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JEE Main202313 Apr 2023Morning ShiftChemistryStates of MatterActual

A certain quantity of real gas occupies a volume of 0 . 15 dm 3 at 100 atm and 500 K when its compressibility factor is 1 . 07 . Its volume at 300 atm and 300 K (When its compressibility factor is 1 . 4 ) is _ _ _ _ _ _ × 10 - 4 dm 3 (Nearest integer)

Correct answer

0

Step-by-step solution

The compressibility factor (Z) is given as: Z   =   PV nRT n = PV ZRT Z 1 = 1 . 07 ,   P 1 = 100 atm ,   V 1 = 0 . 15 L ,   T 1 = 500 K Z 2 = 1 . 4 ,   P 1 = 300 atm ,   V 2 = ? ,   T 2 = 300 K Z 1 Z 2 = P 1 V 1 × T 2 T 1 × P 2 V 2 1 . 07 1 . 4 = 100 × 0 . 15 × 300 500 × 300 × V 2 V 2 = 0 . 03925   dm 3

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