Problem 2.1.3

A gas follows van der Waals equation of state. What is the specific volume of that gas at 80 bars and 100 Kelvin? The gas has the following values for van der Waals constants. First constant, a, is equal to 0.141 meter⁶ Pascal per mol². Second constant, b is equal to 3.913 times 10^-5 cubic meter per mole.

Solution:

Three classes of equations of state are available in the literature.

1. The virial equation of state is a polynomial series in inverse volume which is explicit in pressure. P is equal to RT over V plus RTB over V² plus RTC over V³ plus …. The parameters B, C,… are called the second, third, … virial coefficients and are functions only of temperature for a pure fluid. The virial equation may also be written as a power series in either V or P. For non-polar molecules BP_c over RT_c is equal to B⁽⁰⁾ plus omega B⁽¹⁾. In this relationship, B⁽⁰⁾ and B⁽¹⁾ are functions of reduced temperature.
2. Then we have equations which are cubic in volume. These equations can represent both liquid and vapor behavior of nonpolar molecules over limited ranges of temperature and pressure, and they remain relatively simple from a computational point of view.
3. Then there is generalized version of the Benedict-Webb-Rubn equation, which is applicable over broader ranges of temperatures and pressure than are the cubic equations.

The term “cubic equation of state” implies an equation which, if expanded, would contain volume terms raised to either the first, second, or third power. Many of the two-parameter cubic equations can be expressed by the equation. P is equal to RT over (V minus b) minus a over (V² plus ubV plus wb²).

The value of u and w are zero for the van der Waals equation. P is equal to RT over (V minus b) minus a over V^2.

There are several approaches which have been used to set the values of the two parameters, a and b. One approach is to choose a and b so that the two critical point conditions partial P over partial V at constant T_c is equal to zero, and partial P² over partial V² at constant T_c is equal to zero are satisfied. These conditions are only applicable to pure components.

Following this approach, b is found to be equal to RT_C over 8P_c and a is equal to 27/64 R²T_c² over P_c.

The value of u is equal to and w is zero for the Redlich Kwong equation. P is equal to RT over (V minus b) minus a over (V² plus bV).

Following this approach, b is found to be equal to 0.08664RT_C over P_c and a is equal to 0.42748 R²T_c^2.5 over (P_cT^1/2).

PVT relationship for a gas that follows van der Waals equation is given as a product of P plus a over v² and (v minus b) and this product is equal to RT. The term a is a measure of the attractive force between the molecules. The term b is due to the finite volume of the molecules and to their general incompressibility. Replacing lower case v with upper case V over n, results in the equation for n number of moles of the gas. One can see that this is a cubic equation in volume. For a given temperature, pressure, constant a and b, there are three roots, but we are looking one which is real, is the lowest, and has a positive value. We could use trial and error approach to find this value. Any way the starting point is the volume assuming ideal behavior and that is given by RT over P. We could follow Newton’s method. Let‘s write the equation as a function, and we are interested to find the roots of that function equal to zero. Let’s obtain its derivative. Choose the ideal gas behavior and obtain the starting value of volume. Calculate the value of the function at this value of V; also obtain the value of the derivative. Next iteration value is obtained by subtracting ratio of function and derivative values from the previous iteration value. We could see that the value of the function approaches to almost zero in just three iterations. Real volume is 5.38 ×10^-5 cubic meter.

Compressibility factor is equal to 0.518 as ideal gas volume is 1.04 ×10^-4.

A table is provided for you from CRC Handbook that provides van der Waals’ constants for gases.