Lecture 7: Thermo and Entropy
• Reading: Zumdahl 10.2, 10.3
• Outline
– Isothermal processes
– Isothermal gas expansion and work
– Reversible Processes
Isothermal Processes
• Recall: Isothermal means ∆T = 0.
• Since ∆E = nCv∆T, then ∆E = 0 for an isothermal
process.
• Since ∆E = q + w:
q = -w (isothermal process)
Heat goes into a machine and it does work (q, and –w
are both positive).
Example: Isothermal Expansion
• Consider a mass connected to a ideal gas
contained in a “piston”. Piston is
submerged in a constant T bath such that
∆T = 0 during expansion.
Isothermal Expansion: Start
∆h
∆V = A∆h
PV = PV
1 2 = nRT
∆h
• Initially,
V = V1
P = P1
• Pressure of gas is equal to that created by mass:
P1 = force/area = M1g/A
where A = piston area
g = gravitational acceleration (9.8 m/s2)
Isothermal Expansion: One Step
(Not Reversible or Irreversible)
• One-Step Expansion. We change the weight to
M1/4, then
Pext = (M1/4)g/A = P1/4
Internal pressure is larger than the external pressure
so the gas will expand. The work done is defined
by the pressure on the yielding side (external).
• The mass will be lifted until the internal pressure
equals the external pressure. In this case
Vfinal = 4V1
• Because nRT = PV = PV = P V
1 1
final
final
• q=-w = Pext∆V = P1/4 (4V1 - V1) = 3/4 P1V1
Two Step Expansion
• In this expansion we go in two steps:
Step 1: M1 to M1/2
Step 2: M1/2 to M1/4
• Step 1:
Pext = P1/2, Vfinal = 2V1
• -w1 = Pext∆V = P1/2 (2V1 - V1) = 1/2 P1V1
Two Step Expansion: Second Step
• In Step 2 ( Move M1/2; then move M1/4 ):
Pext = P1/4, Vfinal = 4V1
• -w2 = Pext∆V = P1/4 (4V1 - 2V1) = 1/2 P1V1
• -wtotal = -(w1 + w2) = P1V1/2 +P1V1/2 = P1V1
−w
2 steps
total
= PV
1 1 = nRT >
3 PV
4 1 1
= −w
1step
total
Get more work (but use more heat) to move the weights in 2
steps than a single step. In the two step the second weight is
raised to exactly the same height and the single step and an
additional weight is moved half the distance.
Graph: Two Step Expansion
• Graphically, we can
envision this two-step
process on a PV diagram:
←←AA∆∆hh11 →
→
← A∆h2 →
• Work is given by the area under the “PV” curve.
Graph: Two Step and One Step Expansion
• See both expansion processes on a PV diagram:
One Step Work Part
• Work is given by the area under the “PV” curve.
Graphical: Infinite Step Expansion
Two Step:
Irreversible
Reversible
Infinite Step Expansion
• Imagine that we perform a process in which we
change the weight “infinitesimally” between
expansions.
• A very small weight is removed and the pressure
inside the box is nearly the same as the external
pressure.
• Instead of determining the sum of work performed
at each step to get wtotal, we integrate:
V final
V final
V final
nRT
w = − ∫ Pex dV = − ∫ PdV = − ∫
dV
Vinitial
Vinitial
Vinitial V
Many Step Expansion
Break up ¾ of the weight into many (N) small weights and
take each small weight off one at a time. Consider a single
step in the middle of the process, and then add up all steps.
N ∆ m = 34 M
m1 = m 2 + ∆ m
PV
1 1 = P2V 2 = nR T
V 2 = V1 + ∆ V
mi = M − (i − 1)∆m
∆m
( − wi ) = nRT ∑
∑
i =1
i =1 M − (i − 1) ∆m
N
m1 g
∆mg
− wTotal
= P2 +
A
A
∆mg 

