VKC Notes · Gaseous state · Civil Services P-I · IFoS P-I · Geo-Scientist P-II

4.1 Real Gases: van der Waals Equation, Critical Constants, Boyle Temperature and Corresponding States

★★★ 2021–2025: 28 Mains parts, 215 marks (Civil Services 5 parts / 50 m · IFoS 4 / 40 m · Geo-Scientist 19 / 125 m) and 5 Geo-Scientist Prelims questions. Geo-Scientist Paper II opens with it every year (Q1(a), Q1(b) and often Q2): the surest marks of that paper.

Syllabus lines. Civil Services: “Equation of state for real gases, intermolecular interactions and critical phenomena and liquefaction of gases”. IFoS: “Equation of state for real gases, intermolecular interactions, liquefaction of gases and critical phenomena”. Geo-Scientist: “Real gases, deviation of gases from ideal behaviour; compressibility factor; van der Waals equation of state and its characteristic features. Existence of critical state. Critical constants in terms of van der Waals constants. Law of corresponding states and significance of second virial coefficient. Boyle temperature.”

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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes

4.1.1 Compressibility factor and deviations ★★★

(4.1.1)

A perfect gas has point molecules and no forces between them: Z = 1 at every p and T. A real gas has (i) a finite size, so repulsion at short range makes it harder to compress (Z > 1), and (ii) attraction at longer range, which pulls molecules together (Z < 1) Geo 2021 Q1(b) 5m Geo Pre 2025 Q49.SN
Every Z–p curve starts at Z = 1 as p → 0 (molecules far apart). At moderate p attraction dominates and most gases dip below 1; at high p repulsion dominates and all curves rise with nearly the same slope. The deeper the dip, the larger a and the easier the gas liquefies: CO₂ > CH₄ > N₂ > H₂ Geo Pre 2023 Q45.SN

Ideal behaviour needs both corrections to vanish: low pressure (large V_m: V_m ≫ b and a/V_m² → 0) and high temperature (kinetic energy ≫ attraction), best near the Boyle temperature Geo 2021 Q1(a) 5m.SN
H₂ and He are “exceptional”: two electrons, very low polarisability, so a is tiny (He 0.0346, H₂ 0.248 bar L² mol⁻²). At room temperature b > a/RT, the initial slope is positive and Z > 1 at every pressure. Room temperature is far above their Boyle temperatures (He ≈ 23 K, H₂ ≈ 110 K); cooled below T_B they too show the dip Geo 2025 Q1(a) 5m Geo 2023 Q1(b) 5m.SN

Fig. 4.1.1 Z against p for four gases, computed from the van der Waals equation with their own a and b.

Draw it in 60 s: horizontal dashed line Z = 1; H₂ a straight line rising from it; N₂ a shallow dip; CO₂ a deep dip near 100 bar; all rising together at the right.

Solved PYQ Geo 2025 Q2 10m

A gas at 350 K and 12 atm has a molar volume 12% larger than the perfect-gas value. (a) Z? Which forces dominate? (b) V_m?

Z = V_m/V_m(perfect) = 1.12: Z > 1, the gas is harder to compress than a perfect gas, so repulsive forces (molecular size) dominate.SN
V_m(perfect) = RT/p = (0.08206 × 350)/12 = 2.393 L mol⁻¹; V_m = 1.12 × 2.393 = 2.68 L mol⁻¹.

4.1.2 van der Waals equation: a and b ★★★

(4.1.2)

b, the excluded volume (co-volume): the space per mole that other molecules cannot enter; the free volume is V − nb. a, attraction: a molecule striking the wall is pulled back by the bulk, so the observed pressure is lower than ideal by an²/V² (internal pressure; both the number of molecules hitting and the pull on each ∝ n/V) Geo 2023 Q1(a) 5m.SN
Units: a in bar L² mol⁻² (or Pa m⁶ mol⁻²), b in L mol⁻¹ (m³ mol⁻¹). Larger a → stronger attraction → higher T_c, easier liquefaction Geo Pre 2024 Q44; larger b → larger molecule.SN

Fig. 4.1.4 Why b is four times the volume of the molecules.

Draw it in 60 s: two touching circles of radius r; a dashed circle of radius 2r round the first one; write 8v per pair, 4v per molecule.

