Gaseous & Liquid States

This topic couples the ideal-gas laws with real-gas behaviour (van der Waals), kinetic molecular theory, and liquid-state properties. Advanced favours deviations from ideality, critical constants, and Maxwell-distribution speed comparisons expressed numerically.

Key results

• Ideal gas: PV = nRT; Dalton's partial pressures P_total = ΣP_i; Graham's diffusion rate ∝ 1/√M.
• KMT speeds: u_rms = √(3RT/M) > u_avg = √(8RT/πM) > u_mp = √(2RT/M).
• Real gas (van der Waals): (P + an²/V²)(V − nb) = nRT; a corrects for attraction, b for molecular volume.
• Compressibility factor Z = PV/nRT: Z < 1 (attraction dominant), Z > 1 (repulsion/high P); Z = 1 for ideal.
• Critical constants: Tc = 8a/27Rb, Pc = a/27b², Vc = 3b.

Quantity	Relation
rms speed	√(3RT/M)
Graham's law	rate ∝ 1/√M
Compressibility	Z = PV/nRT
Critical temperature	8a/27Rb

Exam Tricks & Tips

• 🎯 Z < 1 signals dominant attractions (easier to compress); Z > 1 signals repulsions; near-ideal behaviour is at high T and low P.
• 🎯 The van der Waals a measures attraction, b measures size — a larger a means a gas liquefies more easily (higher Tc).
• 🎯 Graham's law (rate ∝ 1/√M) explains lighter gases diffusing faster — U-235 vs U-238 enrichment is a classic application.
• 🎯 rms : average : most-probable speeds are in the ratio √3 : √(8/π) : √2 — order is rms > avg > mp.
• 🎯 At the Boyle temperature a real gas behaves ideally over a range of pressures (a and b effects cancel).
• ❌ Common mistake: using STP molar volume 22.4 L for a real gas under non-ideal conditions — it applies to ideal gases at STP only.

Expected exam pattern

1 question, typically a real-gas/Z or KMT-speed numerical, or a conceptual on van der Waals constants and critical behaviour.

Quick recap

Ideal PV = nRT; real gas (P + an²/V²)(V − nb) = nRT. Z = PV/nRT flags attraction (<1) or repulsion (>1). Speeds rms > avg > mp; Graham rate ∝ 1/√M. Critical constants come from a and b.

Cover the answer, recall, then check. 11 cards on the gaseous state for JEE Advanced.

Q1. Van der Waals equation.
A1. (P + an²/V²)(V − nb) = nRT.

Q2. Meaning of the constants a and b.
A2. a corrects for intermolecular attraction; b corrects for finite molecular volume.

Q3. Compressibility factor and its ideal value.
A3. Z = PV/nRT; Z = 1 for an ideal gas.

Q4. What does Z < 1 indicate?
A4. Attractive forces dominate — the gas is more compressible than ideal.

Q5. Graham's law of diffusion.
A5. Rate of diffusion ∝ 1/√M (lighter gases diffuse faster).

Q6. Order of rms, average, and most-probable speeds.
A6. u_rms > u_avg > u_mp (ratio √3 : √(8/π) : √2).

Q7. Dalton's law of partial pressures.
A7. P_total = ΣP_i; each gas exerts pressure independently.

Q8. Critical temperature in van der Waals terms.
A8. Tc = 8a/27Rb.

Q9. Conditions under which a real gas approaches ideal behaviour.
A9. High temperature and low pressure.

Q10. What is the Boyle temperature?
A10. The temperature at which a real gas obeys the ideal-gas law over a range of pressures (a and b effects cancel).

Q11. Most-probable speed formula.
A11. u_mp = √(2RT/M).

Gases are the simplest state to model quantitatively, and their deviations from ideality reveal intermolecular forces. JEE Advanced tests the ideal gas law fluently, the kinetic-theory speed distribution, and the van der Waals corrections — especially the interpretation of the a and b constants.

Core concept: an ideal gas obeys PV = nRT because its molecules are point masses with no interactions; real gases deviate, corrected by the van der Waals equation.

Gaseous state

Beginner — the gas laws

PV = nRT combines Boyle (PV const at fixed T), Charles (V ∝ T at fixed P) and Avogadro (V ∝ n). Dalton's law: total pressure = sum of partial pressures, with partial pressure = mole fraction × total. Graham's law of effusion: rate ∝ 1/√M, so lighter gases effuse faster (rate₁/rate₂ = √(M₂/M₁)).

