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JEE Advanced Chemistry States of Matter 2023 JEE Advanced 2023 (Paper 1)

JEE Advanced Chemistry Question (2023) — Solution

Question

A gas has a compressibility factor of 0 . 5 and a molar volume of 0 . 4   dm 3   mol - 1 at a temperature of 800   K and pressure x   atm . If it shows ideal gas behaviour at the same temperature and pressure, the molar volume will be   y   dm 3   mol - 1 . The value of   x / y is _________ [Use: Gas constant, R = 8 × 10 - 2   L   atm   K - 1   mol - 1 ]

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

Compressibility factor  ( Z )= V real   V ideal = 0 . 5 V real   = 0 . 4 dm 3   mol - 1 = 0 . 4   L / mol ∴    V ideal   = 0 . 4 0 . 5 = 0 . 8   L / mol ∴     y = 0 . 8   L / mol Using ideal gas equation : PV = nRT P = 1 × 8 × 10 - 2 × 800 0 . 8 x = 80   atm ∴    x y = 80 0 . 8 = 100  

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Related: Chemistry — States of Matter · All PYQ Banks