Thermodynamics
Ch 18 – The Laws of Thermodynamics
18.1 Temperature and the Zeroth Law of Thermodynamics
The Zeroth Law of
Thermodynamics
If two objects are
separately in thermal
equilibrium with a 3rd
object, then the first
two objects are in
thermal equilibrium
with each other.
Thermal equilibrium: If two objects are in thermal equilibrium then neither
object will exchange energy by heat or electromagnetic radiation if they are
placed in thermal contact.
Temperature is a property that determines whether an object is in thermal
equilibrium.
Thermal contact: If in thermal contact two objects will exchange energy
between them if a temperature difference.
18.2 The 1st Law of Thermodynamics
The 1st law of thermodynamics (an energy conservation equation
specifying that) the only type of energy that changes in the
system are the internal energy, Eint or U.
ΔU = Q - W
Internal Energy is a state variable just like P, V, and T
(State variable do not depend upon path, just state)
thus Q - W is INDEPENDENT of path
An isolated system is a system that no
Non-isolated, but is cyclic
(starts and ends in the same state)
energy
is transferred by heat and the work done on ΔU = 0 Joules
the system is zero. (Doesn’t interact with
Thus Q (energy added to system) is equal to the Work done
environment)
by the system.
Q = Work
Q = 0 Joules
ΔUi = ΔUf
Example
Fill in the missing entries for an ideal gas that is taken through the three processes.
Q
W
U
AB
-53 J
(a)
(b)
BC
-280 J
-130 J
(c)
CA
(e)
150 J
(d)
(a)
WAB = PΔV
WAB = 0
(b)
(d)
ΔUCA
ΔUCA
ΔUCA
ΔUAB = QAB – WAB
ΔUAB = -53 – 0
ΔUAB = -53 J
(c)
ΔUBC = QBC – WBC
ΔUBC = -280 – (-130)
ΔUBC = -150 J
ΔU = 0
= -ΔUAB - ΔUBC
= -(-53) – (-150)
= 203 J
(e)
ΔUCA = QCA – WCA
203 = QCA – 150
QCA = 353 J
Fire Syringe: TH-B-FS
18.3 Thermal Processes
ΔEint (or U) = Q – W
An adiabatic
process is one
that no energy
enters or leaves
from the system
by heat.
ΔEint = -Work
As gas expands
adiabatically the
temperature reduces
accordingly
For a cloud in deep space that is spurting
out material very fast, at that point the
temperature can go below the ambient
background radiation temperature of 3
Kelvin
Q = 0 Joules
Isobaric (Constant
Pressure)
W = P ΔV
P is a constant
A plot of P vs V for an ideal gas yields a
hyperbolic curve (an isotherm)
Isothermic (constant Temp)
ΔEint = 0 Joules
Q=W
W = -∫
P
dV
W = -∫ (nRT/V) dV
Work = -nRT ln(Vf/Vi)
Work = nRT ln(Vi/Vf)
Work = 0
ΔU = Q
Isovolumetric
If energy is added by heat to a system,
all of the transferred energy remains
in the system as increased internal
energy
Work and Heat in Thermodynamics
Using a macroscopic approach, the state
of the system is described using P, V, T,
and internal energy, so these are called
“state variables.”
Work = force applied through a
displacement
Work = F ◦ Δr
Work = -F Δy
Work = -PA Δy
Work = - P ΔV
“Transfer variables” only involve transfer of energy.
So no “change” or transfer of energy…no transfer
variables.
The work done on a gas as it is taken from an initial state to a
final state depends on the path between these states
Diffusion in Air: TH-D-DA
18.4 Specific Heats for an Ideal Gas: Constant Pressure, Constant Volume
Q = n CVΔT
The energy added to the gas by heat at a constant volume only the internal energy, U, of the gas
increases.
(No work is done; Area under curve is zero since a vertical line.)
ΔU = Q
+W
ΔU = n CVΔT + W
(but W = 0)
ΔU = n CVΔT
Constant Volume
U
= n CVT
Etrans
= n CVT
3/2 nRT = n CVT
CV = 3/2 R
From the previous chapter the theorem
of equipartition of energy gives us
three degrees of freedom
This result is if the only internal energy is from translation motion
…thus monatomic atoms. Thus we predict CV = 3/2 R for all
monatomic atoms, CV = 3/2 (8.314 J/ K*mole)
CV = 12.5 J/ K*mole
We find that this is true experimentally
Example: Calculate the change in internal energy
of 3.00 mol of helium gas when its temperature is
increased by 2.00 K.
n = N / NA
and
kB = R / NA
Etranslational = 3/2 nRT
Eint = 3/2 n R
ΔT
Eint = 3/2 (3)(8.314)(2K)
Eint = 74.8 Joules
Q = n CPΔT
The energy added to the gas by heat for
constant pressure not only increases the
internal energy, U, but also must account for
the energy transferred out of the system by
the increased volume, Work (area under the curve.)
