SME
Physical Chemistry I: Thermodynamics & Kinetics
Laws of thermodynamics
Use this for “2.50 mol of an ideal gas is compressed reversibly at 350 K from 32.5 L to 12.4 L. Find w, q and ΔS of the gas.”
Diagram
PV = nRT = 2,479 kPa·L at both ends: P₁ = 248 kPa, P₂ = 124 kPa.
w = −nRT ln(V₂/V₁)
w
=
−2,479
×
ln(
20
10
)
=
−1,718
J
The shaded area is 1,718 J.
The gas expands: it does work, so w is negative.
q = +1,718 J.
This is an example. Type your own values to start a new problem.
w (J)
=
−
n (mol)
R
T (K)
ln
V₂ (L)
V₁ (L)
q
Heat taken in by the gas
calculated · J
ΔS
Entropy change of the gas
calculated · J/K
Formulas
w
=
−
n
×
8.314
×
T
×
ln(
V₂
V₁
)
−1,718
=
−1
×
8.314
×
298.15
×
ln(
20
10
)
q
=
−
w
+1,718 = −(−1,718)
ΔS
=
q
T
+5.763
=
1,718
298.15
Assumptions
•
An ideal gas changed slowly (reversibly) at one temperature, so its internal energy stays the same: ΔU = q + w = 0 and q = −w.
•
w is the work done on the gas, w = −∫P dV with P = nRT ÷ V, which integrates to −nRT ln(V₂ ÷ V₁); R = 8.314 J/(mol·K).
•
The heat comes in at one temperature, so ΔS = q ÷ T.
•
Temperatures are in kelvins (°C + 273.15).
Step-by-step
WE KNOW
n = 1 mol, T = 298.15 K, V₁ = 10 L, V₂ = 20 L
FIND
work done on the gas (w), heat taken in by the gas (q), entropy change of the gas (ΔS)
Step 1 · Find work done on the gas (w)
Use w = −n × 8.314 × T × ln(V₂ ÷ V₁)
Integrate
−P
dV
with
P
=
nRT
V
from
V₁
to
V₂
:
the
work
on
the
gas
is
−nRT
ln(
V₂
V₁
).
w
=
−1
×
8.314
×
298.15
×
ln(
20
10
)
w
=
−1
×
8.314
×
298.15
×
ln(2)
w
=
−1
×
8.314
×
298.15
×
0.6931
w
=
−8.314
×
298.15
×
0.6931
w
=
−2,478.8191
×
0.6931
w
=
−1,718
J
→
w
<
0:
the
gas
expands
and
does
work
on
its
surroundings.
Step 2 · Find heat taken in by the gas (q)
Use q = −w
ΔU
=
0
at
constant
temperature,
so
the
heat
in
balances
the
work:
q
=
−
w
.
q
=
−(−1,718)
q
=
+1,718
J
Step 3 · Find entropy change of the gas (ΔS)
Use ΔS = q ÷ T
The
heat
is
taken
in
reversibly
at
one
temperature,
so
divide
it
by
T
.
ΔS
=
1,718
298.15
ΔS = +5.763 J/K
Check
Put the numbers back in. Both sides should match.
−1,718
=
−1
×
8.314
×
298.15
×
ln(
20
10
)
✓
+1,718.1865 = −(−1,718.1865) ✓
+5.7628
=
1,718.1865
298.15
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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.”
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?”