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JEE Main 20169 AprilOnline
Exam
JEE Main
Year
2016
Date
9 April
Mode
Online
Subject
Chemistry
Chapter
States of Matter

JEE Main 2016 · 9 April · Online · Chemistry · States of Matter

At very high pressures, the compressibility factor of one mole of a...

At very high pressures, the compressibility factor of one mole of a gas is given by:

Options

  1. A1+PbRT1 + \dfrac{Pb}{RT}
  2. BPbRT\dfrac{Pb}{RT}
  3. C1−PbRT1 - \dfrac{Pb}{RT}
  4. D1 - \dfrac{b}{\left(VRT\right)}

Correct answer

A. 1+PbRT1 + \dfrac{Pb}{RT}

Step-by-step solution

(P+aV2)(V−b)=RT\left(P + \dfrac{a}{{V}^{2}}\right)\left(V - b\right) = RT

at very high pressure P≫aV2P ≫ \dfrac{a}{{V}^{2}}

P(V−b)=RTP\left(V - b\right) = RT

PV−Pb=RTPV - Pb = RT

∴Z=1+PbRT∴ Z = 1 + \dfrac{Pb}{RT} compressibility factor
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