4.4 Implementation
4.4.2 Model and Parameter Choices
You had learnt earlier about the origin of the deviations of gases from ideal gas behaviour. What are these two assumptions again'' Firstly. the volume of a molecule is by no means neg!Jgible and cannot be ignored under all conditions Secondly. there certainly exists intcr.molecular mteraction between molecules at close distances. Vander Waals modified the ideal gas equation by taking into account the dbove shortcomings.
Shortcomings: Two corrections were applied considering the fact that the molecules of a real gas have definite volume: and (ii) the pressure exerted as a result ofthe number of molecules strikmg the wall and attractmg a molecule from behmd.
Let us take the two corrections one b' one
Volume corrections: Van der Waals realized that the molecules of a real gas have dctlmte volume Therefore_ the entirvolume (V) of the contamer is not available for the free movement of the gas molecules. The volume available for the motion of the molecules can be given by (V-nb). where n IS the number of moles of the gas and - b- the
correction in volume for one mole of the gas. The quanti tv ·b- is known as co-volume.
Hence. corrcctd volume :::: V,.ur;
=V-nb.. (32)
Pressure correction: Van der Waals applied pressure correctiOn by taking into account the intern1olccular forces. The pressure of a gas is due to the collision of the gas molecules on the nalls of the container. Consider two identical molecules in a gas such that one is somewhere in
The van dcr Waals constm1t 'h 'is cqual1o the exluded volume of one mole of a gas. H can be sl10wn that 'h 'is equal to four times the actual volume of the molecules Tt1c constant 'b ·has the units. m3 moi- 1
V
It can be seen that a molecule in the middle of the container is attracted on all side by the other molecules surrounding it. However. in case of a molecule that just strikes the wall. there is a net backward drag on the molecule and it will strike the wall with a somewhat weakened impact. Hence.
the observpressure (p) of a gas will be less than the pressure exerted by an ideal gas. A presssure correction is therefore needed to be applied. The correction term in pressure (Dp) is proportional to two factors. viz.,
• the number of molecules striking the wall per unit areas and
• the number of molecules attracting a molecule from behind.
EaGD-Qf the above factors is proportional to the square of concentration of the gas i.e..
Dp a. (concentration)2
But the J;onc. :ntration of the gas = Number of moles (n) Volume of the container (V) Hence it can be written that.
p a.
.K
V2
i.e., D p =
n
2a
2 3.3
where ·a· is a paran1eter characteristic of a gas. Hence the corrected pressure <Pcorrl is given by.
Pwrr = p + n2a ... 3.4
\(1
If the corrected pressure and the corr 'Ctcd volume of the gas are substituted in the ideal gas equation (Module 2 Unit 2). we obtain
{P+ )V-nb)=nRT ··· .. 3.5
This equation is known as Van dcr Waals equation. Since for one mole of gas. V=V m (i.e.. molar volume) and n = I. hence. Eq. 3.5 becomes
...3.6
Van derWaals equation (Eq. 3.5 or 3.6) is quite important and is applicable over a much wider range of p-V-T data than the ideal gas eguation. The quantities ·a· and 'b. are called the Vander Waals constants or parameters. The quantities ·a· and 'b· arc obtained empirically by fitting values on;, experimental p-V-T data to Eq. 3.5. It may be pointed that 'b' is a measure of the moleclj]ar size and
·a' is related to the intermolecular mteraction. Table 4.1 gives the values of the parameters ·a·and ·b·
of some selected gases. It can be seen that 'b· increases as the size of the molecule increases whereas ·a· has large value for an easily compressible gas. The values of the critical constants p,. V, and T, are also given in Table 4.1.
Table 4.1: Van der Waals Parameters and Critical Constants of Some Gases Gas ·a'/Pa m6 mol·' 106 x ·b '/m3 mol·'
w·'x
PJPa I06 x VJm3moi·1 TJKHe Ar
o,
H, N,co,
H20 NH 3 CH4
0.003457 0.1373 0.02476 0.1378 0.1408 0.3639 0.5536 0.4225 0.2283
23.70 32.19 26.61 31.83 39.13 42.67 30.49 37.07 42.78
2.20 48.64
12.97 50.76 33.94 73.66 220.89
112.5 46.41
57.8 73.3 65.0 78.0 90.1 94.0 55.3 72.5 99.0
5.21 150.7 33.2 154.8 126.3 304.2 647.4 405.5 191.1 Explanation of the behaviour of gases using Van der Waals equation
Many a time, either one or both the correction terms could become negligible. Let us study these cases.
