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Nov 2, 2018

Van der Waals Equation

Van der Waals Equation:

In 1873, Van der Waals modified the Ideal Gas Equation, 
PiVi = RT
By incorporating the size effect and inter molecular attraction effect of the real gases. These above two effects are discussing under the Volume Correction and Pressure Correction of the ideal gas equation.

Volume Correction:

Molecules are assumed to be a hard rigid spheres and in a real gas, the available space for free movement of the molecules becomes less then V
let us take the available space for free movement of 1 mole gas molecules ,
Vi = (V-b)
where V is the molar volume of the gas and b is the volume correction factor.
Vi = molar volume of the ideal gas where the gas molecules are regarded as point masses.
Let us take r is the radius of the molecule and σ = 2r is the diameter using the molecule as a rigid sphere.
Volume Correction
Representation of Gas Molecules
Then it can be shown that the volume correction term or effective volume of one mole gas molecules,
= 4×N₀ (4/3) πr³
When two molecules encounter each other the distance between the centers of the two molecules would be σ . They can not approach beyond this distance. Thus the sphere of radius σ will occupy a space unavailable for a pair of molecules.
This excluded volume = (4/3) π σ₃ for a pair of molecules. 
Thus the effective volume for a single molecule,
= (1/2) × (4/3) πσ³ = (2/3) πσ³
The effective volume for Avogadro number of molecules (present in 1 mole gas),
Volume Correction
The effective volume for Avogadro number of molecules
Thus b measures and helps to calculate the radius ( or diameter) of the gas molecule.
The equation becomes, 
Pi(V-b) = RT

Pressure Correction:

Pressure of the gas is developed due to the wall collision of the gas molecules.
But due to inter-molecular attraction, the colliding molecules will experience an inward pull and so pressure exerted by the molecules in real gas will be less than that if there had not been inter molecular attraction as in ideal gas(Pi).
Thus, PiP
or, Pi = P+Pa
Where Pa is the pressure correction term originating from attractive forces. Higher the inter molecular attraction in a gas, greater is the magnitude of Pa.
Pressure depends on both the frequency of molecular collisions with the walls and the impulse exerted by each collision. Both contributions are diminished by the attractive forces and the strength with which these operate is roughly proportional to the density of the molecules. Therefore the average pressure exerted by the molecules is decreased by Pa is proportional to the square of the density.
Pa ∝ (1/V₂
Since, density ∝ 1/V
or, Pa = a/V₂
Where a = constant for the gas that measures the attractive force between the molecule.
Thus, Pi = P + (a/V₂)
Using the two corrections we have Van der Waals equation for 1 mole real gas is,
(P + a/V₂ )( V-b ) = RT
To convert the equation for n moles volume has to change as it is the only extensive property in the equation. Let V be the volume of n moles gas hence,V =V/n. Putting this term in the Van der Waals equation, we have,
Van der Waals equation
Van der Waals equation for n Moles Real Gas
Problem:
Calculate the pressure of 2 moles of Nitrogen gas occupying 10 lit volume at 27°C. Given, a = 1.4 atm lit² mol⁻² and b=0.04 lit mol⁻¹ using Van der Waals equation. Also calculate the pressure of the gas using ideal gas equation and find the extent of deviation from ideal behavior.

Answer:

P = 4.904 atm, Pi = 4.92 atm and Deviation=0.325%

Let us consider two hypothetical cases to show the size effect and attraction effect on the pressure of the gas:

(a)For the real gas a=0 (that is no inter molecular attraction ) but b≠0 (size considers):

We have Van der Waals Equation,
(P + an²/V² )(V - nb) = nRT but a=0
Hence, P = nRT/(V-nb) ) 〉Pi 
Since, Pi = nRT/V only.
It means that the molecular size effect (repulsive interaction) creates higher pressure then that observed by the ideal gas where molecules have no volume.

(a)For the real gas a≠0 (existence of attractive force) but b=0 (no size consider):

We have Van der Waals Equation,
(P + an²/V² )(V - nb) = nRT but b=0(no size).
Hence, P = nRT/Vーan²/V²ㄑPi 
Since, Pi=nRT/V only. 
Thus, inter molecular attraction effect reduces the pressure of the real gases.

Units of a and b :

From the Van der Waals Equation,
                                       (P + an²/V² )(V - nb) = nRT Where, Pa = an²/V²
Where Pa is the pressure correction term, called often the internal pressure of the gas.
Then, a = Pa × (V²/n²)
Thus the unit of a = atm lit² mol⁻²
Again, nb = Unit of volume (say liter) 
Hence, the unit of b = lit mol⁻¹
Problem:
Find the Units and Dimensions of  Van dar Walls constant 'a' and 'b' from Van der walls Equation.
Answer:
In SI system,unit of 'a' = N m⁻² m⁶ mol⁻² = N m⁴ mol⁻²
In CGS system,unit of 'a' = dyne cm⁻² cm⁶ mol⁻² = dyne cm⁴ mol⁻²
In SI system,unit of 'b' = m³ mol⁻¹
In CGS system,unit of 'b' = cm³mol⁻¹

