Introduction to Van Der Waals Equation
In 1873, Johannes Diderik Van Der Waals derived the Van Der Waals equation. The equation is considered as the updated version of the ideal gas law that states that there are some point masses present in gases that undergo perfectly elastic collisions. However, the real gas law is incapable of explaining the behaviour of real gases. Due to this reason, the Van Der Waals equation was derived to define the physical state of a real gas.
The full name of Van Der Waals was Johannes Diderik van der Waals. He was a Dutch thermodynamicist and theoretical Physicist, who had won the Nobel prize in physics in the year 1910. And as you may have already understood by now, the Van der Waals equation is named after him, because he was the one who gave it.
The Van der Waals equation is the equation of state of thermodynamics, it is based on the theory that suggests that the particles with non-zero volumes, makes the fluids. Van der Waals derived it in 1873, based on the traditional set of derivations, which are also derived from the Van der Waals. The equation explains the condensation of the gases to the liquid phase. The equation was so impactful that another famous scientist of the time James Maxwell said that the name of Van Der Waals is going to be famous in molecular science.
Also, if you like to learn about the Van der Waals forces, then follow this link.
What is Van Der Waals Equation?
Van Der Waals equation is an equation that is used to relate the relationship existing between the pressure, volume, temperature, and amount of real gases. For a real gas containing ‘n’ moles, the real gas equation derivation is as follows.
(P + \[\left ( \frac{an^{2}}{V^{2}} \right )\] ) V−nb = nRT
Where,
P is the pressure,
V is the volume,
T is the temperature,
n is the number of moles of gases,
‘a’ and ‘b’ are the constants that are specific to each gas.
The equation can be written as:-
1. Cube power of volume:
V³ - b+(RT/P) V² + ( a / P) V - ab / P = 0
1. Reduced equation (Law of corresponding states) in terms of critical constants:
( 𝜋 + 3 / φ²) ( 3φ- 1 ) = 8τ
Where:
𝜋 = \[\frac{P}{P_{c}}\]
φ = \[\frac{V}{V_{c}}\]
τ = \[\frac{T}{T_{c}}\]
The units of Van Der Waals constants are:
For unit ‘a’ = atm lit² mol⁻²
For unit ‘b’ = litre mol⁻¹
Derivation of Van Der Waals Equation For Real Gases
It is easy to derive Van Der Waals equation for real gases but only if the right steps are followed. Any mistakes committed in the process to deduce Van Der Waals equation of state can be crucial and affect the whole process.
Let us discuss the process to derive the Van Der Waals gas equation.
In the case of a real gas when students are using Van Der Waals equation, the volume of a real gas is considered as (Vm - b), where b can be considered as the volume occupied by per mole.
Therefore, when the ideal gas law gets substituted with V = Vm - b, it is given as :
P(Vm - b) = nRT
Due to the presence of intermolecular attraction P was modified as follows.
( \[\frac{P+a}{V^{2}}\] ) ( Vm - b) = RT
( \[\frac{P+an^{2}}{V^{2}}\] ) ( V - nb) = nRT
Where,
Vm: molar volume of the gas
R: universal gas constant
T: temperature
P: pressure
V: volume
Thus, it is possible to reduce Van Der Waals equation to ideal gas law as PVm = RT.
Van Der Waals Derivation For One Mole of Gas
To derive the Van Der Waals equation or to deduce the Van Der Waal equation of state for one mole of gas can be turned into an easy process if the right steps are followed.
The steps for derivation of the real gas equation for one gas are as follows.
p = RT / V = (RT / v) p = RT / Vm - b
C = Na - Vm (proportionality between particle surface and number density)
a'C² = a' ( Na / Vm)² = a / Vm²
p = RT / (Vm - b) - a / Vm² => [ [ p + (a / Vm²)] ] Vm−b = RT
[ [ p + ( n²a / V²) ] ] V−nb = nRT
( substituting nVm = V )
Derivation of Real Gas Equation When Applied to Compressible Fluids
Van Der Waals equation derivation, when applied to compressible fluids, can be understood if the concepts are clear.
Compressible fluids like polymers have fluctuating specific volumes and this can be expressed as follows.
(p + A)(V - B) = CT
Where,
P: pressure
V: Specific volume
T: Temperature
A, B, C: Parameters
All this above information is used to derive the Van Der Waals equation of state.
Merits and Demerits of Van Der Waals Equation of State
Merits:
It can predict the behaviour of gas much better and accurately than the ideal gas equation.