When 'b' is negligible:
If'b' is very small, then equation 3.6 becomes.
(P +
v Jvm
RT . . ... 3.7 i.e., pVm = RT - JLVm or z = pVm = I -_a_
RT RTVm ... 3.8
Thi;shows that under these conditions. pVm will be less than RT or z will be less than unity. Equation 3.8 will be valid for substances like water vapour for which 'a' is large and 'b· is comparativelv small (SeTable 4.1 ). Also for gases such as N 2• CH4 and C02 (Fig. 4.1) at moderately low pressures. Vm is large such that (Vm- b) is nearly equal to Vm· Hence. equation 3.8 is applicable for such gases at moderately low pressures.
When a is negligible:
If 'a' is negligible we have p(Vm-b)=RT
i.e., pVm = RT + pb pVm pb or z=- =I+- RT
RT
... 3.9
3.10
Hence, pVm will be greater than RT or z will be greater than unity. Particularly, this is tme for hydrogen (Fig. 4.1) and noble gases for which the value of ·a' is small. This IS also tme for all the gases at h1gh
)
When 'a' and 'b' are both negligible:
When pressure is very low or the temperature is very high. p is small but V m is very large. In this case.
the correction terms,
.JL
and bare both negligible in comparison to p and Vm·Vm2
Hence, at very low pressures or high temperatures. the gases obey ideal gas equation and their z value is nearly equal to unity.
Let us now illustrate the use of Equation 3.5 in the calculation of pressure of2.000 mol of methane at I.OOOK x 103 occupyingavolumeof5.000x I0·2 m'-
Re-arranging quation 3.5. we can write.
_ nRT p- (V- nb)
From Table 4.1, a= 0.2283 Pa m6 mol-2 b = 42.78 x 10-<> m-' mol" 1
substituting the values of the parameters we get,
p = 2 000 mol x 8 314 J moi·1
K'
x I 000 x 103K _ (2.000 mol)2 x (0.2283 Pa m6 moi·2)(5.000 x 10·2m3 - 2.000 mol x 42.78 x 10·6 m3 mo1"1) (5.000 x 10-2 m3 2 p = 3.328 X 105 Pa
Applying van der Waals equation to methane at 1.000 x 103 K, the pressure calculated is 3.328 x
10
5 Pa.Let us also calculate the pressure of methane using the same values of n, T and V but assuming ideal behaviour.
_ n RT _ 2.000 mol X 8.314 j moi-'K- 1 X 1.000 X 10-'K p-
V-
5.000 x J0·2 m3=
3 3226 x 105PaIt is interesting to see that the pressure values of methane obtained by Van der Waals equation and ideal gas equation at I.000 x 103 K are more or less same. This indicates that the methane behaves ideally at
1.000 X 103 K.
Virial Equation of State
A number of attempts have been made to propose equation-of state for ideal gases. These are supposed to represent the p-V-T data over as wide range as possible. However, from practical consideration, it is desirable that the equation of state should have only a few adjustable parameters. It should be simple from mathematical point of view.
The most gener:tl cquabon of state \L proposed by Karnmerlingh -- Onnes and is known as virial equation of state. In this equation. the pressure is represented as power series of
I
vm
p , RT ,B(T) + C(T) + _
V' "'
TI1e coefficients B(T). C(T) ... are known as vi rial coefficients. It may be noted that these depend on temperature. By having sufficient number of tem1s m this equation, p-V-T data can be represented to desired accuracy.
In the next section, we shall introduce the critical phenomena and then study the relationship bcl\vcen Van der Waals constants and critical constants. Before that. work out the following Exercise:
Exercise I
Calculate the pressure of 2.000 mol of methane at 298.2 K using the other data from Table 4.1 and assuming that it obeys Van der Waals equation. Also calculate its value. if methane were to behave ideally at 298.22 K.