The constants "a" and "b" in the van der waals equation are "empirical coefficients". What exactly does that mean?

Show that the units for "a" and "b" are appropriate for the the Van der Waals equation?

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http://en.wikipedia.org/wiki/Van_der_Waals_equation

http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/waal.html

http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/waal.html

http://chemistry.about.com/od/workedchemistryproblems/a/Ideal-Gas-Vs-Non-Ideal-Gas-Example-Problem.htm

The constants "a" and "b" in the Van der Waals equation are often referred to as empirical coefficients. This means that these values are determined experimentally rather than being derived from theoretical principles.

In the Van der Waals equation, "a" accounts for the attractive forces between gas molecules, while "b" accounts for the volume occupied by the individual gas molecules.

To show that the units for "a" and "b" are appropriate for the Van der Waals equation, let's consider the units for each coefficient:

1. Coefficient "a":
The Van der Waals equation is represented as (P + a * (n / V)^2) * (V - n * b) = nRT
The left side of the equation represents the pressure and volume correction due to the attractive forces between gas molecules.

The units for pressure (P) are typically in Pascals or atmospheres, the units for volume (V) are typically in liters, and the ideal gas constant (R) has units of J/(mol · K). The value of "n" represents the number of moles.

Rearranging the equation, the units of "a" can be determined as follows:
P = Pascals or atmospheres
V = liters
n = moles
T = Kelvin

(a * (n / V)^2) * (V - n * b) must have the same units as "P" in order for the equation to be dimensionally consistent. Simplifying, we have:
(Pascals or atmospheres) = (moles)^2 / (liters^2) * (liters - (moles * liters / mol))
(Pascals or atmospheres) = moles / liters

Hence, the units for "a" are given as (Pascals · liters^2) / (mol^2) or (atmospheres · liters^2) / (mol^2).

2. Coefficient "b":
The right side of the Van der Waals equation represents the volume correction due to the actual volume occupied by the gas molecules.

The units for volume (V) are typically in liters, and the units for "n" remain in moles.

To determine the appropriate units for "b," we need to make sure that the term (V - n * b) has the same units as "V":
liters = liters - (moles * liters / mol)

This equation suggests that the units for "b" are given as liters / mol.

In summary, the units for the empirical coefficients "a" and "b" in the Van der Waals equation are:
- "a" has units of (Pascals · liters^2) / (mol^2) or (atmospheres · liters^2) / (mol^2).
- "b" has units of liters / mol.

In the Van der Waals equation, "a" and "b" are empirical coefficients. The term "empirical" indicates that these coefficients are determined experimentally rather than being derived from first principles. They are introduced to account for the deviations from ideal gas behavior observed in real gases.

The coefficient "a" adjusts for the attractive forces between gas molecules, representing the strength of intermolecular attractions. It is the measure of the attractive forces that leads to the decrease in pressure compared to what would be expected for an ideal gas. "a" has units of pressure multiplied by volume squared (Pa·m^6/mol^2 or atm·L^2/mol^2).

The coefficient "b" accounts for the volume occupied by the gas molecules themselves, which causes the gas molecules to be effectively smaller than they would be in an ideal gas. "b" is the volume excluded by each mole of gas particles. It has units of volume (m^3/mol or L/mol).

To verify the units for "a" and "b" in the Van der Waals equation, we can examine the equation itself:

[P + (a(n/V)^2)](V - nb) = nRT

Looking at the left-hand side of the equation:

[P] has units of pressure (Pa or atm)
[a(n/V)^2] has units of (pressure) * (volume^2) / (volume^2) = pressure (Pa or atm)
[V - nb] has units of volume (m^3 or L)

On the right-hand side of the equation:

n has units of amount of substance (moles)
R is the ideal gas constant with units that depend on the choice of pressure unit
T is the temperature with units of Kelvin (K)

Therefore, the units for the Van der Waals equation are appropriate, as both sides of the equation have consistent units for their terms.