SME
Engineering Thermodynamics · Problem type
Polytropic boundary work
Use this for “Gas at 100 kPa is compressed from 0.8 m³ to 0.2 m³ along PVⁿ = constant with n = 1.3. Find P₂ and the work.”
Chart
Shaded: ∫ from 0.8 to 0.2 of P(x) dx = −137.5
This is an example. Type your own values to start a new problem.
P₂ (kPa)
=
P₁ (kPa)
×
(
V₁ (m³)
÷
V₂ (m³)
)
W
Boundary work
calculated · kJ
Formulas
P₂
=
P₁
×
(
V₁
V₂
)
n
606.3
=
100
×
(
0.8
0.2
)
1.3
W
=
P₂
×
V₂
−
P₁
×
V₁
1 −
n
−137.5
=
606.3 × 0.2 − 100 × 0.8
1 − 1.3
Assumptions
•
A slow (quasi-static) process along PVⁿ = constant; n = 1 is isothermal for an ideal gas.
•
W = ∫P dV is the work done by the gas: negative when it is compressed.
•
Pressures in kPa and volumes in m³, so the work comes out in kJ.
Step-by-step
WE KNOW
P₁ = 100 kPa, V₁ = 0.8 m³, V₂ = 0.2 m³, n = 1.3
FIND
final pressure (P₂), boundary work (W)
Step 1 · Find final pressure (P₂)
Use P₂ = P₁ × (V₁ ÷ V₂)ⁿ
P₁V₁ⁿ = P₂V₂ⁿ along the path, so divide by V₂ⁿ.
P₂
=
100
×
(
0.8
0.2
)
1.3
P₂
=
100
×
4
1.3
P₂
=
100
×
6.063
P₂
=
606.3
kPa
Step 2 · Find boundary work (W)
Use W = (P₂ × V₂ − P₁ × V₁) ÷ (1 − n)
W
=
∫P
dV
with
PVⁿ
=
C
gives
P₂V₂
−
P₁V₁
1 −
n
.
1.3
≠
1:
n
≠
1
W
=
606.3 × 0.2 − 100 × 0.8
1 − 1.3
W
=
121.26 − 80
−0.3
W
=
41.257
−0.3
W
=
−137.5
kJ
Check
Put the numbers back in. Both sides should match.
606.3
=
100
×
(
0.8
0.2
)
1.3
✓
−137.5
=
606.3 × 0.2 − 100 × 0.8
1 − 1.3
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
First law
Main lesson
Turbine power by the steady-flow energy balance
Use this for “2 kg/s of steam enters a turbine at 3230 kJ/kg and leaves at 2600 kJ/kg, losing 20 kW of heat. Find the power.”
Nozzle exit speed
Use this for “Steam enters a nozzle at 30 m/s and its enthalpy drops 50 kJ/kg. How fast does it leave?”
Mixing chamber outlet enthalpy
Use this for “2 kg/s of water at 335 kJ/kg mixes with 3 kg/s at 84 kJ/kg. Find the outlet enthalpy.”