The_Cambridge_Handbook_of_Physics_Formulas

5.3 Gas laws
111
Van der Waals gas
Equation of state
(
p+ a )
Vm
2 (V m −b)=RT (5.67)
p
V m
R
T
a,b
pressure
molar volume
molar gas constant
temperature
van der Waals’ constants
Critical point
T c =8a/(27Rb) (5.68)
p c = a/(27b 2 ) (5.69)
V mc =3b (5.70)
T c
p c
V mc
critical temperature
critical pressure
critical molar volume
Reduced equation
of state
(
p r + 3 )
Vr
2 (3V r −1) = 8T r (5.71)
p r
V r
T r
= p/p c
= V m /V mc
= T/T c
Dieterici gas
Equation of state p = RT
V m −b ′ exp ( −a
′
RTV m
)
(5.72)
p pressure
V m molar volume
R molar gas constant
T temperature
a ′ ,b ′ Dieterici’s constants
5
Critical point
Reduced equation
of state
T c = a ′ /(4Rb ′ ) (5.73)
p c = a ′ /(4b ′2 e 2 ) (5.74)
V mc =2b ′ (5.75)
p r =
T (
r
2V r −1 exp 2− 2 )
V r T r
(5.76)
T c critical temperature
p c critical pressure
V mc critical molar volume
e =2.71828...
p r = p/p c
V r = V m /V mc
T r = T/T c
pr
1.4
1.2
1
0.8
0
0
0.9
Van der Waals gas
1.1
1.0
0.6 0.8
0.4 0.6
0.4
0.2 0.8
1
T r =1.2
pr
2
1.8
1.6
1.4
1.2
1
Dieterici gas
T r =1.2
1.1
1.0
0.9
0.8
0
0.2
2 3 4 5
0
1 2 3 4 5
V r
V r