Intermediate — kinetic theory and molecular speeds

Kinetic theory gives PV = ⅓ mN v_rms². Three speeds: most probable v_mp = √(2RT/M), average v_avg = √(8RT/πM), root-mean-square v_rms = √(3RT/M), in ratio √2 : √(8/π) : √3 ≈ 1 : 1.128 : 1.224. Average KE per mole = (3/2)RT depends only on temperature, not identity — so at the same T, all gases have equal average KE but different speeds.

Advanced — real gases and van der Waals

Real gases deviate at high P and low T. The van der Waals equation: (P + an²/V²)(V − nb) = nRT. Here a corrects for intermolecular attraction (larger a → more easily liquefied, higher critical temperature) and b corrects for finite molecular volume. The compressibility factor Z = PV/nRT: Z < 1 means attraction dominates (real V smaller than ideal), Z > 1 means repulsion/finite-size dominates (high pressure). At the Boyle temperature a gas behaves ideally over a range. Critical constants relate to a, b: T_c = 8a/27Rb, P_c = a/27b².

Worked example

Two gases, He (M = 4) and CH₄ (M = 16), effuse through a pinhole. Compare their rates. By Graham's law, rate(He)/rate(CH₄) = √(M_CH₄/M_He) = √(16/4) = √4 = 2. Helium effuses twice as fast. If a mixture starts equimolar, the escaping gas is enriched in the lighter component — the principle behind isotope separation. Note the inverse-square-root: quadrupling the molar mass only halves the rate.

How JEE Advanced tests this

Partial-pressure and mole-fraction problems with reacting/effusing mixtures; Graham's-law effusion and enrichment; comparing v_mp, v_avg, v_rms and average KE across gases at the same/different temperatures; interpreting Z-vs-P curves and van der Waals a/b to rank gases by liquefiability; and critical-constant relations.

Exam tricks & shortcuts

• All gases at the same T have equal average KE = (3/2)RT.
• Speed ratio v_mp : v_avg : v_rms = √2 : √(8/π) : √3 (≈ 1 : 1.13 : 1.22).
• Larger van der Waals a ⇒ stronger attraction ⇒ higher T_c ⇒ easier to liquefy.
• Mnemonic: "a attracts, b is bulk."

Key formulas

• Ideal gas: PV = nRT; combined P₁V₁/T₁ = P₂V₂/T₂.
• Kinetic: v_rms = √(3RT/M); v_avg = √(8RT/πM); v_mp = √(2RT/M).
• Dalton: P_total = ΣPᵢ; partial pressure Pᵢ = xᵢ P_total.
• van der Waals (real gas): (P + an²/V²)(V − nb) = nRT.
• Compressibility factor: Z = PV/nRT (Z = 1 ideal).
• Graham's diffusion: rate ∝ 1/√M.

Real gases deviate from ideality (via a, b constants); Z measures the deviation from PV = nRT.

Worked Example

Problem: Calculate the pressure exerted by 2.0 mol of CO₂ confined to 5.0 L at 300 K using (a) the ideal gas equation and (b) the van der Waals equation, and comment on the difference. (a = 3.6 atm·L²·mol⁻², b = 0.043 L·mol⁻¹, R = 0.0821 L·atm·mol⁻¹·K⁻¹.)

Solution:
(a) Ideal gas: P = nRT/V = (2.0)(0.0821)(300)/5.0 = 49.26/5.0 = 9.85 atm.

(b) Van der Waals: [P + a n²/V²](V − nb) = nRT, so
P = nRT/(V − nb) − a n²/V².
V − nb = 5.0 − 2.0 × 0.043 = 5.0 − 0.086 = 4.914 L.
First term = 49.26/4.914 = 10.02 atm.
Correction term = a n²/V² = 3.6 × (2.0)²/(5.0)² = 3.6 × 4/25 = 0.576 atm.
P = 10.02 − 0.576 ≈ 9.45 atm.

The van der Waals pressure is lower than ideal because intermolecular attractions (the a-term) dominate over the finite-volume correction at this density.

Answer: (a) P_ideal ≈ 9.85 atm; (b) P_vdW ≈ 9.45 atm.