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Vander waals equations






Deviation from Ideal Gas Behaviour

Real Gases: -
  • A gas which do not follow ideal gas behaviour under all conditions of temp. and pressure is called real gas.
  • Deviation from ideal behaviour with respect to pressure can be studied by plotting pressure versus volume curve at constant temp. (Boyle’s law)
 
Compressibility factor (z): -
$z =\frac {PV}{nRT}$
For ideal gases z = 1 and for real gases z ≠ 1.
If z > 1 then real gases show +ve deviation from ideal behaviour and if z < 1 the gas show –ve deviation from ideal behaviour
 
compressibility factor chart

Vander waals Equation -

Reason for Deviation: -
  1. There is no force of attraction between the molecules of gas but it is not true as gaseous particles have force of attraction present between them. So, correction is made in pressure  $P + \frac {a}{V^2}$
  2. Volume occupied by the gas molecules is negligible as compound to total volume of gas but the volume occupied by the gas doesn’t has negligible value. So, the correction in Volume is V – b
    Where, ‘a’ and ‘b’ are Vander Wall gas constant.
From above two points Vander Wall derived the equation as
 $(P +\frac {a}{V^2})(V-b) =nRT$
So, for ‘n’ moles of gas
$ (P +\frac {an^2}{V^2})(V-nb) =nRT$
The Vander Waals constant (a) Signify the magnitude of intermolecular forces of attraction between the gas particles while (b) signifies the effective size of gas molecules.
VANDER WAAL’S EQUATION
Important Note
1. Real Gases shows ideal gas behaviour when pressure and temperature is such that intermolecular forces are negligible.  Real gases show ideal gas behaviours when the pressure approaches zero. At low pressure, all gases have Z=1
2. Gases show ideal gas behaviour when volume occupied is so large that volume occupied by the molecules can be neglected. i.e. when pressure is low
3. Real gases show ideal gas behaviour at low pressure and high temperature
 

Example

Question 1
A 39 dm3 cylinder contains 212 gm oxygen gas at 21o C. What mass of oxygen will be released to reduce the pressure in the cylinder to 1.24 bar. (r = 0.0821 dm3 bar K-1 mol-1
Answer
Volume = 34 dm3
Temperature = 21o C = 273 + 21 K = 294
Pressure = 1.24bar
P V =nRT
1.24×34=n×.0821×294
n=1.7 mole
Now, $Moles =\frac {Weight }{Molar Mass}$       
 1.7 =w/32
 w=545.4 gram
Weight of air released:
=212 – 54.4 gm= 157.6 gm
Question 2
What is the SI unit of expression $\frac {PV^2T^2}{N}$
Answer
 $\frac {PV^2T^2}{N} =Nm^-2(m^3)^2K^2(mol)^-1=Nm^4K^2(mol)^-1$
 
 




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