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JEE Main 20149 AprilOnline
Exam
JEE Main
Year
2014
Date
9 April
Mode
Online
Subject
Chemistry
Chapter
States of Matter

JEE Main 2014 · 9 April · Online · Chemistry · States of Matter

Van der Waal's equation for a gas is stated as, P = nRT V - nb - a...

Van der Waal's equation for a gas is stated as,

P=nRTV−nb−a(nV)2\text{P} = \dfrac{\text{nRT}}{\text{V} - \text{nb}} - \text{a}{\left(\dfrac{\text{n}}{\text{V}}\right)}^{2}.

This equation reduces to the perfect gas equation, P=nRTV\text{P} = \dfrac{\text{nRT}}{\text{V}} When,

Options

  1. Atemperature is sufficiently high and pressure is low.
  2. Bboth temperature and pressure are very low.
  3. Cboth temperature and pressure are very high.
  4. Dtemperature is sufficiently low and pressure is high.

Correct answer

A. temperature is sufficiently high and pressure is low.

Step-by-step solution

At very high temperatures & low pressures, the characteristic constant behaviour can be understood from virial expansion

Z=PVmRT=1+(b−aRT)1Vm+b2Vm2+....\text{Z} = \dfrac{{\text{PV}}_{\text{m}}}{\text{RT}} = 1 + \left(\text{b} - \dfrac{\text{a}}{\text{RT}}\right) \dfrac{1}{{\text{V}}_{\text{m}}} + \dfrac{{\text{b}}^{2}}{{\text{V}}_{\text{m}}^{2}} + \text{....}

Vm{V}_{m}, molar volume large i.e b2Vm2→o\dfrac{{\text{b}}^{2}}{{\text{V}}_{\text{m}}^{2}} \to \text{o}

i.e Third term in virial expansion can be ignored.

In qualitative terms, the real gas will show ideal behaviour when the intermolecular forces of attraction will be negligible which is possible at high temperature and low pressure.
Also (b−aRT)→o\left(\text{b} - \dfrac{\text{a}}{\text{RT}}\right) \to \text{o}

i.e. Z≈1\text{Z} \approx 1 and gas behaves ideally.

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