Q21 / 90 · JEE Main 2023 (13 Apr Shift 1) · Chemistry

A certain quantity of real gas occupies a volume of 0.15 dm3 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 ______×10-4dm3(Nearest integer)

Correct answer: 392

Solution

The compressibility factor (Z) is given as:

Z = PVnRT

n=PVZRT

Z1=1.07, P1=100atm, V1=0.15L, T1=500KZ2=1.4, P1=300atm, V2=?, T2=300K

Z1Z2=P1V1×T2T1×P2V2

1.071.4=100×0.15×300500×300×V2

V2=0.03925 dm3

Hence, the correct answer is 392.

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