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JEE Main 202313 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
13 April
Shift
Morning Shift
Subject
Chemistry
Chapter
States of Matter

JEE Main 2023 · 13 April · Morning Shift · Chemistry · States of Matter

A certain quantity of real gas occupies a volume of 0 . 15 dm ^ 3...

A certain quantity of real gas occupies a volume of 0.15dm30 . 15{dm}^{3} at 100atm100atm and 500K500K when its compressibility factor is 1.07.1 . 07 . Its volume at 300atm300atm and 300K300K (When its compressibility factor is 1.41 . 4) is ______ \times {10}^{- 4}{dm}^{3}(Nearest integer)

Correct answer

0

Step-by-step solution

The compressibility factor (Z) is given as:

Z=PVnRTZ = \dfrac{PV}{nRT}

n=PVZRTn = \dfrac{PV}{ZRT}

Z1=1.07,P1=100atm,V1=0.15L,T1=500KZ2=1.4,P1=300atm,V2=?,T2=300K{Z}_{1} = 1 . 07, {P}_{1} = 100atm, {V}_{1} = 0 . 15L, {T}_{1} = 500K{Z}_{2} = 1 . 4, {P}_{1} = 300atm, {V}_{2} = ? , {T}_{2} = 300K

Z1Z2=P1V1×T2T1×P2V2\dfrac{{Z}_{1}}{{Z}_{2}} = \dfrac{{P}_{1}{V}_{1} \times {T}_{2}}{{T}_{1} \times {P}_{2}{V}_{2}}

1.071.4=100×0.15×300500×300×V2\dfrac{1 . 07}{1 . 4} = \dfrac{100 \times 0 . 15 \times 300}{500 \times 300 \times {V}_{2}}

V2=0.03925dm3{V}_{2} = 0 . 03925{dm}^{3}

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