Give Physical arguments explaining why the critical pressure and temperature should increase Vander Waals values?
Answers
Answer:
Van der Waals equation was derived by Johannes Diderik van der Waals in the year 1873. The equation is basically a modified version of the Ideal Gas Law which states that gases consist of point masses that undergo perfectly elastic collisions. However, this law fails to explain the behaviour of real gases. Therefore, Van der Waals equation was devised and it helps us define the physical state of a real gas.
More significantly, the Van der Waals equation takes into consideration the molecular size and molecular interaction forces (attractive and repulsive forces). Sometimes, it is also referred to as Van der Waals equation of state
Van der Waals equation is an equation relating the relationship between the pressure, volume, temperature, and amount of real gases. For a real gas containing ‘n’ moles, the equation is written as;
\left( P+\frac{a{{n}^{2}}}{{{V}^{2}}} \right)\left( V-nb \right)=n\,R\,T\,(P+
V
2
an
2
)(V−nb)=nRT
Where, P, V, T, n are the pressure, volume, temperature and moles of the gas. ‘a’ and ‘b’ constants specific to each gas.
The equation can further be written as;
Cube power of volume: {{V}^{3}}-\left( b+\frac{RT}{P} \right){{V}^{2}}+\frac{a}{P}V-\frac{ab}{P}=0V
3
−(b+
P
RT
)V
2
+
P
a
V−
P
ab
=0
Reduced equation (Law of corresponding states) in terms of critical constants:
\left( \pi +\frac{3}{{{\varphi }^{2}}} \right)\left( 3\varphi -1 \right)=8\tau :\,\,where\,\,\pi =\frac{P}{Pc},\varphi =\frac{V}{Vc}\,\,\,and\,\,\tau =\frac{T}{Tc}(π+
φ
2
3
)(3φ−1)=8τ:whereπ=
Pc
P
,φ=
Vc
V
andτ=
Tc
T
Units of Van der Waals Constants
Unit of “a” and is atm lit² mol⁻²
Unit of “b” litre mol⁻¹
Also Read: Ideal Gas Law
Van der Waals Equation Derivation
Van der Waals equation derivation is based on correcting the pressure and volume of the ideal gases given by Kinetic Theory of Gases. Another derivation is also used that is based on the potentials of the particles. Nonetheless, both derivations help us establish the same relationship.
Van der Waals Equation of State for Real Gases – Derivation
Kinetic theory of ideal gases assumes the gaseous particles as –
Point masses without any volume,
Independent having no interactions and
Undergo perfectly elastic collisions.
In practice, Van der Waals assumed that, gaseous particles –
Are hard spheres.
Have definite volume and hence cannot be compressed beyond a limit.
Two particles at close range interact and have an exclusive spherical volume around them.
Volume Correction in Van der Waals Equation
As the particles have a definite volume, the volume available for their movement is not the entire container volume but less.
Volume in the ideal gas is hence an over-estimation and has to be reduced for real gases.
Volume of the real gas VR = Volume of the container/ideal gas (VI) – Correction factor(b)
VR = VI – b.
If the particles are independent, then,
Total volume of the particle = number of particle x volume of one particle =\left( n\frac{4}{3}\pi {{r}^{3}} \right)=(n
3
4
πr
3
)
But, the particles are not independent, they do interact.
Van der Waal considered two hard-sphere particles can come as close as to touch each other and they will not allow any other particle to enter in that volume as shown in the diagram.
Van der Waal two hard-sphere particles
The two sphere model, has a total radius of ‘2r’ (r is the radius of the sphere particle) and Volume of \frac{4}{3}\pi 2{{r}^{3}}=8\times \frac{4}{3}\pi {{r}^{3}}=8\times
3
4
π2r
3
=8×
3
4
πr
3
=8× volume of single particle.
Then, each of the two particles has a sphere of influence of 4 times the volume of the particle.
Volume correction for each particle is not volume of the particle but four times of it =b=4\times \frac{4}{3}\pi {{r}^{3}}=b=4×
3
4
πr
3
Volume correction for ‘n’ particles =nb=4n\times \frac{4}{3}\pi {{r}^{3}}=nb=4n×
3
4
πr
3
Volume (V) of the real gas = Vi – nb