SME
Thermal & Statistical Physics · Problem type
Heat capacities of an ideal gas
Use this for “3.00 mol of helium warms by 40 K. How much heat does it take at constant volume, how much at constant pressure, and how much work does it do at constant pressure?”
Diagram
First law: ΔU = Q − W = 2,910 − 831.4 = 2,079 J
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.
Q
V
=
ΔU
=
f
2
×
n (mol)
R
ΔT (K)
=
QV (J)
CV
Heat capacity at constant volume
worked out (not typed) · J/(mol·K)
CP
Heat capacity at constant pressure
worked out (not typed) · J/(mol·K)
QP
Heat at constant pressure
calculated · J
W
Work at constant pressure
calculated · J
Formulas
CV
=
f
2
×
8.314
20.785
=
5
2
×
8.314
CP = CV + 8.314
29.099 = 20.785 + 8.314
QV
=
n
×
CV
×
ΔT
+2,079 = 2 × 20.785 × 50
QP
=
n
×
CP
×
ΔT
+2,910 = 2 × 29.099 × 50
W
=
n
×
8.314
×
ΔT
+831.4 = 2 × 8.314 × 50
Assumptions
•
Equipartition: each degree of freedom holds ½RT per mole, so CV = (f ÷ 2)R; R = 8.314 J/(mol·K).
•
f = 3 for a monatomic gas, 5 for a diatomic one near room temperature, 7 once its vibration joins in.
•
Constant volume: no work, so QV = ΔU. Constant pressure: W = PΔV = nRΔT and QP = ΔU + W.
Step-by-step
WE KNOW
n = 2 mol, f = 5, ΔT = +50 K
FIND
heat capacity at constant volume (CV), heat capacity at constant pressure (CP), heat at constant volume (QV), heat at constant pressure (QP), work at constant pressure (W)
Step 1 · Find heat capacity at constant volume (CV)
Use CV = f ÷ 2 × 8.314
Each
degree
of
freedom
holds
1
2
RT
per
mole,
so
warming
by
1
K
takes
1
2
R
for
each.
CV
=
5
2
×
8.314
CV = 2.5 × 8.314
CV = 20.785 J/(mol·K)
Step 2 · Find heat capacity at constant pressure (CP)
Use CP = CV + 8.314
At constant pressure the gas also pushes out by nRΔT, so each mole takes R more per kelvin.
CP = 20.785 + 8.314
CP = 29.099 J/(mol·K)
Step 3 · Find heat at constant volume (QV)
Use QV = n × CV × ΔT
At constant volume no work is done, so the heat QV all goes into internal energy: QV = ΔU = nCVΔT.
QV = 2 × 20.785 × 50
QV = 41.57 × 50
QV = +2,079 J
Step 4 · Find heat at constant pressure (QP)
Use QP = n × CP × ΔT
At constant pressure the heat raises the internal energy and pays for the work: nCPΔT.
QP = 2 × 29.099 × 50
QP = 58.198 × 50
QP = +2,910 J
Step 5 · Find work at constant pressure (W)
Use W = n × 8.314 × ΔT
At
constant
pressure
W
=
PΔV,
and
PV
=
nRT
turns
PΔV
into
nRΔT.
W
=
2
×
8.314
×
50
W
=
16.628
×
50
W
=
+831.4
J
Check
Put the numbers back in. Both sides should match.
20.785
=
5
2
×
8.314
✓
29.099 = 20.785 + 8.314 ✓
+2,078.5 = 2 × 20.785 × 50 ✓
+2,909.9 = 2 × 29.099 × 50 ✓
+831.4 = 2 × 8.314 × 50 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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