Verified Solution • 2025–26 Session moderate

QJEEM22Chdbdb: JEE Main 2022 Chemistry - Class 11 JEEM22Chdbdb Solution: For a real gas at temperature and high pressure (99 bar) the…

For a real gas at 25∘C25^{\circ} \mathrm{C} temperature and high pressure (99 bar) the value of compressibility factor is 2, so the value of Vander Waal's constant 'b' should be __________ ×10−2 L mol−1\times 10^{-2} \mathrm{~L} \mathrm{~mol}^{-1} (Nearest integer)

(Given R=0.083 L\mathrm{R}=0.083 \mathrm{~L} bar K−1 mol−1\mathrm{K}^{-1} \mathrm{~mol}^{-1} )

Step-by-Step Verified Solution

For 1 mole at high pressure

P(V−b)=RTP(V-b)=R T

PV−Pb=RT\mathrm{PV}-\mathrm{Pb}=\mathrm{RT}

PVRT=1+PbRT\frac{\mathrm{PV}}{\mathrm{RT}}=1+\frac{\mathrm{Pb}}{\mathrm{RT}}

Z=1+PbRTZ=1+\frac{\mathrm{Pb}}{\mathrm{RT}}

1=99( b)0.083×2981=\frac{99(\mathrm{~b})}{0.083 \times 298}

b=0.083×29899≃0.249≃25×10−2b=\frac{0.083 \times 298}{99} \simeq 0.249 \simeq 25 \times 10^{-2}

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