SME
Engineering Thermodynamics · Problem type
Entropy change of an ideal gas
Use this for “Air goes from 290 K and 100 kPa to 450 K and 600 kPa. Find its entropy change per kilogram.”
Diagram
Δs
=
cₚ
ln(
T₂
T₁
)
−
R
ln(
P₂
P₁
)
=
0.1981
kJ/(kg·K):
it
depends
on
the
end
states
only,
so
any
path
between
them
(dashed)
gives
it.
This is an example. Type your own values to start a new problem.
Δs (kJ/(kg·K))
=
1.005
×
ln(
T₂ (K)
÷
T₁ (K)
)
−
0.287
×
ln(
P₂ (kPa)
÷
P₁ (kPa)
)
Formulas
Δs
=
1.005
×
ln(
T₂
T₁
)
−
0.287
×
ln(
P₂
P₁
)
+0.1981
=
1.005
×
ln(
500
300
)
−
0.287
×
ln(
300
100
)
Assumptions
•
Air is an ideal gas with constant cₚ = 1.005 kJ/(kg·K) and R = 0.287 kJ/(kg·K).
•
Temperatures in kelvins and absolute pressures; only their ratios matter.
•
Δs depends on the end states only, not on the path between them.
•
cₚ stays near 1.005 kJ/(kg·K) from about 250 K to 1,000 K.
Step-by-step
WE KNOW
T₁ = 300 K, P₁ = 100 kPa, T₂ = 500 K, P₂ = 300 kPa
FIND
entropy change (Δs)
Step 1 · Find entropy change (Δs)
Use Δs = 1.005 × ln(T₂ ÷ T₁) − 0.287 × ln(P₂ ÷ P₁)
Heating
adds
cₚ
ln(
T₂
T₁
);
squeezing
takes
away
R
ln(
P₂
P₁
).
Δs
=
1.005
×
ln(
500
300
)
−
0.287
×
ln(
300
100
)
Δs = 1.005 × ln(1.6667) − 0.287 × ln(3)
Δs = 1.005 × 0.51085 − 0.287 × 1.0986
Δs = 0.5134 − 0.3153
Δs = +0.1981 kJ/(kg·K)
Check
Put the numbers back in. Both sides should match.
+0.1981
=
1.005
×
ln(
500
300
)
−
0.287
×
ln(
300
100
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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