(4.1.3)

Excluded volume = 4 × own volume: two hard spheres of radius r cannot bring their centres closer than 2r; round one centre a sphere of radius 2r, volume 8v, is closed to the other centre. This exclusion belongs to the pair, so per molecule it is 4v, and per mole b = 4N_Av IFoS 2022 Q4(a) 10m Geo 2022 Q2(b) 5m.SN
He, b = 23.70 cm³ mol⁻¹: d³ = 3b/(2πN_A) = 1.879 × 10⁻²⁹ m³, d = 266 pm (radius 133 pm; van der Waals radius of He 140 pm). Using b = N_A(4/3)πr³ by mistake gives a diameter of 422 pm Geo 2021 Q10(b) 5m.
CO₂, 7.0 mol in 10.4 L at 47 °C (a = 3.59 atm L² mol⁻², b = 0.0427 L mol⁻¹): p_ideal = nRT/V = 17.68 atm; p_vdW = 18.21 − 1.63 = 16.58 atm; difference 1.10 atm. The repulsion term alone would raise p by 0.53 atm; attraction lowers it by 1.63 atm IFoS 2023 Q3(c) 10m.

Solved PYQ CSE 2023 Q4(a) 20m

V_m = 5.0 × 10⁻⁴ m³ mol⁻¹ at 273 K and 3.0 × 10⁶ Pa; a = 0.50 m⁶ Pa mol⁻². Find b, Z, and comment on the interactions.

a/V_m² = 0.50/(2.5 × 10⁻⁷) = 2.0 × 10⁶ Pa, so p + a/V_m² = 5.0 × 10⁶ Pa.
V_m − b = RT/(p + a/V_m²) = 2269.7/5.0 × 10⁶ = 4.54 × 10⁻⁴; b = 4.6 × 10⁻⁵ m³ mol⁻¹.
Z = pV_m/RT = 1500/2269.7 = 0.66. By Eq. 4.1.3, Z = 1.10 − 0.44: size alone would make Z = 1.10, attraction pulls it down by 0.44.
Z < 1, so attraction dominates; B = b − a/RT = −1.7 × 10⁻⁴ m³ mol⁻¹ < 0 agrees. The gas is far below its T_B (≈ 1300 K) and near its T_c (≈ 390 K).SN

4.1.3 Virial equation and Boyle temperature ★★★

(4.1.4)

Virial equation (Kamerlingh Onnes): a power series; B, C… depend only on the gas and T. B comes from interactions of pairs of molecules, C from triples. B = initial slope of Z against 1/V_m; B′ = (∂Z/∂p)_T as p → 0 Geo 2024 Q1(b) 5m.SN

Multiply Eq. 4.1.2 by V_m/RT: Z = (1 − b/V_m)⁻¹ − a/RTV_m.
Expand (1 − x)⁻¹ = 1 + x + x² + … with x = b/V_m ≪ 1: Z = 1 + (b − a/RT)/V_m + b²/V_m² + …
Compare term by term Geo 2021 Q2 10m:SN

(4.1.5)

Sign of B: low T, attraction wins, B < 0 and Z dips below 1; high T, B > 0. At one temperature B = 0: the gas obeys Boyle’s law (pV_m constant) over a range of low pressures. That is the Boyle temperature Geo 2024 Q2 10m Geo 2021 Q11(a) 10m.SN

(4.1.6)

Fig. 4.1.2 Z of N₂ (van der Waals) at four temperatures: the dip disappears at T_B = a/Rb.

Draw it in 60 s: four curves from the point (0, 1): one dipping deeply (low T), one dipping slightly, one leaving flat (T_B), one rising straight away (high T).

O₂ (a = 1.36 dm⁶ atm mol⁻², b = 0.0318 dm³ mol⁻¹): T_B = 1.36/(0.08206 × 0.0318) = 521 K; measured 406 K, so the model overestimates by about 28% Geo 2022 Q1(b) 5m. H₂ (a = 22.0 kPa dm⁶ mol⁻², b = 0.024 dm³ mol⁻¹, R = 8.3): 110 K = −163 °C Geo Pre 2023 Q43.SN
Order Ar < CH₄ < C₆H₆: a rises much faster (1.34, 2.27, 18.6 L² atm mol⁻²) than b (0.032, 0.043, 0.119 L mol⁻¹), so a/b and T_B rise: about 509, 643, 1897 K from 27T_c/8 CSE 2024 Q3(a)(ii) 5m.
Significance: above T_B a gas (H₂, He at room temperature) has Z > 1 at all p; below it (N₂ at 300 K is near; NH₃ far below) Z first dips. NH₃ (T_B ≈ 995 K with van der Waals constants) is strongly non-ideal at room temperature Geo 2021 Q11(a).
Largest initial slope: slope = b/RT − a/R²T². Setting d/dT = 0 gives −b/RT² + 2a/R²T³ = 0, so T = 2a/Rb = 2T_B and the maximum slope = b²/4a IFoS 2021 Q3(a) 10m.SN
For a gas with p(V_m − b) = RT (no attraction): Z = 1 + bp/RT > 1 and ln(f/p) = bp/RT, so fugacity f > p: repulsion raises the escaping tendency Geo 2025 Q1(f) 5m. Joule–Thomson inversion temperature Geo Pre 2025 Q56: Eq. 4.1.6.

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