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__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

__V__and

**olume Correction****of the ideal gas equation.**

__Pressure Correction__####
__Volume Correction:__

__Molecules are assumed to be a hard rigid spheres and in a__the available space for free movement of the molecules becomes

**real gas**,__less then V__.

let us take the available space for free movement of

__1 mole gas molecules ,__**Vi = (V-b)**

**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__

**Ïƒ = 2r**

__is the diameter using the molecule as a rigid sphere.__

Representation of Gas Molecules |

__one mole gas__

__molecules,__

b = 4×N₀ (4/3) Ï€r³

__When two molecules encounter each other the distance between the centers of the two molecules would be Ïƒ .__They can not approach beyond this distance. Thus the sphere of radius Ïƒ

__will occupy a space unavailable for a pair of molecules.__

This excluded volume = (4/3) Ï€ Ïƒ₃

__for a pair of molecules.____Thus the effective volume for a single molecule,__

= (1/2) × (4/3) Ï€Ïƒ³ =

**(2/3) Ï€Ïƒ³**__The effective volume for Avogadro number of molecules (present in 1 mole gas),__

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**

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__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, Pi 〉P

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

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 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:__

__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.

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/Vãƒ¼an²/V²ã„‘Pi

Since, Pi=nRT/V only.

Since, Pi=nRT/V only.

__Thus, inter molecular attraction effect reduces the pressure of the real gases.__

####
__Units of a and b :__

**(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 c**

**m⁴ mol⁻²**

__In SI system,unit of 'b'__=

**m³ mol⁻¹**

__In CGS system,unit of 'b'__=

**cm³mol⁻¹**

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__Significance of a and b:__

__‘a’ term originates from the inter molecular attraction and__

**Pa**= an²/V² .**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,

__= 3.95 atm lit² mol⁻² while,__

**a of CO₂**__= 0.22 atm lit² mol⁻².__

**a of H₂**__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(Ïƒ)__.

__The grater the value of ‘b’ larger the size of the gas molecule.__

Thus,

**= 0.04 lit mol⁻¹ while,**

__b of CO₂__**= 0.02 lit mol⁻¹.**

__b of H₂__####
__Calculation of Boyle Temperature (TB):__

Mathematical condition for Boyle’s temperature is,
TB = [d(PV)/dP]T When P→0

(P + a/V²)(V - b) = RT

or, P = (RT/V - b) - a/V²

Hence, PV = {RTV/(V - b)} - a/V

Thus, TB = [d(PV)/dP]T = [RT/(V - b) - {RTV/(V - b)²} + a/V²](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/V²

or, TB = (a/Rb) {(V - b)/V}² Since P → 0, V is large.

Thus, (V - b)/V ≃ 1

Hence,

or, P = (RT/V - b) - a/V²

Hence, PV = {RTV/(V - b)} - a/V

Thus, TB = [d(PV)/dP]T = [RT/(V - b) - {RTV/(V - b)²} + a/V²](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/V²

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.

**TðŸ‡§ = 427 K**

__Answer:__**Problem:**

Comment on the possibility of defining a Boyle Temperature if a=0 and a=b=0.

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__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.__

When, a = 0, Van der Waals Equation,

__Differentiating with respect of pressure at constant temperature,__

__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

__That is at any temperature this becomes zero and so all the temperature are Boyle Temperatures.__

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__Explanation of Amagat’s Curves:__

__We have Amagat Curves,__

(a) Amagat Curve For Different Gases |

(b) Amagat Curve For CO₂ |

(P + a/V₂ )(V - b) = RT

or, PV - Pb + (a/V) + (a/V₂) = RT

or, PV - Pb + (a/V) + (a/V₂) = RT

Neglecting the small term (

PV = RT + Pb - (a/V) **a/V₂)**, we get ,
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.__

__For__

**CO₂**‘**a**’ is very high since the gas is easily liquefied.**{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__

**H₂**‘**a**’ 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〈 TÐ² that is T〈 a/Rb or,b〈 a/RT and {b−(a/RT)} = (-)ve.

That is the value of

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 = TÐ² = 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**TÐ²**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

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.