Van der Waal's Equation
Van der Waal's Equation
Van der Waals equation is an equation of state for a fluid composed of particles that have a non-zero size and a pairwise attractive force (i.e., van der Waals forces) between them. It was derived by Johannes Diderik van der Waals in 1873, who received the Nobel Prize in 1910 for his work on the equation of state for gases and liquids.
Ideal Gas Law
Before diving into the Van der Waals equation, let's recall the Ideal Gas Law, which is a good approximation for the behavior of real gases under many conditions, although it has its limitations. The Ideal Gas Law is given by:
[ PV = nRT ]
Where:
- ( P ) is the pressure of the gas,
- ( V ) is the volume of the gas,
- ( n ) is the number of moles of the gas,
- ( R ) is the universal gas constant,
- ( T ) is the temperature of the gas in Kelvin.
Van der Waals Equation
The Van der Waals equation modifies the Ideal Gas Law to account for the finite size of molecules and the attractive forces between them. The equation is written as:
[ \left( P + \frac{an^2}{V^2} \right) (V - nb) = nRT ]
Where:
- ( a ) is a measure of the average attraction between particles,
- ( b ) is the volume occupied by one mole of particles.
Explanation of the Corrections
Pressure Correction: Real particles exert attractive forces on each other, which means that the pressure exerted by the gas on the walls of its container is less than it would be if the particles did not attract each other. The term ( \frac{an^2}{V^2} ) adds an amount to the observed pressure to account for this attraction.
Volume Correction: Real particles have a finite volume, which means that the space available for particles to move is less than the total volume of the container. The term ( nb ) subtracts the volume occupied by the particles from the total volume of the container.
Differences Between Ideal Gas and Van der Waals Gas
| Property | Ideal Gas | Van der Waals Gas |
|---|---|---|
| Particle Size | Negligible (point particles) | Finite size |
| Particle Interaction | None (no forces between particles) | Attractive forces between particles |
| Equation of State | ( PV = nRT ) | ( (P + \frac{an^2}{V^2})(V - nb) = nRT ) |
| Compressibility Factor | ( Z = \frac{PV}{nRT} = 1 ) | ( Z = \frac{PV}{nRT} \neq 1 ) |
| Applicability | High temperatures, low pressures | Closer to real gases' behavior |
Examples
Example 1: Calculating Pressure Using Van der Waals Equation
Suppose we have 1 mole of a gas at a temperature of 300 K in a 1-liter container. The Van der Waals constants for the gas are ( a = 3.592 \, \text{L}^2 \cdot \text{atm} \cdot \text{mol}^{-2} ) and ( b = 0.04267 \, \text{L} \cdot \text{mol}^{-1} ). Calculate the pressure exerted by the gas using the Van der Waals equation.
Solution:
Given:
- ( n = 1 \, \text{mol} )
- ( T = 300 \, \text{K} )
- ( V = 1 \, \text{L} )
- ( a = 3.592 \, \text{L}^2 \cdot \text{atm} \cdot \text{mol}^{-2} )
- ( b = 0.04267 \, \text{L} \cdot \text{mol}^{-1} )
- ( R = 0.0821 \, \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} )
Plugging these values into the Van der Waals equation:
[ \left( P + \frac{(3.592)(1)^2}{(1)^2} \right) (1 - (0.04267)(1)) = (1)(0.0821)(300) ]
[ \left( P + 3.592 \right) (0.95733) = 24.63 ]
[ P + 3.592 = \frac{24.63}{0.95733} ]
[ P = \frac{24.63}{0.95733} - 3.592 ]
[ P \approx 22.39 \, \text{atm} ]
Example 2: Understanding the Impact of ( a ) and ( b )
Consider two gases, Gas X and Gas Y, with Van der Waals constants ( a_X ), ( b_X ) and ( a_Y ), ( b_Y ) respectively. If ( a_X > a_Y ) and ( b_X = b_Y ), Gas X has stronger intermolecular forces than Gas Y. If ( a_X = a_Y ) and ( b_X > b_Y ), Gas X has larger molecules than Gas Y.
Understanding Van der Waals equation is crucial for predicting the behavior of real gases, especially under conditions of high pressure and low temperature, where deviations from the Ideal Gas Law are significant. It is also essential for understanding phenomena such as critical points and phase transitions in fluids.