SME
Thermal & Statistical Physics · Problem type
Adiabatic expansion
Use this for “2.00 mol of a diatomic ideal gas at 400 K expands slowly with no heat in or out, from 4 L to 12 L. Find its final temperature and the work it does.”
Diagram
P₂
=
P₁
(V₁/V₂)
γ
P₂
=
249
×
(
10
20
)
1.667
=
78.6
kPa
T₂ = T₁(V₁/V₂)γ − 1 = 189 K, so the gas cools.
W
=
P₁V₁ − P₂V₂
γ − 1
=
+1,384
J
The shaded area is 1,385 J.
The gas expands: it does work, so W is positive.
Dashed: the isotherms T₁ = 300 K, T₂ = 189 K.
This is an example. Type your own values to start a new problem.
T₂ (K)
=
T₁ (K)
V₁ (L)
V₂ (L)
−
1
n
Amount of gas
example · mol
CV
Molar heat capacity
worked out (not typed) · J/(mol·K)
W
Work done by the gas
calculated · J
Formulas
T₂
=
T₁
×
(
V₁
V₂
)
γ
− 1
189
=
300
×
(
10
20
)
5
3
− 1
CV
=
8.314
γ
− 1
12.471
=
8.314
5
3
− 1
W
=
n
×
CV
×
(
T₁
−
T₂
)
+1,384 = 1 × 12.471 × (300 − 189)
Assumptions
•
Adiabatic: no heat crosses (Q = 0), so ΔU = −W and the gas cools as it does work.
•
An ideal gas: γ = 5/3 for a monatomic gas, 7/5 for a diatomic one; CV = R ÷ (γ − 1), R = 8.314 J/(mol·K).
•
Slow (quasi-static), so TVγ−1 stays the same; W is the work done by the gas; temperatures in kelvins.
Step-by-step
WE KNOW
n = 1 mol, γ = 5/3, T₁ = 300 K, V₁ = 10 L, V₂ = 20 L
FIND
final temperature (T₂), molar heat capacity (CV), work done by the gas (W)
Step 1 · Find final temperature (T₂)
Use T₂ = T₁ × (V₁ ÷ V₂)ᵞ⁻¹
Along
a
slow
adiabat
T
V
γ
− 1
stays
the
same,
so
T
changes
by
the
volume
ratio
to
the
power
γ
−
1.
T₂
=
300
×
(
10
20
)
5
3
− 1
T₂
=
300
×
0.5
2
3
T₂
=
300
×
0.63
T₂
=
189
K
→
T₂
<
T₁
:
the
gas
pays
for
its
work
out
of
its
own
energy,
so
it
cools.
Step 2 · Find molar heat capacity (CV)
Use CV = 8.314 ÷ (γ − 1)
CP
−
CV
=
R
and
CP
CV
=
γ
together
give
CV
=
R
γ
− 1
.
CV
=
8.314
5
3
− 1
CV
=
8.314
2
/3
CV = 12.471 J/(mol·K)
Step 3 · Find work done by the gas (W)
Use W = n × CV × (T₁ − T₂)
Q
=
0,
so
the
work
the
gas
does
is
all
taken
from
its
internal
energy:
W
=
−ΔU
=
nCV(
T₁
−
T₂
).
W
=
1
×
12.471
×
(300
−
189)
W
=
12.471
×
111
W
=
+1,384
J
Check
Put the numbers back in. Both sides should match.
189
=
300
×
(
10
20
)
5
3
− 1
✓
12.471
=
8.314
5
3
− 1
✓
+1,384.4287 = 1 × 12.471 × (300 − 188.9882) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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