SME
Engineering Thermodynamics
First law
Use this for “0.5 kg of air at 200 kPa is heated at constant pressure from 300 K to 500 K. Find the work, ΔU and the heat added.”
Diagram
First law: ΔU = Q − W = 100.5 − 28.7 = 71.8 kJ
Q is + when heat flows into the gas; W is + when the gas does work by pushing the piston out.
More energy came in than went out as work: the internal energy rises and the gas warms.
This is an example. Type your own values to start a new problem.
W (kJ)
=
P (kPa)
×
(
V₂ (m³)
−
V₁ (m³)
)
ΔU (kJ)
=
m (kg)
×
0.718
×
(
T₂ (K)
−
T₁ (K)
)
Q (kJ)
=
ΔU (kJ)
+
W (kJ)
Formulas
V₁
=
m
×
0.287
×
T₁
P
0.2153
=
0.5
×
0.287
×
300
200
V₂
=
m
×
0.287
×
T₂
P
0.3588
=
0.5
×
0.287
×
500
200
W
=
P
×
(
V₂
−
V₁
)
+28.7 = 200 × (0.3588 − 0.2153)
ΔU
=
m
×
0.718
×
(
T₂
−
T₁
)
+71.8 = 0.5 × 0.718 × (500 − 300)
Q
=
ΔU
+
W
+100.5 = 71.8 + 28.7
Assumptions
•
Air is an ideal gas: R = 0.287 and cᵥ = 0.718 kJ/(kg·K), both constant.
•
The piston moves slowly at constant pressure, so the boundary work is P(V₂ − V₁).
•
Heat into the air and work done by it are +: ΔU = Q − W.
Step-by-step
WE KNOW
m = 0.5 kg, P = 200 kPa, T₁ = 300 K, T₂ = 500 K
FIND
starting volume (V₁), final volume (V₂), boundary work (W), internal energy change (ΔU), heat added (Q)
Step 1 · Find starting volume (V₁)
Use V₁ = m × 0.287 × T₁ ÷ P
The
ideal-gas
law
PV
=
mRT,
divided
by
P
.
V₁
=
0.5
×
0.287
×
300
200
V₁
=
0.1435
×
300
200
V₁
=
43.05
200
V₁
=
0.2153
m³
Step 2 · Find final volume (V₂)
Use V₂ = m × 0.287 × T₂ ÷ P
The
ideal-gas
law
PV
=
mRT,
divided
by
P
.
V₂
=
0.5
×
0.287
×
500
200
V₂
=
0.1435
×
500
200
V₂
=
71.75
200
V₂
=
0.3588
m³
Step 3 · Find boundary work (W)
Use W = P × (V₂ − V₁)
W
=
∫
P
dV;
with
P
held,
it
is
P
times
the
change
in
volume
(kPa
×
m
3
=
kJ).
W
=
200
×
(0.3588
−
0.2153)
W
=
200
×
0.1435
W
=
+28.7
kJ
Step 4 · Find internal energy change (ΔU)
Use ΔU = m × 0.718 × (T₂ − T₁)
An ideal gas’s internal energy depends on T alone: mcᵥ times the rise.
ΔU = 0.5 × 0.718 × (500 − 300)
ΔU = 0.5 × 0.718 × 200
ΔU = 0.359 × 200
ΔU = +71.8 kJ
Step 5 · Find heat added (Q)
Use Q = ΔU + W
The
first
law
ΔU
=
Q
−
W
:
the
heat
raises
U
and
pays
for
the
work.
Q
=
71.8
+
28.7
Q
=
+100.5
kJ
Check
Put the numbers back in. Both sides should match.
0.2153
=
0.5
×
0.287
×
300
200
✓
0.3588
=
0.5
×
0.287
×
500
200
✓
+28.7 = 200 × (0.3588 − 0.2153) ✓
+71.8 = 0.5 × 0.718 × (500 − 300) ✓
+100.5 = 71.8 + 28.7 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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