+
P
 2
 V1 = P2 (V1 + ∆ V )
A 

∆m
 ∆mg 
− w1 = P2 ∆ V = 
 V1 = nR T
m1
 A 
P1 =
N
 N

1
= nRT ∑ 4

 i =1 3 N − (i − 1) 
The work as N becomes large;
many small steps
We have divided the weights a bit differently (from the two
step) but we can compare 1, 2 , 3 and an infinite number
of steps, by doing the sum for different values of N:
N
N
1
∑
4 N − (i − 1)
i =1 3
1
3
4
2
1000
0.975
1.385
Reversible Isothermal Expansion:
(Infinite number of steps; can go back and forth)
• Add up areas (integrate) from V1 to V2
 V2 
nRT
V2
= − ∫ PdV = − ∫
dV = −nRT(ln(V )) |V1 = −nRT ln 
 V1 
V1
V1 V
V2
w total
V2
qrev = − wrev
 V2 
= nRT ln  
 V1 
Compare Work for different steps
q1step =
• One step
q2 steps = PV
• Two
• Infinite Number of Steps:
3 PV
4
qrevers = PV ln ( 4 ) = 1.386 PV
qreversible > q2 > q1
One Step Compression
State of the system after the 1 Step Expansion:
Vacuum
1M
4 1
Vinitial = 4V1
Pinitial = 14 P1
3M
4 1
What do we have to do to get the system back to its original
state before 1-step expansion? [Pick up Weight]
System did work: Raised M/4; to restore it we have to raise
3M/4 the same distance. We must do more work to
restore the system than we got from it in the first place.
Two Step Compression
State of the system after the 2 Step Expansion:
Vacuum
1M
4 1
1M
4 1
Vinitial = 4V1
Pinitial =
1P
4 1
1M
2 1
What do we have to do to get the system back to its original
state before 2-step expansion? [Pick Weights Up]
Compression in two steps:
first: place on mass M1/4, second: then M1/2 on pan.
Two Step Compression:
Compare work done by system with work we need to
do to restore the system
• To expand the system did total work (P1V1)
• To compress: In first step: we need to move weight M1/4
up to the pan and use that pressure to compress the gas
2 Step
step1
P
=
1M g
2 1 A
= 12 P1
2 Step
∆Vstep
1 = 2V1 − 4V1 = −2V1
2 Step
2 Step
2 Step
1 P ⋅ 2V = PV
wstep
=
−
P
∆
V
=
1
step1
step1
1
1 1
2 1
Second step is similar and the same work is done on the system.
However, we do not know how much of the work we did ended up as
heat into the reservoir or heat into the weights as they dropped. We
just know that to compress it back to (approximately) its original
state it took twice as much compression work (not counting the extra
work we did to move the weights) to compress as it did to expand.
Compression/Expansion
The system is the gas in the box and does not include the weights
in the environment. The weights help us keep track of the work.
• In two step example:
wexpan. = -P1V1
wcomp. = 2P1V1
wtotal = P1V1
qtotal = -P1V1
• We have undergone a
“cycle” where the
system returns to the
starting state.
• Now, ∆E = 0 (state fxn)
qtotal = − wtotal ≠ 0
A Thermodynamic Engine: Defining Entropy
Imagine doing all steps reversibly, then compression is simpler to
understand. The system variables (P,T) are always only
incrementally different from the external values.
Use two different heat reservoirs at two different temperatures.
Let’s consider the four-step cycle
illustrated:
1
4
2
3
V2
V1
Volume
–
–
–
–
1:
2:
3:
4:
Isothermal expansion
Isochoric cooling
Isothermal compression
Isochoric heating
First Step of Four Cycle
• Step 1: Isothermal Expansion
at T = Thigh from V1 to V2
1
4
2
3
• Do expansion reversibly. Then:
V2
V1
Volume
• Now ∆T = 0; therefore, ∆E = 0
and q = -w
 V2 
q1 = − w1 = nRThigh ln 
 V1 
Step 2
• Step 2: Isochoric Cooling to T =
Tlow.
1
4
• Now ∆V = 0; therefore, w = 0
2
3
V2
V1
Volume
• q2 = ∆E = nCv∆T
= nCv(Tlow-Thigh)
Step 3
• Step 3: Isothermal Compression
at T = Tlow from V2 to V1.
1
4
2
• Now ∆T = 0; therefore, ∆E = 0
and q = -w
3
V2
V1
Volume
• Do compression reversibly, then
 V1 
q3 = − w3 = nRTlow ln 
 V2 
Step 4
• Step 4: Isochoric Heating to T =
Thigh.
1
4
• Now ∆V = 0; therefore, w = 0
2
3
V2
V1
Volume
• q4 = ∆E = nCv∆T
= nCv(Thigh-Tlow) = -q2
All 4 steps of the engine, added together
qtotal = q1 + q2 + q3 + q4
1
4
2
qtotal = q1 + q3
3
V2
V1
Volume
qtotal
qtotal
qtotal
 V2 
 V1 
= nRThigh ln   + nRTlow ln  
 V1 
 V2 
 V2 
 V2 
= nRThigh ln   − nRTlow ln  
 V1 
 V1 
 V2 
= nR∆T ln   = − wtotal
 V1 
Defining Entropy
1
4
2
qtotal
 V2 
 V2 
= nRThigh ln  − nRTlow ln 
 V1 
 V1 
3
V2
V1
Volume
Because this quantity is zero when
summed over all changes which
brings the system back to where it
started, is zero it must depend on the
state of the system and so is a state
function.
q1
q3
0=
+
Thigh Tlow
final
dqrev qrev
∆S = ∫
=
T
initial T
Thermodynamic definition of entropy: The heat transferred under reversible
conditions divided by the temperature. What if the transfer is not reversible?
Then the amount of heat transferred is not the same but the entropy is the same.
Calculating Entropy
final
∆S =
∆T = 0
dqrev
∫ T
initial
∆V
= nRT
V
 V final 
dqrev
∆S = ∫
= nRln

 Vinitial 
initial T
qrev = nCv ∆T
 Tfinal 
dqrev
∆S = ∫
= nCv ln

T
T
 initial 
initial
qrev
final
final
∆V = 0
 Tfinal 
dqrev
∆S = ∫
= nC p ln

T
T
 initial 
initial
final
∆P = 0
qrev = nCP ∆T
Calculating Entropy
• Example: What is ∆S for the heating of a mole of a monatomic
gas isochorically from 298 K to 350 K?
 Tfinal 
dqrev
∆S = ∫
= nCv ln

 Tinitial 
initial T
final
3/2R
 350K 
∆S = (1mol) 3 R ln

2
 298K 
( )
∆S = 2 J
K
Connecting with Lecture 6
• From this lecture:
 V final 
dqrev qrev nRT  V final 
∆S = ∫
=
=
ln
 = nRln

T
T
 Vinitial 
 Vinitial 
initial T
final
• Exactly the same as derived in the previous lecture!
 V final 
 V final 
∆S = Nk ln

 = nRln
 Vinitial 
 Vinitial 