# 9.6 Non-ideal gas behavior

 Page 1 / 5
By the end of this section, you will be able to:
• Describe the physical factors that lead to deviations from ideal gas behavior
• Explain how these factors are represented in the van der Waals equation
• Define compressibility (Z) and describe how its variation with pressure reflects non-ideal behavior
• Quantify non-ideal behavior by comparing computations of gas properties using the ideal gas law and the van der Waals equation

Thus far, the ideal gas law, PV = nRT , has been applied to a variety of different types of problems, ranging from reaction stoichiometry and empirical and molecular formula problems to determining the density and molar mass of a gas. As mentioned in the previous modules of this chapter, however, the behavior of a gas is often non-ideal, meaning that the observed relationships between its pressure, volume, and temperature are not accurately described by the gas laws. In this section, the reasons for these deviations from ideal gas behavior are considered.

One way in which the accuracy of PV = nRT can be judged is by comparing the actual volume of 1 mole of gas (its molar volume, V m ) to the molar volume of an ideal gas at the same temperature and pressure. This ratio is called the compressibility factor (Z)    with:

$\text{Z}=\phantom{\rule{0.2em}{0ex}}\frac{\text{molar volume of gas at same}\phantom{\rule{0.2em}{0ex}}T\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}P}{\text{molar volume of ideal gas at same}\phantom{\rule{0.2em}{0ex}}T\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}P}\phantom{\rule{0.2em}{0ex}}={\left(\frac{P{V}_{m}}{RT}\right)}_{\text{measured}}$

Ideal gas behavior is therefore indicated when this ratio is equal to 1, and any deviation from 1 is an indication of non-ideal behavior. [link] shows plots of Z over a large pressure range for several common gases.

As is apparent from [link] , the ideal gas law does not describe gas behavior well at relatively high pressures. To determine why this is, consider the differences between real gas properties and what is expected of a hypothetical ideal gas.

Particles of a hypothetical ideal gas have no significant volume and do not attract or repel each other. In general, real gases approximate this behavior at relatively low pressures and high temperatures. However, at high pressures, the molecules of a gas are crowded closer together, and the amount of empty space between the molecules is reduced. At these higher pressures, the volume of the gas molecules themselves becomes appreciable relative to the total volume occupied by the gas ( [link] ). The gas therefore becomes less compressible at these high pressures, and although its volume continues to decrease with increasing pressure, this decrease is not proportional as predicted by Boyle’s law.

At relatively low pressures, gas molecules have practically no attraction for one another because they are (on average) so far apart, and they behave almost like particles of an ideal gas. At higher pressures, however, the force of attraction is also no longer insignificant. This force pulls the molecules a little closer together, slightly decreasing the pressure (if the volume is constant) or decreasing the volume (at constant pressure) ( [link] ). This change is more pronounced at low temperatures because the molecules have lower KE relative to the attractive forces, and so they are less effective in overcoming these attractions after colliding with one another.

what is the meaning of intermolecular force
is the force of attraction that exist between two or more molecules
Johnson
What is a primary standard solution ?
Duval
a known solution
Fiko
Characteristic of a primary standard solution
Duval
pauli's exclusion is based on what?
What is greatest modification made in dalton's atomic theory?
Types of electrolytes
Strong, weak and non-electrolytes
Grace
welcome
Alieu
thanks what's this platform all about
Nnamdi
list 6 subatomic particles and their mass, speed and charges
combination of acid and base
that salt
Talhatu
calculate the mass in gram of NaOH present in 250cm3 of 0.1mol/dm3 of its solution
The mass is 1.0grams. First you multiply the molecular weight and molarity which is 39.997g/mol x 0.1mol/dm3= 3.9997g/dm3. Then you convert dm3 to cm3. 1dm3 =1000cm3. In this case you would divide 3.9997 by 1000 which would give you 3.9997*10^-3 g/cm3. To get the mass you multiply 3.9997*10^-3 and
Kokana
250cm3 and get the mass as .999925, with significant figures the answer is 1.0 grams
Kokana
nitrogen, phosphorus, arsenic, antimony and Bismuth
What is d electronic configuration of for group 5
Can I know d electronic configuration of for group 5 elements
Miracle
2:5, 2:8:5, 2:8:8:5,...
Maxime
Thanks
Miracle
Pls what are d names of elements found in group 5
Miracle
define define. define
what is enthalpy
total heat contents of the system is called enthalpy, it is state function.
Sajid
background of chemistry
what is the hybridisation of carbon in formic acid?
sp2 hybridization
Johnson
what is the first element
HYDROGEN
Liklai
Element that has positive charge and its non metal Name the element
Liklai
helium
oga
sulphur
oga
hydrogen
Banji
account for the properties of organic compounds
properties of organic compounds
mercy