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__Critical Constants of a Gas:__

__A gas can be liquefied by lowering temperature and increasing of pressure.__But

__influence of temperature is more important__. Most of gases are liquefied at ordinary pressure by suitable lowering of the temperature. But a gas can not be liquefied unless its temperature is below of a certain value depending upon the nature of the gas whatever high the pressure may be applied.

This temperature of the gas is called its Critical Temperature (

**Tc**).

__A gas can only be liquefied when the temperature is kept below__.

**Tc**of the gas####
__Definitions of Critical Constants:__

__Definitions of Critical Constants:__

__Critical Temperature (Tc):__Critical Temperature (

**Tc**) is the

__maximum temperature at which a gas can be liquefied__, that is the temperature above which a liquid cannot exist.

__Critical Pressure (Pc):__Critical Pressure (

**Pc**)

__is the maximum pressure required to cause liquefaction at the temperature__(

**Tc**).

__Critical Volume (Vc):__Critical Volume (

**Vc**) is

__the volume occupied by one mole of a gas at critical temperate__(

**Tc**) and

__Critical Pressure__(

**Pc**).

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__Andrews Isotherms:__

In 1869, __Andrews Isotherms:__

__Thomas Andrews__carried out an experiment in which

**P - V**relations of c

__arbon dioxide gas were measured at different temperatures__.

Andrew's Graphs of P Vs V |

__Following are observed from this graph:__

(1) At

__high temperature__, such as T₄,__the isotherms look like those of an ideal gas__.
(2)At

(3) At still higher temperatures we get a similar type of curve as discussed in the above point, except that the horizontal portion is reduced. The Pressure corresponding to this portion being higher then at lower temperatures.

(4) At temperatures

Thus for every gas, there is a limit of temperature above which it cannot be liquefied, no matter what the pressure is.

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It appears from the Amagat curve at

The continuity of the states from the gas to liquid can be explains from the following isotherm

The gas at

No where in the process liquid would appear. At

__low temperatures__, the__curves have altogether different appearances__. Consider, for examples, a typical curve**abcd**. As the__pressures increases__, the__volume of the gas decrease__s (curve**a**to**b**). At point**b**liquefaction commences and the volume decreases rapidly as the gas converted to a liquid with much higher density. This conversion takes place at constant pressure**P**. At the point**C**, liquefaction is complete and thus the**cd**is__evidence of the fact that the liquid cannot be easily compressed__. Thus, we note that**ab**represents the__gaseous state__,__bc__,__liquid and vapour in equilibrium__, and**cd**shows the__liquid state only__.(3) At still higher temperatures we get a similar type of curve as discussed in the above point, except that the horizontal portion is reduced. The Pressure corresponding to this portion being higher then at lower temperatures.

(4) At temperatures

**Tc**the horizontal portion is reduced to a mere point. At temperatures higher then**Tc**there is no indication of qualification at all.Thus for every gas, there is a limit of temperature above which it cannot be liquefied, no matter what the pressure is.

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__Continuity of State:__

It appears from the Amagat curve at __Continuity of State:__

**T**there is discontinuity or break during the transformation of gas to liquid. __But it is not so__.

The continuity of the states from the gas to liquid can be explains from the following isotherm

**pqrs**at**T**Schematic Representation of the Continuity of State |

**A**is heated to**B**at constant volume (**V**) along**AB**. Then the gas is gradually cooled at constant**P**along**BC**, the volume is reduced considerably. The gas is again cooled at constant**V**until the point**D**is reached.No where in the process liquid would appear. At

**D**, the system is highly compressed gas. But from the curve, this point is the representation of the liquid state. Hence there is hardly a distinction between the liquid state and the gaseous state.__There are no line of separation between the two phases. This is known as the principle of continuity of state__.####
__Critical Phenomena and Van der Walls Equation:__

For one mole of a gas the __Critical Phenomena and Van der Walls Equation:__

__Van der Waals equation__,

**(P + a/V²)(V - b) = RT**

This equation can be written as,

**V³ - (b + RT/P)V² + (a/P)V - ab/P = 0**
This equation has three roots in

**V**for given values of**a**,**b**,**P**and**T**. It found that either all the three roots are real or one is real and the other two are imaginary.Van der Waals Isotherms |

__main characteristics of the above isotherm are as follows__:

(a) At

__higher temperature and higher volume region__, the isotherms looks much like the__isotherms for a Ideal gas__.
(b) At the

