SME
General Chemistry I · Problem type
Real gases: the van der Waals pressure
Use this for “Find the pressure of 1.00 mol of CO₂ in 0.500 L at 300 K with the van der Waals equation (a = 3.59 L²·atm/mol², b = 0.0427 L/mol). Compare it with the ideal pressure.”
Diagram
P(ideal)
=
nRT
V
=
1
×
0.08206
×
300
0.5
=
49.2
atm
P
=
nRT
V − nb
−
an²
V
2
=
53.8
−
14.4
P = 39.5 atm
Z
=
PV
nRT
=
0.8017:
below
1,
so
attraction
wins
for
CO₂.
The band is nb, 8.54% of V; each dashed pair is attraction.
This is an example. Type your own values to start a new problem.
P (atm)
=
n (mol)
×
R
×
T (K)
÷
(
V (L)
−
n (mol)
×
b (L/mol)
)
−
a (L²·atm/mol²)
×
(
n (mol)
)
2
÷
(
V (L)
)
2
Pid
Ideal pressure
calculated · atm
Pfree
Pressure in the free volume
calculated · atm
Pₐₜₜ
Pressure lost to attraction
calculated · atm
Z
Compressibility factor
calculated
Formulas
Pid
=
n
×
0.08206
×
T
V
49.24
=
1
×
0.08206
×
300
0.5
Pfree
=
n
×
0.08206
×
T
V
−
n
×
b
53.83
=
1
×
0.08206
×
300
0.5 − 1 × 0.0427
Pₐₜₜ
=
a
×
n
2
V
2
14.36
=
3.59
×
1
2
0.5
2
P
=
Pfree
−
Pₐₜₜ
39.47 = 53.83 − 14.36
Z
=
P
Pid
0.8017
=
39.47
49.24
Assumptions
•
b is the room one mole of the molecules takes up itself, so they move in only V − nb. a measures their pull on each other, which takes an² ÷ V² off the push on the walls.
•
R = 0.08206 L·atm/(mol·K), so the volume is in liters, the pressures in atm and the temperature in kelvins. Read a and b for your gas from a table.
•
The compressibility factor Z = PV ÷ nRT is the real pressure over the ideal one: Z < 1 means attraction wins, Z > 1 means the molecules’ own volume wins.
Step-by-step
WE KNOW
n = 1 mol, V = 0.5 L, T = 300 K, a = 3.59 L²·atm/mol², b = 0.0427 L/mol
FIND
ideal pressure (Pid), pressure in the free volume (Pfree), pressure lost to attraction (Pₐₜₜ), van der Waals pressure (P), compressibility factor (Z)
Step 1 · Find ideal pressure (Pid)
Use Pid = n × 0.08206 × T ÷ V
The
ideal
gas
law
solved
for
P
:
the
pressure
if
the
molecules
had
no
size
and
no
pull.
Pid
=
1
×
0.08206
×
300
0.5
Pid
=
24.618
0.5
Pid = 49.24 atm
Step 2 · Find pressure in the free volume (Pfree)
Use Pfree = n × 0.08206 × T ÷ (V − n × b)
The
ideal
gas
law
in
the
room
the
molecules
leave
free,
V
−
nb:
less
room,
more
hits
on
the
walls.
Pfree
=
1
×
0.08206
×
300
0.5 − 1 × 0.0427
Pfree
=
0.08206
×
300
0.4573
Pfree
=
24.618
0.4573
Pfree = 53.83 atm
Step 3 · Find pressure lost to attraction (Pₐₜₜ)
Use Pₐₜₜ = a × n² ÷ V²
A
molecule
about
to
hit
the
wall
is
pulled
back
by
its
neighbors.
Both
counts
grow
with
n
V
,
so
the
loss
goes
as
a
(
n
V
)
2
.
Pₐₜₜ
=
3.59
×
1
2
0.5
2
Pₐₜₜ
=
3.59
0.25
Pₐₜₜ = 14.36 atm
Step 4 · Find van der Waals pressure (P)
Use P = Pfree − Pₐₜₜ
The van der Waals equation: the push in the free volume, less what the attraction takes off.
P
=
53.83
−
14.36
P
=
39.47
atm
Step 5 · Find compressibility factor (Z)
Use Z = P ÷ Pid
Z
=
PV
nRT
is
the
real
pressure
over
the
ideal
one
at
the
same
n
,
V
and
T
.
Z
=
39.47337
49.236
Z
=
0.8017
Check
Put the numbers back in. Both sides should match.
49.24
=
1
×
0.08206
×
300
0.5
✓
53.83
=
1
×
0.08206
×
300
0.5 − 1 × 0.0427
✓
14.36
=
3.59
×
1
2
0.5
2
✓
39.47 = 53.83 − 14.36 ✓
0.8017
=
39.47337
49.236
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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