PHYSICAL CHEMISTRY IN BRIEF

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CHAP. 2: STATE BEHAVIOUR [CONTENTS] 51

2.2.6 Van der Waals equation of state


p =

nRT
V−nb

−a

(
n
V

) 2
=

RT

Vm−b


a
Vm^2

,

z =

Vm
Vm−b


a
RT Vm

. (2.23)

Parametersa andbare the constants of the van der Waals equation of state. Their values
depend on the kind of gas being described. They can be obtained from experimental data on
state behaviour, or they can be estimated from the critical quantities of substances using the
relations


a=

27

64

(RTc)^2
pc

, b=

1

8

RTc
pc

, (2.24)

which follow from equations (2.5) and (2.23).
For the second virial coefficient, the van der Waals equation yields


B=b−

a
RT

=

1

8

RTc
pc

(
1 −

27

8

Tc
T

)

. (2.25)


For the Boyle temperature, the van der Waals equation of state yields


TB=

a
Rb

=

27

8

Tc= 3. 375 Tc. (2.26)

Example
Using the van der Waals equation of state calculate the pressure and compressibility factor of a
substance whose critical temperature is 800 K and critical pressure is 8.314 MPa, at a temperature
of 1000 K and a molar volume of 1100 cm^3 mol−^1.
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