1.3 Real Gases 21
1.3 Real Gases
Most gases obey the ideal gas law to a good approximation when near room temperature
and at a moderate pressure. At higher pressures one might need a better description.
Several equations of state have been devised for this purpose. Thevan der Waals
equation of stateis
(
P+
an^2
V^2
)
(V−nb)nRT (1.3-1)
The symbolsaandbrepresent constant parameters that have different values for dif-
ferent substances. Table A.3 in Appendix A gives values of van der Waals parameters
for several substances.
The van der Waals equation of state is
named for Johannes Diderik van der
Waals,1837–1923, a Dutch physicist
who received the 1910 Nobel Prize in
physics for his work on equations of
state.
We solve the van der Waals equation forPand note thatPis actually a function of
only two intensive variables, the temperatureTand the molar volumeVm, defined to
equalV/n.
P
nRT
V−nb
−
an^2
V^2
RT
Vm−b
−
a
Vm^2
(1.3-2)
This dependence illustrates the fact thatintensive variables such as pressure cannot
depend on extensive variablesand that theintensive state of a gas or liquid of one
substance is specified by only two intensive variables.
EXAMPLE 1.8
Use the van der Waals equation to calculate the pressure of nitrogen gas at 273.15 K and
a molar volume of 22.414 L mol−^1. Compare with the pressure of an ideal gas at the same
temperature and molar volume.
Solution
P
(
8 .134JK−^1 mol−^1
)
( 273 .15 K)
0 .022414 m^3 mol−^1 − 0 .0000391 m^3 mol−^1
−
0 .1408 Pa m^3 mol−^1
(
0 .022414 m^3 mol−^1
) 2
1. 0122 × 105 Pa 0 .9990 atm
For the ideal gas
P
RT
Vm
(
8 .134JK−^1 mol−^1
)
( 273 .15 K)
0 .022414 m^3 mol−^1
1. 0132 × 105 Pa 1 .0000 atm
Exercise 1.7
a.Show that in the limit thatVmbecomes large, the van der Waals equation becomes identical
to the ideal gas law.
b.Find the pressure of 1.000 mol of nitrogen at a volume of 24.466 L and a temperature of
298.15 K using the van der Waals equation of state. Find the pressure of an ideal gas under
the same conditions.