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ExamsJEE AdvancedChemistry

Match the properties in Column-I with the correct descriptions in Column-II for a van der Waals gas. Column-I: (P) Free volume (Q) Critical temperature (R) Boyle's temperature (S) Compressibility factor in the high-pressure region. Column-II: (1) Depends on the nature of the gas (2) Constant for a particular gas (3) Depends on pressure for a particular gas (4) Depends on the radius of the gas molecule.

  1. P -> 1,3,4; Q -> 1,2,4; R -> 1,2,4; S -> 1,3,4
  2. P -> 1,3; Q -> 1,2; R -> 3; S -> 2,4
  3. P -> 4,2; Q -> 4; R -> 3,4; S -> 4,2
  4. P -> 3,4; Q -> 2,1; R -> 3,2; S -> 3,1

Correct answer: P -> 1,3,4; Q -> 1,2,4; R -> 1,2,4; S -> 1,3,4

Solution

P (Free volume): V_f = V - b. Here b depends on molecular radius (4) and is specific to the gas (1); V_f changes with volume/pressure (3). So P -> 1, 3, 4. Q (Critical temperature): T_c = 8a/(27Rb); a and b are gas-specific, so T_c depends on gas nature (1), is constant for a given gas (2), and depends on molecular size via b (4). So Q -> 1, 2, 4. R (Boyle temperature): T_B = a/(Rb); same reasoning: R -> 1, 2, 4. S (Compressibility factor at high P): Z ≈ 1 + Pb/(RT); for a given gas (fixed b and T), Z depends on P (3) and on gas nature/radius (1 and 4 via b). So S -> 1, 3, 4.

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