SME
Physical Chemistry I: Thermodynamics & Kinetics · Problem type
Reversible adiabatic expansion
Use this for “1.50 mol of a diatomic ideal gas at 400 K expands reversibly and adiabatically from 5 L to 15 L. Find its final temperature and w.”
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 negative.
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)
R
Cᵥ
For an ideal gas γ = Cₚ ÷ Cᵥ = 1 + R ÷ Cᵥ, so the picture’s T₂ = T₁(V₁/V₂)γ − 1 is this equation.
n
Amount of gas
example · mol
Cᵥ
Molar heat capacity at constant volume
example · J/(mol·K)
w
Work done on the gas
calculated · J
Formulas
T₂
=
T₁
×
(
V₁
V₂
)
8.314 ÷ Cᵥ
189
=
300
×
(
10
20
)
8.314 ÷ 12.47
w
=
n
×
Cᵥ
×
(
T₂
−
T₁
)
−1,384 = 1 × 12.47 × (189 − 300)
Assumptions
•
No heat crosses (q = 0), so ΔU = w: the gas cools as it expands and warms as it is squeezed.
•
An ideal gas with constant Cᵥ: 12.47 J/(mol·K) (3R/2) for a monatomic gas, 20.79 (5R/2) for a diatomic one.
•
Reversible, so Cᵥ dT = −RT dV ÷ V all the way, which integrates to T₂ = T₁(V₁ ÷ V₂)R/Cᵥ; R = 8.314 J/(mol·K).
•
w is the work done on the gas; temperatures are in kelvins.
Step-by-step
WE KNOW
n = 1 mol, Cᵥ = 12.47 J/(mol·K), T₁ = 300 K, V₁ = 10 L, V₂ = 20 L
FIND
final temperature (T₂), work done on the gas (w)
Step 1 · Find final temperature (T₂)
Use T₂ = T₁ × (V₁ ÷ V₂)8.314 ÷ Cᵥ
Along
a
reversible
adiabat
T
Cᵥ/R
×
V
stays
the
same,
so
T
changes
by
the
volume
ratio
to
the
power
R
Cᵥ
.
T₂
=
300
×
(
10
20
)
8.314 ÷ 12.47
T₂
=
300
×
0.5
0.6667
T₂
=
300
×
0.6299
T₂
=
189
K
→
T₂
<
T₁
:
the
gas
pays
for
its
work
out
of
its
own
energy,
so
it
cools.
Step 2 · Find work done on the gas (w)
Use w = n × Cᵥ × (T₂ − T₁)
q = 0, so the work on the gas is all of its change in internal energy, nCᵥΔT.
w
=
1
×
12.47
×
(189
−
300)
w
=
12.47
×
(−111)
w
=
−1,384
J
Check
Put the numbers back in. Both sides should match.
189
=
300
×
(
10
20
)
8.314 ÷ 12.47
✓
−1,384 = 1 × 12.47 × (189 − 300) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Laws of thermodynamics
Main lesson
Enthalpy and entropy of heating
Use this for “2.50 mol of liquid ethanol (Cₚ = 112.3 J/(mol·K)) warms at constant pressure from 293.15 K to 333.15 K. Find ΔH and ΔS.”
Reaction enthalpy at another temperature (Kirchhoff)
Use this for “A reaction has ΔH = −57.2 kJ/mol at 298.15 K and ΔCₚ = +12.6 J/(mol·K). Find ΔH at 400 K.”
State or path function?
Use this for “A gas goes from state 1 to state 2 by two different routes. Which of q, w, ΔU and ΔS are the same for both?”