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In this section we would examine how critical temperature (T,), critical pressure (Pel and critical volume (V,) are related to Van der Waals constants.

3.4.1 Critical Constants and Van derWaals Constants

The Van der Waals equation (Eq. 3.6) can be made the basis of a theoretical consideration. The curves in Fig. 4.5, known as Vander Waals isotherms, sho\\ the isotherms 'calculated on the basis of this equation.

Expanding Eq. 3.6 we get,

(p + a!V.,²⁾ (V.,- b)= RT ··· 3.11 i.e., pV.,- pb +a!V.,- abN., ² = RT

Multiplying the equation throughout by Vm2/p, we obtain, V³ .,- bV.,'+ aV.,/p--ab/p = RTV.,²/p

ie., V.,³ - V.,² (b + RT/p) +alp V.,- ab/p = 0

.3.12 3.13

The cubic equation will yield three values for Vm corresponding to a given presure and temperature.

All the three valus t>f V m may be real or one may be real and the other two may be complex conjugates. Isotherms III and IV do yield three values ofV., in certain ranges of p and V This is true in general for all isotherms below the critical tempemture. Curve JI corresponds to critical temperature and curves at higher temperature (such as curve I) approach the isotherm representing the Boyle's law. These theoretical curves are similar to those obtained by Andrews for C0₂ but a maJor difference is the wave like portion BCDEF in the theoretical curves. If experiments are performed without perturbance, then portions BC and EF are realisable, these portions represent the supersaturated vapour and superheated liquid, respectively.

Fig. 4.5: fim .rier Waals isotherms

The wave like portion decreases as temperature increases "<.t the critical temperature, it is reduced to a point which means that all the three roots ofEq. 3.13 are identical and equal to the critical volume, V,.

We can obtain a cubic equation bv raising it to power three, i.e., (V.,- Vel ³ = 0 ··· ··· ... ··· ... 3.14 or V.,..-3V, Vm' + 3V/ V.,-V, ³ = 0 !··· ...3.!5

"'- a I

This equation should be identical with the expanded form of the Van dcr Waals equation (Eq. 3.13) at critical temperature and pressure.

. ... 3.16 No"· comparing the coefficients of equal powers of Vm in Eqs. 3.15 and 3.16. we obtam -3 V,= -(b + RTJpJ or 3V,

=

b + RT/p,

3V/ =alp,

and- V/

=

-ab/p, or V,³

=

^ab/p,

From Eqs. 3.1 X and 3.19. we obtain : =

v

b or V, = 3b

3

Substituting the value of V, in Eq. 3 I X.

a p,

=

27b'

From Eqs. 3.17 and 3.20. "e get RT, =(3V -b)= Xb

g,; ^I.

... .3.17 ... 3.18 ... 3.19

... 3.20

... 3.21

orT, = Sb. p, /R

ou.

27b2. R.

(using Eq. 3 21)

= 8a 27Rb

3.22 Hence. the values of p,. V, and T, can be calculated from the Vander Waals constants

3 4.2 Determination of Critical Constants

Let us study experimental method of determination of critical constants.

Critical temperature

Aim: To detennine critical ccmpcraturc.

Apparatus: capillary tube. alumimum block. water. Bunsen burner, thermometer.

Procedure:

• Select and evacuate a capillary tube that is able to withstand high pressure.

• Fill it up with the liquid and seal up.

• Place it in an aluminium block having a wmdow

• Heat the set up system gradually.

• Observe closely the mcmscus of the liquid through the window.

Repeat it a number oftimes by varying the temperature in both directions.

Observation

You will observe that initially the liquid is in equilibrium with the vapours and a distinct boundary can be

'

-experiment a number of times by varying the temperature in both· directions, you can then take the mean as the experimental value of critical temperature.

Critical pressure

The gas under observation is taken in a high pressure vessel at the critical temperature. Initially the gas pressure is kept low. Slowly the gas is compressed at constant temperature. As soon as the vessel inside is covered with mist, it indicates the formation of some liquid and this pressure corresponds to P,. Since the pressure is generally much higher.than what an ordinary barometer can measure. special pressure gauges are to be used.

Critical Volume

Critical volume is determined indirectly based on the findings of Cailletc and Mathies. According to them, a plot of the mean values of the densities of a liquid and its saturated vapour against temperature is a straight line (Fig. 4.6).

Fig. 4.6: Plot o(densities o(vapour (1). liquid (11) and their mean values (ill) against temperature It is clear that the density of the vapour (curve I) increases with increase in temperature because the evaporation is higher at higher temperature. But the density of the liquid decreases as temperature increases (curve II). At the critical temperature. the two densities must be equal. However, due to flunctuations it is very difficult to measure the density at T,. Hence, the two curves, I and II are extrapolated to intersect the combined curves I and II at C. The density at

C.

known as critical density.

represents the density at T,. From this. the critical volume is calculated using the relationship.

V=

Molar mass

'

Crillcal density

Some of the experimental values of the critical constants are already given in Table 4.1.

3.4.3 Test for Van derWaals Equatioq

The calculation of the compressibility factor at the critical point (zJ based on experimental p,. V, and T< values can be a test for Van der Waals equation. Theoretically the values of Z,can be derived as tollows:

Z ⁼ p,V,

RT, ... . ... 3.23

Z,= (a) (3b) _.I. (P a V 36 T Sa ) 27b2 R 8a/27Rb c = 2ibz' c = , c = Z7Rb

=

^3/8

=

^0.375

For most gases, the value of Z, obtained from the experimental values of the critical constants lies between 0.2- 0.4. This variation from the theoretical value of0.375 indicates the appropriate nature of Van der Waals equation. Why don't you apply these principles in solving the following exercises'' Exercise 2

lndane gas supplied for household usc is mostly a mixture ot propane and butane. Are the critical temperatures of these two gases higher than 298 K''

4.0 Conclusion

In this unit. we have been able to examine the reasons for c al gases deviation from ideal gas laws at high pressures and low temperatures. Similarly the nature and types of intermolecclar forces have been discussed.

Also above critical kmperature. a substance can exist only as a gas for the mere fact that the kinetic energies of the molecules are so great that intennolecular attractions are negligible at very high pressures.

However, the conditions for liquefaction of gases have been discussed. The volume and pressure corrections in Van der Waals equation are also explained.

50 Summary

In this Unit. we have learnt that:

• Real gases behave non-Ideally at very high pressures.

• The pressure necessary to liquefy a gas at its critical temperature is called its critical pressure.

• The volume occupied by one mole of a tluid at its critical temperature and critical pressure is called its critical volume

g The intermolecular forces include Van der Waals forces. metallic bonding, hydrogen bonding etc

• The deviation occurs due to the presence of intermolecular forces.

• The Van der Waals c1uation 'corrects' the obseCYed pressure and the obserYed volume of a real gas to those of an Ideal gas as follows:

I ,

t

I p + 'lJl V- nb)=nRT

V')

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