__temperature lower than__**Tc**the__isotherm exhibits a maximum and a minimum__. For certain values of pressure, the equation gives three roots of volume as**V₁**,**V₂**and**V₃**at pressure**P₁**. The section AB and**ED**of the__Van der walls curve__at**T₂**can be realized experimentally.**ED**represents__Supersaturated or Super cooled vapour__and**AB**__represents super heated liquid__. Both these states are__meta-stable__. These states are unstable in the seance that slight disturbances are sufficient to cause the system to revert spontaneously into the stable state with the two__phases present in equilibrium__.
(c) The section

**BCD**of the__Van der Walls isotherms__cannot be__realized experimentally__. In this region the slope of the**P - V**curve is positive.__Increasing or decreasing the volume of such a system would increase or decrease of pressure__. The line**BCD**represents the__meta- stable state__.####
__Expression of Critical Constants in Terms of Van der Waals Constants:__

__Expression of Critical Constants in Terms of Van der Waals Constants:__

Again with

__increase of temperature__the minimum and maximum points come close to each other and at the critical point (**C**) both__maximum and minimum coalesce__. The__slope__and__curvature__both becomes__zero at this point__.
Thus the

__mathematical condition of critical point is__,**(dP/dV)á´› = 0**and

**(d²P/dV²)á´› = 0**

__Van der Waals equation__for 1 mole gas is

(P + a/V²)(V - b) = RT

or,

**P = RT/(V - b) - a/V²**__Differentiating Van der Waals equation__with respect to

**V**at constant

**T**,

We get

__Slope__= (dP/dV)á´› = - {RT/(V - b)²} + 2a/V³
And

__Curvature__= (d²P/dV²)á´› = {2RT/(V - b)³} - 6a/V⁴
Hence at the critical point,

- {RTc/(Vc - b)²} + 2a/Vc³ = 0

or, RTc/(Vc - b)² = 2a/Vc³

and {2RTc/(Vc - b)³} - 6a/Vc⁴ = 0

and {2RTc/(Vc - b)³} - 6a/Vc⁴ = 0

or, 2RTc/(Vc -b)³ = 6a/Vc⁴

Thus, (Vc - b)/2 = Vc/3

or,

__Vc = 3b__
Putting the value of

**Vc**=**3b**in RTc/(Vc - b)² = 2a/Vc³.
We have, RTc/4b² = 2a/27b³

or,

__Tc = 8a/27Rb__
Again the

__Van der Walls equation at the critical state is__**Pc**= RTc/(Vc - b) - a/Vc²

Putting the value of

**Vc**and**Tc**,
We have

__Pc = a/27b²__

__Problem:__
The Critical constants for water are 647 K, 22.09 MPa and 0.0566 dm³mol⁻¹. Calculate the value of Van der Waals constants a and b.

__Answer:__
a = 212.3 KPa dm⁶ mol⁻² and b = 0.0189 dm³ mol⁻¹

Full Details

__Click Here__

__Compressibility Factor at the Critical State (Zc):__
The Critical coefficient is defined as RTc/PcVc

Thus the value of Critical Coefficient = {R × (8a/27Rb)}/{(a/27b²) × 3b} = 8/3 = 2.66

Thus the value of Compressibility factor at the critical state (Zc) = PcVc/RTc = 3/8 = 0.375

__Van der Waals constants in terms of critical constants:__
Van der Waals constants can be determind from critical constants Tc and Pc of the gas. Vc in the expression is avoided due to difficulty in its determination.

We have, b = Vc/3 but PcVc/RTc = 3/8 or, Vc = (3/8) × (RTc/Pc)

hence,

hence,

**b = (1/8)(RTc/Pc)**
Again, a = Pc × 27b² = 3 × Pc × (3b)² = 3 PcVc² = 3Pc × (3RTc/8Pc)²

∴ a = (27/64)(R²Tc²/Pc)

__Problem:__
Calculate Van der Waals constants for Ethylene. (Tc = 280.8 K and Pc = 50 atm)

a = 0.057 lit mol⁻¹ and b = 4.47 lit² atm mol⁻²

Full Details

Argon has (Tc = - 122°C, Pc = 48 atm). What is the radius of the Argon atom?

1.47 × 10⁻⁸ cm

Full Details

__Answer:__a = 0.057 lit mol⁻¹ and b = 4.47 lit² atm mol⁻²

Full Details

__Click Here____Problem:__Argon has (Tc = - 122°C, Pc = 48 atm). What is the radius of the Argon atom?

__Answer:__1.47 × 10⁻⁸ cm

Full Details

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