Etranslational = N (½mvave2)
Etranslational = 3N(½kBT)
Etranslational = 3/2 NkBT
We see that T+ΔT is the same for both
processes (constant pressure and constant
volume), but the constant pressure path has a
change of volume which represents the work that
is also done on the system
ΔU
ΔU
n CVT
CP – CV
CP – 3R/2
CP
=
Q
= nCPΔT
= nCPΔT
=R
=R
= 5R/2
- W
- P ΔV
- nRΔT
We know ΔU for both
constant pressure and
constant volume is the same,
n CVΔT, due to both go to
the same isotherm.
A ratio of these two specific heats is useful in
solving problems, thus
γ = CP/CV = 5R/2 / 3R/2
γ = 1.67
18 Adiabatic Processes for an Ideal Gas
Adiabatic Process is a process where no energy is transferred to surrounding environment. These
processes are either quick processes or insulated processes and is represented by PVγ = constant
Example
Air in a thundercloud expands as
it rises. If its initial temp is
300 K, and no energy is lost by
thermal conduction on expansion,
what is its temperature when the
initial volume has doubled?
PVγ
PVγ
T
Vγ-1
300 Vγ-1
(V/2V)1.40-1
= constant
= PfVfγ
= Tf Vfγ-1
= Tf (2V)γ-1
= Tf /300
γ
γ
P V =
Pf
Vf
γ
γ
(nRT/V)V
= (nRTf/Vf)Vf
γ
γ
(T/V) V
= (Tf/Vf) Vf
γ
γT V -1
= T f Vf 1
γ for diatomic molecules, 1.40
Tf = 227 K
The Equipartition of Energy
As we’ve seen the
model for molar
specific heat agrees
only with monatomic
gases as expected.
We need to include
(a) translational KE
AND
(b) rotational
As we can see the distance the
diatomic atom is from the y-axis so
the Inertial,
KR = ½Iyω2 = 0 Joules,
but Inertia is far from zero for
rotation about the x & z-axes
Also our value of
AND
CP – CV = R does prove (c) vibrational (see below)
to be accurate for all
gases.
Thus the total inertia for any
diatomic molecule will just be two of
the three axes.
(We can rotate the axis so that Inertia will always be
zero about one of the axes.)
To write Eint with both (a) translational and (b) rotational
energy yields
U
=
ETrans
+ ERot
U
= 3N(½kbT)
+ 2N(½kbT)
U
= 5N(½kbT)
U
= 5/2 nRT
ΔEint = n CV ΔT
(solve for CV)
CV = ( ΔEint / ΔT) / n
CV = ( 5/2 nRT / ΔT) / n
CV = 5/2 R
CP – CV = R
CP – 5/2 R = R
CP = 7/2 R
U
=
QV +
PΔV
But with QV, PΔV = 0 J.
So QV = U
ΔEint = n CV ΔT
(each degree of freedom contributes on average ½k bT per
molecule, for an additional two degrees of freedom for
rotation)
We know from the previous section
monatomic atoms
γmono = 1.67
And for diatomic atoms
γdi = 1.40
The results agree well with experimental data.
CV = 1/n (ΔEint/ΔT)
γ = CP / CV
γ = 7R/2 / 5R/2
γ = 1.40
To write U with both (a) translational and (b) rotational energy (c) and vibrational yields
U
U
U
U
=
ETrans
+ ERot
+ EVib
= 3N(½kBT) + 2N(½kBT) + 2N(½kBT)
= 7/2 nRT
= 7N(½kBT)
(2 more degrees of freedom for
vibration; each atom is vibrating)
CV = 1/n (dEint/dT)
CV = 7/2 R
monatomic atoms,
γmono = 1.67
CP – CV = R
CP – 7/2 R = R
CP = 9/2 R
diatomic atoms,
γdi = 1.40
γ = CP / CV
γ = 9R/2 / 7R/2
γ = 1.29
And for polyatomic atoms,
γpoly = 1.29
To explain molar specific heat of a solid at high temps, we use small displacements of an atom from its
equilibrium position which is approximated by simple harmonic motion.
