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In a Van der Wall's equation, unit of 'a' is-
1. dyne × cm4
2. dyne × cm2
3. dyne × cm3
4. dyne × cm

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Correct Answer - Option 1 : dyne × cm4

Concept:

Real Gas: It will obey the gas rule only at high temperature and low pressure.

  • The molecules are not having a negligible volume.
  • They can be repulsive and attractive between the molecules.
  • It obeys Van der Waals equation: \(\left( {P + \frac{a}{{{V^2}}}} \right)\left( {v - b} \right) = RT\)

Where P = pressure, V = volume, R = universal gas constant and T = temperarure.

Explanation:

  • Van der Waals equation: 

\(⇒ \left( {P + \frac{a}{{{V^2}}}} \right)\left( {v - b} \right) = RT\)

As pressure can be added only to pressure, therefore, a/v2 represents pressure P i.e.,

\(⇒ \frac{a}{V^2}=P\) or,

⇒ a = PV2

\(\implies a = \frac{Force}{Area} ×( Volume)^2\) (Pressure is Force by area)

Now, the Cgs unit of Force is dyne, of volume is cm 3 and area is cm 2 .

So, the value of a is

\(\implies unit \ of \ a = \frac{dyne}{cm^2} ×(cm^3)^2\)

\(\implies unit \ of \ a = \frac{dyne}{cm^2} ×(cm^6)\)

⇒ unit of a = Dyne Cm (6 - 2) = Dyne × cm4

So, the correct option is dyne × cm4

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