It is also applicable to fluids in spite of gases.
The arrangement is made in a manner of cubic equation in volume. The cubic equation can give three volumes which can be used for calculating the volume at and below the critical temperatures.
Demerits:
It can only get accurate answers for real gases which are above the critical temperature.
Below critical temperature results also get accepted.
In the transition phase of gas, the equation is a failure.
Conclusion
More importantly, the Van Der Waals equation tends to take into consideration the molecular size and the molecular interaction forces which can be attractive or repulsive forces. Sometimes it is also known as Van Der Waals equation of state. In this article, students will learn how to derive the Van Der Waals equation of state.
FAQs on Van Der Waals Equation Derivation
1. What is the Van der Waals equation and what do its terms represent?
The Van der Waals equation is a modified version of the ideal gas law that describes the behavior of real gases. For 'n' moles of a gas, the equation is written as (P + an²/V²)(V - nb) = nRT. In this equation:
- P is the pressure of the gas.
- V is the volume of the container.
- n is the number of moles of the gas.
- R is the universal gas constant.
- T is the absolute temperature.
- The term 'a' corrects for intermolecular attractive forces, and 'b' corrects for the finite volume occupied by the gas molecules themselves.
2. How is the Van der Waals equation derived from the ideal gas law?
The derivation of the Van der Waals equation involves applying two main corrections to the ideal gas law (PV = nRT) to account for the properties of real gases:
- Volume Correction: The ideal gas law assumes molecules are point masses with zero volume. In reality, molecules occupy space. The term 'nb' (excluded volume) is subtracted from the total volume V, giving the actual space available for movement as (V - nb).
- Pressure Correction: The ideal gas law ignores intermolecular forces. In a real gas, attractive forces between molecules reduce the pressure they exert on the container walls. This is corrected by adding the term an²/V² to the observed pressure P.
3. What is the physical significance of the Van der Waals constants 'a' and 'b'?
The Van der Waals constants 'a' and 'b' are unique for each gas and represent key physical properties:
- Constant 'a': This constant is a measure of the magnitude of intermolecular attractive forces between gas molecules. A higher value of 'a' indicates stronger attraction, making the gas easier to liquefy.
- Constant 'b': This constant, also known as the excluded volume or co-volume, represents the effective volume occupied by the gas molecules themselves. It is approximately four times the actual volume of one mole of molecules. A larger 'b' value corresponds to larger molecules.
4. Why was the ideal gas law insufficient for describing real gases?
The ideal gas law fails for real gases, especially at high pressures and low temperatures, because it is based on two flawed assumptions of the Kinetic Theory of Gases:
- Assumption 1: Negligible Molecular Volume. It assumes gas molecules are point masses and their volume is negligible compared to the container's volume. This is untrue at high pressure when molecules are close together.
- Assumption 2: No Intermolecular Forces. It assumes there are no attractive or repulsive forces between molecules. This is incorrect as real gas molecules exhibit weak attractions (Van der Waals forces), which become significant at low temperatures when molecular motion slows down.
5. Under what specific conditions does a real gas behave most like an ideal gas?
A real gas behaves most like an ideal gas under conditions of very low pressure and very high temperature. According to the Van der Waals equation, (P + an²/V²)(V - nb) = nRT:
- At low pressure, the volume (V) is very large. This makes the pressure correction term (an²/V²) and the volume correction term (b, relative to V) so small that they become negligible.
- At high temperature, the kinetic energy of the molecules is very high, overcoming the weak intermolecular forces of attraction, making the 'a' term insignificant.
6. How do the values of constants 'a' and 'b' explain why some gases are easier to liquefy than others?
The ease of liquefaction of a gas is directly related to the strength of its intermolecular forces, which is represented by the Van der Waals constant 'a'. A gas with a higher 'a' value has stronger attractive forces between its molecules. These strong forces make it easier for the gas to transition into the liquid phase when pressure is applied and temperature is lowered. For example, ammonia (NH₃) has a much higher 'a' value than hydrogen (H₂) and is therefore much easier to liquefy.
7. What are the main limitations of the Van der Waals equation?
While the Van der Waals equation is a significant improvement over the ideal gas law, it has some limitations:
- It provides only a qualitative improvement and is not perfectly accurate for all real gases across all conditions.
- The constants 'a' and 'b' are not truly constant but show some variation with temperature.
- The equation is less accurate in the critical region and fails to accurately model the liquid-gas phase transition itself, where it predicts unrealistic oscillations (Van der Waals loops).
