Compressibility factor how to calculate, examples and exercises

The compressibility factor Z, or compression factor for gases, is a dimensionless value (without units) that is introduced as a correction in the equation of state of ideal gases. In this way, the mathematical model more closely resembles the observed behavior of gas..

In the ideal gas, the equation of state that relates to the variables P (pressure), V (volume) and T (temperature) is: P.V _ideal = n.R.T with n = number of moles and R = ideal gas constant. Adding the correction for the compressibility factor Z, this equation becomes:

P.V = Z.n.R.T

Article index

• 1 How to calculate compressibility factor?
• 2 Examples
  • 2.1 The compressibility factor in ideal gases, air, hydrogen and water
• 3 Solved exercises
  • 3.1 Exercise 1
  • 3.2 Exercise 2
• 4 References

How to calculate compressibility factor?

Taking into account that the molar volume is V_cool = V / n, we have the real molar volume:

P . V_real = Z. R. T → Z = PV _real/ RT

Since the compressibility factor Z depends on gas conditions, it is expressed as a function of pressure and temperature:

Z = Z (P, T)

Comparing the first two equations, it can be seen that if the number of moles n is equal to 1, the molar volume of a real gas is related to that of the ideal gas by:

V_real / V_ideal = Z → V _real = Z V_ideal

When the pressure exceeds 3 atmospheres, most of the gases stop behaving as ideal gases and the real volume differs significantly from the ideal.

This was realized in his experiments by the Dutch physicist Johannes Van der Waals (1837-1923), which led him to create a model that was better suited to practical results than the ideal gas equation: the Van equation of state. der Waals.

Examples

According to the equation P.V_real= Z.n.RT, For an ideal gas, Z = 1. However, in real gases, as the pressure increases, so does the value of Z. This makes sense because the higher the pressure the gas molecules have more opportunities to collide, therefore the repulsive forces increase and with it the volume.

On the other hand, at lower pressures, the molecules move more freely and the repulsive forces decrease. Therefore a lower volume is expected. As for the temperature, when it increases, Z decreases.

As Van der Waals observed, in the vicinity of the so-called critical point, the behavior of the gas deviates greatly from that of an ideal gas..

The critical point (T_c, P_c) of any substance are the pressure and temperature values ​​that determine its behavior before a phase change:

-T_c is the temperature above which the gas in question does not liquefy.

-P_c  is the minimum pressure required to liquefy the gas at temperature T_c

Each gas has its own critical point, however, defining the temperature and the reduced pressure T_r And p_r as follows:

P_r = P / P_c

V_r = V / V_c

T_r = T / T_c

It is observed that a confined gas with identical V_r Y T_r exerts the same pressure P_r. For this reason, if Z is plotted as a function of P_r to oneself T_r, every point on that curve is the same for any gas. This is called principle of corresponding states.

The compressibility factor in ideal gases, air, hydrogen and water

Below is a compressibility curve for various gases at various reduced temperatures. Here are some examples of Z for some gases and a procedure to find Z using the curve.

Ideal gases

Ideal gases have Z = 1, as explained at the beginning.

Air

For air Z is approximately 1 in a wide range of temperatures and pressures (see figure 1), where the ideal gas model gives very good results.

Hydrogen

Z> 1 for all pressures.

Water

To find Z for water, you need the critical point values. The critical point of the water is: P_c = 22.09 MPa and T_c= 374.14 ° C (647.3 K). Again it is necessary to take into account that the compressibility factor Z depends on temperature and pressure..

For example, suppose you want to find Z of water at 500 ºC and 12 MPa. Then the first thing to do is to calculate the reduced temperature, for which the degrees Celsius must be converted to Kelvin: 50 ºC = 773 K:

T_r = 773 / 647.3 = 1.2

P_r = 12 / 22.09 = 0.54

With these values ​​we place in the graph of the figure the curve corresponding to T_r = 1.2, indicated by a red arrow. Then we look on the horizontal axis for the value of P_r closer to 0.54, marked in blue. Now we draw a vertical until we intercept the curve T_r = 1.2 and finally it is projected from that point to the vertical axis, where we read the approximate value of Z = 0.89.

Solved exercises

Exercise 1

There is a gas sample at a temperature of 350 K and a pressure of 12 atmospheres, with a molar volume 12% greater than that predicted by the ideal gas law. Calculate:

a) Compression factor Z.

b) Molar volume of gas.

c) According to the previous results, indicate which are the dominant forces in this gas sample.

Data: R = 0.082 L.atm / mol.K

Solution to

Knowing that V _real  is 12% greater than V_ideal :

V_real  = 1.12V_ideal

Z = V _real / V_ideal = 1.12

Solution b

P. V_real = Z. R. T → V_real = (1.12 x 0.082 x 350/12) L / mol = 2.14 L / mol.

Solution c

The repulsive forces are those that predominate, since the volume of the sample increased.

Exercise 2

There are 10 moles of ethane confined in a volume of 4.86 L at 27 ºC. Find the pressure exerted by ethane from:

a) The ideal gas model

b) The van der Waals equation

c) Find the compression factor from the previous results.

Data for ethane

Van der Waals coefficients:

a = 5,489 dm⁶. atm. mole^-two  and b = 0.06380 dm³. mole^-1.

Critical pressure: 49 atm. Critical temperature: 305 K

Solution to

The temperature is passed to kelvin: 27 º C = 27 +273 K = 300 K, also remember that 1 liter = 1 L = 1 dm³.

Then the supplied data is substituted into the ideal gas equation:

P.V = n.R.T → P = (10 x 0.082 x 300 / 4.86 L) atm = 50.6 atm

Solution b

The Van der Waals equation of state is:

Where a and b are the coefficients given by the statement. When clearing P:

Solution c

We calculate the reduced pressure and temperature:

P_r = 35.2 / 49 = 0.72

T_r = 300/305 = 0.98 ≈ 1

With these values ​​we look for the value of Z in the graph of figure 2, finding that Z is approximately 0.7.

 References

1. Atkins, P. 1999. Physical Chemistry. Omega editions.
2. Cengel, Y. 2012. Thermodynamics. 7^ma Edition. Mcgraw hill.
3. Engel, T. 2007. Introduction to Physicochemistry: Thermodynamics. Pearson.
4. Levine, I. 2014. Principles of Physico-chemistry. 6th. Edition. Mcgraw hill.
5. Wikipedia. Compressibility Factor. Recovered from: en.wikipedia.org.