A gas can be liquefied by lowering temperature and increasing pressure. But the influence of temperature is more important.

- Most of the gases are liquefied at ordinary pressure by the suitable lowering of the temperature. But gas cannot be liquefied unless its temperature is below 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 (T

_{c}). Gas can only be liquefied when the temperature is kept below T

_{c}of the gas.

### Definitions of critical constants of a gas

#### Critical Temperature of a gas

- Critical temperature (T

_{c}) is the maximum temperature at which a gas can be liquefied, which is the temperature above which a liquid cannot exist.

#### Critical Pressure of a gas

- Critical pressure (P

_{c}) is the maximum pressure required to cause liquefaction at the temperature (T

_{c}).

#### Critical Volume of a gas

- Critical volume (V

_{c}) is the volume occupied by one mole of a gas at critical temperate (T

_{c}) and critical pressure (P

_{c}).

### Andrews isotherms

- In 1869, Thomas Andrews carried out an experiment in which P - V relations of carbon dioxide gas were measured at different temperatures.

Critical constant of gas |

- At high temperatures, such as T
_{4}, the isotherms look like those of an Ideal gas. - At low temperatures, the curves have altogether different appearances. Consider, for example, a typical curve. As the pressures increase, the volume of the gas decreases (curve A to B).
- At the 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 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 vapor in equilibrium, and CD shows the liquid state only.
- 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 is higher than at lower temperatures.
- At temperatures, T
_{c}the horizontal portion is reduced to a mere point. At temperatures higher then T_{c}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.

### Continuity of state

- It appears from the Amagat curve at 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 explained from the following isotherm ABCD at T₁.

- The gas at 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. Nowhere in the process liquid would appear.

- At D, the system is a 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 is no line of separation between the two phases. This is known as the principle of continuity of the state.

### Critical phenomena and Van der Walls equation

- For one mole of gas the Van der Waals equation,

(P + a/V

or, V

^{2})(V - b) = RTor, V

^{3}- (b + RT/P)V^{2}+ (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.

Critical phenomena |

### Characteristics of the above isotherm

- At higher temperatures and higher volume regions, the isotherms look much like the isotherms for an Ideal gas.
- At the temperature lower than T
_{c}the isotherm exhibits a maximum and a minimum. For certain values of pressure, the equation gives three roots of volume as V_{1}, V_{2, }and V_{3}at pressure P1.

The section AB and ED of the Van der Waals curve at T_{1}can be realized experimentally. ED represents Supersaturated or Supercooled vapor and AB represents superheated liquid. Both of 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. - The section BCD of the Van der Waals 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 in pressure. The line BCD represents the metastable state.

### Critical constants in terms of Van der Waals constants

- Again with the 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 become zero at this point.

Thus the mathematical condition of critical point is,

(dP/dV)

Van der Waals equation for 1 mole gas is,

(P + a/V

or, P = RT/(V - b) - a/V

Differentiating Van der Waals equation with respect to V at constant T,

We get Slope,

(dP/dV)

And Curvature,

(d

Hence at the critical point,

- {RT

or, RT

and {2RT

or, 2RT

Thus, (V

(dP/dV)

_{T}= 0 and (d^{2}P/dV^{2})_{T}= 0Van der Waals equation for 1 mole gas is,

(P + a/V

^{2})(V - b) = RTor, P = RT/(V - b) - a/V

^{2}Differentiating Van der Waals equation with respect to V at constant T,

We get Slope,

(dP/dV)

_{T}= - {RT/(V - b)^{2}} + 2a/V^{3}And Curvature,

(d

^{2}P/dV^{2})_{T}= {2RT/(V - b)^{3}} - 6a/V^{4}Hence at the critical point,

- {RT

_{c}/(V_{c}- b)^{2}} + 2a/V_{c}^{3}= 0or, RT

_{c}/(V_{c}- b)^{2}= 2a/V_{c}^{3}and {2RT

_{c}/(V_{c}- b)^{3}} - 6a/V_{c}^{4}= 0or, 2RT

_{c}/(V_{c}-b)^{3}= 6a/V_{c}^{4}Thus, (V

_{c}- b)/2 = V_{c}/3∴ V

_{c}= 3bPutting the value of V

We have, RT

_{c}= 3b in RT_{c}/(V_{c}- b)^{2}= 2a/V_{c}^{3}.We have, RT

_{c}/4b^{2}= 2a/27b^{3}∴ T

_{c}= 8a/27RbAgain the Van der Walls equation at the critical state,

P

Putting the value of V

P

_{c}= RT_{c}/(V_{c}- b) - a/V_{c}^{2}Putting the value of V

_{c}and T_{c},∴ P

_{c}= a/27b^{2}Problem

- The Critical constants for water are 647 K, 22.09 MPa and 0.0566 dm

^{3}mol

^{-1}. Calculate the value of Van der Waals constants a and b.

We have T

Thus, b = V

∴ b = 0.0189 dm

a = 3 P

= 3 (22.09 × 10

∴ a = 213.3 kPa mol

_{c}= 647 K, P_{c}= 22.09 Mpa = 22.09 × 10^{3}kPa and V_{c}= 0.0566 dm^{3}mol^{-1}Thus, b = V

_{c}/3 = (0.0566 dm^{3}mol^{-1})/3∴ b = 0.0189 dm

^{3}mol^{-1}a = 3 P

_{c}V_{c}^{2}= 3 (22.09 × 10

^{3}kPa)(0.0566 dm^{3}mol^{-1})2∴ a = 213.3 kPa mol

^{-2}### Compressibility factor at the critical state

- The Critical coefficient is defined as,

- RT

_{c}/P

_{c}V

_{c}

Thus the value of Critical Coefficient,

= {R × (8a/27Rb)}/{(a/27b

= 8/3

= 2.66

Thus the value of compressibility factor at the critical state (Z

= 3/8

= 0.375

= {R × (8a/27Rb)}/{(a/27b

^{2}) × 3b}= 8/3

= 2.66

Thus the value of compressibility factor at the critical state (Z

_{C}) = P_{C}V_{C}/RT_{C}= 3/8

= 0.375

### Van der Waals constants in terms of critical constants

- Van der Waals constants can be determined from critical constants T

_{c}and P

_{c}of the gas. V

_{c}in the expression is avoided due to difficulty in its determination.

We have, b = V

or, V

_{c}/3 but, P_{c}V_{c}/RTC_{c}= 3/8or, V

_{c}= (3/8) × (RT_{c}/P_{c})- ∴ b = (1/8)(RT

_{c}/P

_{c})

Again, a = P

= 3 P

= 3P

_{c}× 27b^{2}= 3 × P_{c}× (3b)^{2}= 3 P

_{c}V_{c}^{2}= 3P

_{c}× (3RT_{c}/8P_{c})^{2}∴ a = (27/64)(R

Problem^{2}T_{c}^{2}/P_{c})- Calculate Van der Waals constants for ethylene. (T

_{C}= 280.8 K and P

_{C}= 50 atm).

- a = 0.057 lit mol

^{-1}and b = 4.47 lit

^{2}atm mol

^{-2}

- Argon has (T

_{C}= - 122°C, P

_{C}= 48 atm). What is the radius of the argon atom?

- 1.47 × 10

^{-8}cm