Significance of a and b:

‘a’ term originates from the inter molecular attraction and Pa = an²/V² . 
Thus, ‘a’ is a measure of internal pressure of the gas and so the attractive force between the molecules. Higher the value of ‘a’ grater is the inter molecular attraction and more easily the gas could be liquefied.
Thus, a of CO₂ = 3.95 atm lit² mol⁻² while, a of H₂ = 0.22 atm lit² mol⁻².
Another constant, ‘b’ measures the molecular size and also a measure of repulsive forces. The value of b can also be utilized to calculate the molecular diameter(σ)
The grater the value of ‘b’ larger the size of the gas molecule.
Thus, b of CO₂ = 0.04 lit mol⁻¹ while, b of H₂ = 0.02 lit mol⁻¹.

Calculation of Boyle Temperature (TB):

Mathematical condition for Boyle’s temperature is, 
TB = [d(PV)/dP]T When P0
From the Van der Waals Equation for 1 mole gas,
(P + a/V²)(V - b) = RT
or, P = (RT/V - b) - a/
Hence, PV = {RTV/(V - b)} - a/V
Thus, TB = [d(PV)/dP]T = [RT/(V - b) - {RTV/(V - b)²} + a/](dV/dP)T
= [{RT(V - b) - RTV/(V - b)²} + a/V²] (dV/dP)T
=[{- RTb/(V - b)²}a/V²] (dV/dP)T
But when T = TB[d(PV)/dP]T = 0 and (dV/dP)T ≠ 0
Thus We have, RTBb/(V - b)² = a/
or, TB = (a/Rb) {(V - b)/V}² Since P → 0, V is large.
Thus, (V - b)/V ≃ 1 
Hence, TB = a/Rb

Problem:
Calculate the Boyle temperature of nitrogen gas given a=1.4 atm lit² mol⁻² and b=0.04 lit mol⁻¹.
Answer:  T🇧 = 427 K
Problem:
Comment on the possibility of defining a Boyle Temperature if a=0 and a=b=0.

Answer:
When, a = 0, Van der Waals Equation,
P (V - b) = RT
or, PV = RT + Pb
Differentiating with respect of pressure at constant temperature,
[d(PV)/dP]T = b, but b ≠ 0 hence [d(PV)/dP]T ≠ 0
The gas has no Boyle temperature since at any temperature since at any temperature, [d(PV)/dP]T ≠ 0
Again when a = b = 0, Van der Walls equation becomes PV = RT
or, [d(PV)/dP]T = 0
That is at any temperature this becomes zero and so all the temperature are Boyle Temperatures.

Explanation of Amagat’s Curves:

We have Amagat Curves,
 Amagat Curve
(a) Amagat Curve For Different Gases
Amagat Curve
(b) Amagat Curve For CO₂
Van der Waals equation for 1 mole real gas,
(P + a/V₂ )(V - b) = RT 
or, PV - Pb + (a/V) + (a/V₂) = RT
Neglecting the small term (a/V₂), we get ,
PV = RT + Pb - (a/V) 
Using ideal gas equation for small term, a/V = aP/RT and taking  Z = (PV/RT),
We have, Z = 1 + (1/RT){b - (a/RT} P
This Shows that, Z = (T,P)
This equation can be uses to explain Amagat curve qualitatively at low pressure and moderate pressure region.
(a) For CO₂a’ is very high since the gas is easily liquefied. 
Thus (a/RT) 〉b in the equation and {b - (a/RT)} = - ve . That is slope of Z vs P curve is ( -) for carbon dioxide at moderate pressure region. That is the value of Z decreases with increase of Pressure and it is also found in the curve of carbon dioxide.
For H₂a’ is very small and It is not easily liquefied.
 so a/RT〈 b and slope of Z vs P curve for H₂ becomes (+) ve and the value of Z increases with pressure.
(b)  (i) When T〈 Tв that is T〈 a/Rb or,b〈 a/RT and {b−(a/RT)} = (-)ve.
That is the value of Z decreases with increases with P at the moderate pressure region of the curve of CO₂ , Z〈 1 and more compressible.
(ii) When T = Tв = a/Rb or, b = a/RT hence, {b - (a/RT)} = 0
That is Z = 1 the gas shows ideal behavior. The size effect compensates the effect due to inter molecular attraction of the gas. Z runs parallel to P axis up to certain range of pressure at low pressure region.
(iii) When T 〉TB, that is T 〉a/Rb or, b 〉a/RT and {b - (a/RT} = (+)ve.
That is value of Z increases with increase of P when T 〉TB.
The size effect dominates over the effect due to inter molecular attraction.

For hydrogen and helium, 0°C is grater then their  values and Z vs P slope becomes (+)ve.
At very low pressure P→0 and high temperature,the volume is very large and both the
size effect and attraction effect becomes negligible that is Pb and (a/RT) P are negligible and Z=1 that is the gas obeys ideal behavior.