Ex = ½mvx2 + ½kx2 (review from Ch 7 & 8)
Similarly along the y & z-axes, thus for a total of 6 degrees of freedom.
Eint = Ex
+ Ey
+
Ez
(Remember…energy isn’t a vector)
Eint = 2N(½kBT) + 2N(½kBT) + 2N(½kBT)
Eint = 3NkBT = 3nRT
CV = 1/n (dEint/dT)
CV = 3 R (at lower temperatures classical physics just isn’t a good enough approximation; this is only good at high temps)
Example
A certain molecule has f degrees of freedom. Show that an ideal gas consisting of such molecules has the
following properties:
(1) total internal energy is fnRT/2;
Eint = N(½kBT) for each degree of freedom
theorem of equipartition energy
So if, f, degrees of freedom…then
Eint = fN(½kBT); where n = N/kB
From previous chapter:
kB = R / NA ; n = N / NA; N = nR / kB
Eint = fn(½RT)
(2) molar specific heat at constant vol. is fR/2;
CV = 1/n ( dEint
/ dT)
CV = 1/n (d [fn(½RT)] / dT)
CV = ½fR
(3) molar specific heat at constant pressure;
is (f + 2)R/2;
CP – CV = R
CP = ½fR + R
CP = fR/2 + 2R/2
CP = ½(f + 2)R
(4) specific heat ratio is  = CP/CV = (f + 2)/f
γ=
CP
/ CV
γ = ½(f + 2)R / ½fR
γ = (fR + 2R) / fR
γ = (f + 2) / f
18.5 The 2nd Law of Thermodynamics
When objects of different
temperatures are brought into
thermal contact, the spontaneous
flow of heat that results is
always from the high
temperature object to the low
temperature object. Spontaneous
heat flow never proceeds in the
reverse direction.
A heat engine is a device
A heat engine carries some working substance through a cyclic process
that takes an energy input
through heat an does a
fraction of the input energy
as work as a cyclic process.
A sterling engine
A steam
locomotive
Automotive engine
Example
Suppose a heat engine is
connected to two energy
reservoirs, one a pool of
molten aluminum (660°C)
and the other a block of
solid mercury
(–38.9°C). The engine
runs by freezing 1.00 g
of aluminum and melting
15.0 g of mercury during
each cycle. The Lf of Al is
during
the working substance absorbs energy by HEAT from a high
temp energy reservoir
work is done by the engine
energy is expelled by heat to a lower-temp reservoir
Heat required to melt
15.0 g of Hg
Energy absorbed by 1.00 g of
aluminum
the work output
Q= m
Lf
Q = 0.015 (1.18x104)
Qc = 177 Joules
Q= m
Lf
Q = 0.001(3.97x105)
Qh = 397 J
Wengine = Qh – Qc
Wengine = 397-177
Wengine = 220 J
eff = Wengine / Qh
eff = 220 / 397
eff = 55.4%
The theoretical (Carnot) efficiency is
3.97 x 105 J/kg; Lf of Hg is
What is
the efficiency of this
engine?
1.18 x 104 J/kg.
18.6 Heat Engines and the Carnot Cycle
effmax = 1 – TC/TH
effmax = 1 – (273.15-38.9) / (273.15+660)
effmax = 74.9%
Example
Carnot’s theorem –
No real heat engine
operating between
two energy reservoirs
can be more efficient
than a Carnot engine
operating between
the same two
reservoirs.
What is the coefficient of performance of a
refrigerator that operates with Carnot efficiency
between temperatures –3.00°C and +27.0°C?
COPrefrig =
TC
/ ∆T
COPrefrig = (273.15+27) / 30
COPrefrig = 9
This figure is explained below in Gasoline
engine section
Example
A heat engine
operating between
200°C and 80.0°C
achieves 20.0% of the
maximum possible
efficiency. What
energy input will enable
the engine to perform
10.0 kJ of work?
Intake stroke O  A
The Carnot efficiency of
the engine is
effc =
∆T /
Th
effc = (200-80) / (273.15+200)
effc = 25.3%
Chemical PE In
Positive Work on gas
Compression stroke A PE to Qh
B
Gas expands adiabatically
If 20% efficient we only get
25.3% * 20% = 5.06%
eff
5.06%
Qh
= Wengine/ Qh
= 10 KJ / Qh
= 197 KJ
Combustion B  C
P drops suddenly
V drops V1 to V2
Power stroke C  D
Exhaust opens D 
A
Exhaust stoke A 
O
If we assume an ideal gas, then the
efficiency of the Otto cycle is
eff = 1 – (V1/V2)1-γ
γ = CP/CV; where V1/V2
is the compression ratio
Stirling Engine: TH-F-SC
Example
A gasoline engine has a compression ratio of 6.00 and
uses a gas for which γ = 1.40.
