to strike the container wall is pulled back very slightly by the attraction of its neighbours (a molecule surrounded on all sides instead feels balanced, cancelling attraction), so it hits the wall a little less forcefully than it would with no attraction at all -- the measured pressure of a real gas is therefore somewhat lower than the true "ideal" pressure. This attractive effect scales with the square of the gas density, giving a correction term $\dfrac{an^2}{V^2}$ (where a is the first van der Waals constant, larger for more strongly attracting molecules), so $P_{ideal}=P+\dfrac{an^2}{V^2}$ .

Volume correction. Because real molecules occupy actual space and cannot overlap, the volume genuinely available for them to move in is less than the container volume V. Modelling molecules as hard spheres of radius r, the excluded volume around a colliding pair works out to $8V_m$ (eight times a single molecule's own volume $V_m$ ), giving an excluded volume per molecule of $4V_m$ ; for n moles this totals $nb$ , where $b=4V_m$ is the second van der Waals constant (larger for physically bigger molecules). The corrected, genuinely available volume is $V_{ideal}=V-nb$ .

The van der Waals equation. Substituting both corrections into $PV=nRT$ gives

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

Both a and b are gas-specific constants: a larger 'a' means stronger intermolecular attraction (and, in particular, a gas is more easily liquefied the larger its 'a' value is, since strong attraction is exactly what pulls molecules together into a liquid); a larger 'b' means physically bigger molecules. Since $P'=an^2/V^2$ has units of pressure, 'a' carries units of $\text{atm}\,\text{L}^2\text{mol}^{-2}$ (or $\text{L}^2\text{atm}\,\text{mol}^{-2}$ ), and since $nb$ has units of volume, 'b' carries units of $\text{L}\,\text{mol}^{-1}$ . The van der Waals equation is a substantial improvement over the plain ideal gas equation for describing real-gas behaviour, though it remains an approximate, not exact, description.

"Van der Waals equation of state derivation and constants a and b" and "van der Waals equation class 11 chemistry important questions" are frequent searches tied to the States of Matter chapter of the NCERT/CBSE Class 11 Chemistry curriculum, a topic tested regularly in JEE Main and NEET real-gas questions. Understanding why a larger 'a' value means easier liquefaction is exactly the kind of conceptual link competitive-exam MCQs like to test alongside the raw formula.

Step 1. The van der Waals equation adds a pressure-correction term $\dfrac{an^2}{V^2}$ (from $P'=a\rho^2=a(n/V)^2$ ) and a volume-correction term $nb$ to the ideal gas equation: $\left(P+\dfrac{an^2}{V^2}\right)(V-nb)=nRT$ .

Step 2. Check each option against this exact form. (a) has $\dfrac{a}{n^2V^2}$ -- the n is in the denominator instead of the numerator, which is dimensionally wrong (as $n\to$ larger amounts of gas, the correction should grow, not shrink). (b) has $\dfrac{na}{n^2V^2}=\dfrac{a}{nV^2}$ after simplifying -- also has n in the wrong place. (d) has $\dfrac{n^2a^2}{V^2}$ -- a is wrongly squared.

Step 3. Only (c), $\dfrac{an^2}{V^2}$ , matches the correct pressure-correction term.

✓Final answer

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

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

CBSE 2025Set ANNUAL3 marks
$Q. Can a gas with Vander Waals constant a = 0 be liquefied ? Explain.$
›Reveal solutionSolution

No — a gas with van der Waals constant a = 0 cannot be liquefied, because 'a' measures intermolecular attractive forces, and liquefaction fundamentally requires such attraction to hold molecules together in a condensed liquid phase.

The van der Waals equation of state for a real gas is:

(P + a n^2/V^2)(V - nb) = nRT

Here, the constant 'a' accounts for the intermolecular forces of attraction between gas molecules (the correction term a n^2/V^2 added to the measured pressure P represents the extra 'internal pressure' caused by molecules pulling on each other, which slightly reduces the pressure the molecules actually exert on the container walls). The constant 'b' accounts for the finite volume occupied by the molecules themselves (excluded volume).

Liquefaction of a gas is physically the process of overcoming the kinetic energy of the molecules (by lowering temperature and/or increasing pressure) so that intermolecular ATTRACTIVE forces can pull the molecules close together into the more ordered, closely-packed liquid state. If a = 0, it means the gas molecules exert NO attractive force on each other whatsoever — they behave as if they only ever collide elastically (like an ideal gas with only the finite-size correction b remaining). With no attractive force to draw the molecules together and hold them in a condensed phase, no amount of cooling or compression (short of the finite-size packing limit) can cause the molecules to condense into a liquid.

✓Final answer
$No, a gas with a = 0 cannot be liquefied — since 'a' represents the intermolecular attractive force, a = 0 means there is no attraction at all between molecules, and liquefaction is impossible without such attraction to hold the liquid together.$
CBSE 2018Set ANNUAL3 marks
$Q. Write the significances of Vanderwaal's constants.$
›Reveal solutionSolution
In PV = nRT corrected as (P + an^2/V^2)(V - nb) = nRT, 'a' quantifies intermolecular attraction and 'b' quantifies molecular size (excluded volume).

The van der Waals equation modifies the ideal gas equation to account for real-gas behaviour:

(P + an^2/V^2)(V - nb) = nRT

Significance of 'a' (pressure correction term): Real gas molecules attract one another, which reduces the pressure a gas exerts compared to an ideal gas (since molecules near the container wall are pulled inward by neighbouring molecules, striking the wall with less force). The constant 'a' measures the magnitude of these intermolecular attractive forces - a larger value of 'a' indicates stronger intermolecular attraction, meaning the gas deviates more from ideal behaviour, is easier to liquefy, and generally has a higher critical temperature.

Significance of 'b' (volume correction term), also called the excluded volume or co-volume: Real gas molecules have a finite size and cannot be compressed into zero volume (unlike the point-mass assumption of ideal gases). The constant 'b' accounts for the actual volume occupied by the gas molecules themselves (b is approximately four times the actual volume of the molecules), representing the volume that is NOT available for the free movement of other molecules. A larger 'b' indicates larger molecular size.

✓Final answer
$'a' is a measure of the intermolecular forces of attraction between gas molecules, and 'b' is a measure of the effective size (excluded volume) of the gas molecules.$

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Q.Which of the following is the correct expression for the equation of state of van der Waals gas?

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