Question:

Compressibility factor for $1$ mole of a van der Waals' gas at $0^{\circ} C$ and $100$ atmospheric pressure is found to be $0.5$ the volume of gas molecules is :

Updated On: Aug 15, 2022
  • 2.0224
  • 1.4666
  • 0.8542
  • 0.1119
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The Correct Option is D

Solution and Explanation

Compressibility factor $(Z)=\frac{P V}{n RT}$ For ideal gas $P V=n R T$ $\therefore Z=1$ For real gas $P V \neq n R T$ $\therefore Z \neq 1$ $0.5=\frac{100 \times V}{1 \times 0.82 \times 273}$ $V=0.1119\, L$
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Concepts Used:

States of Matter

The matter is made up of very tiny particles and these particles are so small that we cannot see them with naked eyes.

There are three States of Matter:

The three states of matter are as follows:

Solid State:

  • The solid-state is one of the fundamental states of matter.
  • Solids differ from liquids and gases by the characteristic of rigidity.
  • The molecules of solids are tightly packed because of strong intermolecular forces; they only oscillate about their mean positions.

Liquid State:

  • The molecules in a liquid are closely packed due to weak intermolecular forces.
  • These forces are weaker than solids but stronger than that of gases.
  • There is much space in between the molecules of liquids which makes their flowing ability easy.

Gaseous State:

  • In this state of matter, distances between the molecules are large (intermolecular distance is in the range of 10-7-10-5 cm.
  • The intermolecular forces experienced between them are negligible.
  • Thus, translatory, rotatory and vibratory motions are observed prominently in gases.