(a) What is the efficiency of the engine if it operates in
an idealized Otto cycle?
(b) What If? If the actual efficiency is 15.0%, what
fraction of the fuel is wasted as a result of friction and
energy losses by heat that could by avoided in a
reversible engine?
effotto = 1 – (V2 / V1)γ-1
effotto = 1 – (1 / 6)1.4 - 1
effotto = 51.2%
losses = effotto - effactual
losses = 51.2% - 15%
losses = 36.2%
18.7 Refrigerators, Air Conditioners, and Heat Pumps
A heat pump transfers energy from a cold to a hot reservoir. We must add energy to accomplish this
process. Other common names, air conditioner, refrigerator, etc.
For heating mode
COP(heater) = Qh /
Work
For cooling mode
COP(refrig)
= Qc / Work
Good refrigerators have a coefficient
of performance, COP, of about 6
Clausius statement
Energy does NOT transfer spontaneously by heat from a cold object to a hot object!
Example
a.
COP(refrig)
= Qc / Work
A refrigerator has a coefficient of
5.00
= 120 J / Work
performance equal to 5.00. The
Work
= 24.0 Joules
refrigerator takes in 120 J of
energy from a cold reservoir in each b.
Heat expelled = Heat removed + Work done
cycle. Find
(a) the work required in each cycle
and
(b) the energy expelled to the hot
reservoir.
Qh
Qh
Qh
=
Qc
=
120
= 144 Joules
+ Work
+ 24
18.8 Entropy
Entropy is a measure of
disorder in isolated systems.
Entropy is another state
variable of 2nd law of
thermodynamics
For a Carnot engine operating in a cycle
|Qh|/Th = |Qc|/Tc
Thus
∆S = 0
Example
An ice tray contains 500 g of
liquid water at 0°C. Calculate
the change in entropy of the
water as it freezes slowly and
completely at 0°C.
For a freezing process, change of entropy,
∆S = ∆Q / T
Quasi-Static, Reversible Process for an Ideal Gas
∆S = nCv ln(Tf/Ti) + nR ln(Vf/Vi)
∆S =
-mLf / T
∆S = -.5(3.33x105)/273.15
∆S = -610 J/K
The total entropy of an isolated system that undergoes a change CANNOT decrease
In irreversible processes Entropy always increases
In reversible processes (non-real processes) ∆S is constant
Entropy in free expansion
(no pressure increase)
Entropy Change in Calorimetric Processes
Tf = (m1c1Tc + m2c2Th) / (m1c1 + m2c2)
∆S =
∆Qr
/T
∆S = (nRT/V) ∆V / T
(utilize calculus to solve)
∆S = nRln(Vf/Vi)
Example
The temperature at the surface of the Sun is
approximately 5700 K, and the temp at the
surface of the Earth is approximately 290 K.
What entropy change occurs when 1000 J of
energy is transferred by radiation from the Sun
to the Earth?
∆S =
∆Qr
/T
∆S = m1c1 ln(Tf/Tc) + m2c2 ln(Tf/Th)
∆Ssystem = ∆(Q / T)
∆Ssystem = ∆(1000/290 – 1000/5700)
∆Ssystem = 3.27 J/K
18.9 Order, Disorder, and Entropy
Entropy can be thought of as the increase in disorder in the universe.
If we look at the ultimate fate of the universe in light of the continual increase in entropy, we
might envision a future in which the entire universe would have come to the same temperature. At
this point, it would no longer be possible to do any work, nor would any type of life be possible. This
is referred to as the “heat death” of the universe.
So if entropy is continually increasing, how is life possible? How is it that species can evolve into
ever more complex forms? Doesn’t this violate the second law of thermodynamics?
No – life and increasing complexity can exist because they use energy to drive their functioning.
The overall entropy of the universe is still increasing. When a living entity stops using energy, it
dies, and its entropy can increase rather quickly.
Bromine Diffusion: TH-D-BD
18.10 The Third Law of Thermodynamics
Absolute zero is a temperature that an object can get arbitrarily close to, but never attain.
Temperatures as low as 2.0 x 10-8 K have been achieved in the laboratory, but absolute zero will
remain ever elusive – there is simply nowhere to “put” that last little bit of energy.
This is the third law of thermodynamics:
It is impossible to lower the temperature of an object to absolute zero in a finite number of steps.