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JEE Main 20146 April
Exam
JEE Main
Year
2014
Date
6 April
Subject
Chemistry
Chapter
States of Matter

JEE Main 2014 · 6 April · Chemistry · States of Matter

If Z is the compressibility factor, then Van der Waal's equation at...

If Z is the compressibility factor, then Van der Waal's equation at low pressure can be written as:If Z is the compressibility factor, then Van der Waal's equation at low pressure can be written as:

Options

  1. AZ=1+RTPb\text{Z} = 1 + \dfrac{\text{RT}}{\text{Pb}}
  2. BZ=1−aVRT\text{Z} = 1 - \dfrac{\text{a}}{\text{VRT}}
  3. CZ=1−PbRT\text{Z} = 1 - \dfrac{\text{Pb}}{\text{RT}}
  4. DZ=1+PbRT\text{Z} = 1 + \dfrac{\text{Pb}}{\text{RT}}

Correct answer

B. Z=1−aVRT\text{Z} = 1 - \dfrac{\text{a}}{\text{VRT}}

Step-by-step solution

Van der Waal's Equation of state for one mole

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

At low pressure, the characteristic constant (a) becomes significant whereas (b) becomes negligible. So (b) is ignored.

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

PV+aV=RT\text{PV} + \dfrac{\text{a}}{\text{V}} = \text{RT}

PVRT=RTRT−aVRT=1−aVRT\dfrac{\text{PV}}{\text{RT}} = \dfrac{\text{RT}}{\text{RT}} - \dfrac{\text{a}}{\text{VRT}} = 1 - \dfrac{\text{a}}{\text{VRT}}

At low pressure for most gases z < 1.

If Z is the compressibility factor, then Van der Waal's equation at low pressure can be written as:
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