Vander Waals equation in different forms

(i) At low pressures: ‘ V ’ is large and ‘ b ’ is negligible in comparison with V.

Deviation of gases from ideal behaviour with pressure , t he Vander Waals equation reduces to:

$\large (P + \frac{a}{V^2}) V = R T$

$\large (P V + \frac{a}{V}) = R T$

$\large P V = R T – \frac{a}{V}$

$\large P V < R T $

This accounts for the dip in PV vs P isotherm at low pressures

(ii) At fairly high pressures
The plot of Z vs P for N2 gas at different temperature is shown here.

a/V2 may be neglected in comparison with P.

The Vander Waals equation becomes

P (V – b) = RT

PV – Pb = RT

PV = RT + Pb or PV > RT

This accounts for the rising parts of the PV vs P isotherm at high pressures

(iii) At very low pressures: V becomes so large that both b and become negligible and the Vander Waals equation reduces to PV = RT. This shows why gases approach ideal behaviour at very low pressures.

(iv) Hydrogen and Helium: These are two lightest gases known. Their molecules have very small masses. The attractive forces between such molecules will be extensively small. So a/V2 is negligible even at ordinary temperatures.

Thus PV > RT

Thus Vander Waals equation explains quantitatively the observed behaviour of real gases and so is an improvement over the ideal gas equation.

Vander Waals equation accounts for the behaviour of real gases. At low pressures, the gas equation can be written as,

$\large (P + \frac{a}{V_m^2}) V_m = R T$

$\large Z = \frac{P V_m}{R T}$

Where Z is known as compressibility factor.

Its value at low pressure is less than 1 and it decreases with increase of P. For a given value of Vm  , Z has more value at higher temperature

At high pressures, the gas equation can be written as

P (Vm – b) = RT

Z = P Vm /RT = 1 + Pb/RT

Here, the compressibility factor increases with increase of pressure at constant temperature and it decreases with increase of temperature at constant pressure.

For the gases H2 and He, the above behaviour is observed even at low pressures, since for these gases, the value of ‘a’ is extremely small.

Illustration : One litre of a gas at 300 atm and 473 K is compressed to a pressure of 600 atm and 273 K. The compressibility factors found are 1.072 & 1.375 respectively at initial and final states. Calculate the final volume.

Solution:

P1V1 = Z1nRT1 and P2V2 = Z2nRT2

$\large \frac{P_2 V_2 }{T_2} \times \frac{T_1}{P_1 V_1} = \frac{Z_2}{Z_1}$

$\large V_2 = \frac{Z_2}{Z_1} \frac{T_2}{T_1} \frac{P_1 V_1}{P_2}$

$\large V_2 = \frac{1.375}{1.072} \frac{273}{473} \frac{300 \times 1}{600}$

= 370.1 ml

Also Read:

The Gas Laws
Combined Gas Law
The Kinetics Theory of Gases
Graham’s Law of Effusion & Diffusion
Imperfect or Real Gases
Real Gases and the van der Waals Equation
Mean Free Path
Gas Eudiometry

Next Page → 

← Back Page

Leave a Reply