First Law of Thermodynamics
• Internal Energy
• Heat and Work
• Getting work from a heat engine
– Thermodynamic cycles
• Our Strategy about heat and work, they get
us started and they are measures of
processes or paths. They teach us about
state functions of Energy. We discovery
entropy from them. Then we can bypass
them except for measurement.
2-1
Internal Energy (U)
U(V,T), U(P,T), or U(P,V), can take on a number of
forms
•Kinetic energy of molecules in system.
•the potential energy of the constituents of the system relative to
the center of mass of the system. For example, a crystal consisting
of dipolar molecules will experience a change in its potential
energy as an electric field is applied to the system.
•the internal energy stored in the form of molecular vibration and
rotation.
•the internal energy stored in the form of chemical bonds that can
be released through a chemical reaction.
2-1
First Law of Thermodynamics
The First Law of Thermodynamics is based on our
experience that energy can neither be created nor
destroyed, if both the system and the surroundings are
taken into account.
The internal energy is U or Usystem.
ΔU total = ΔU system + ΔU surroundings = 0
ΔU system = −ΔU surroundings
U can be changed by either heat transfer or by work “flow”
ΔU = q + w
2-2
Work: any quantity of energy that “flows” across the
boundary between the system and surroundings that can be
used to change the height of a mass in the surroundings.
mass
piston
mass
mechanical
stops
pi,Vi
Initial
State
piston
pf,Vf
cylinder
Final
State
cylinder
•Work only appears during a change in state of the system and
surroundings. Only energy, and not work, is associated with the
initial and final states of the systems.
2-3
•Net effect of work is to change U in accordance with the First Law.
If the only change in the surroundings is that a mass in the
surroundings has been raised or lowered, work has flowed across the
system between system and surroundings.
•The quantity of work can be calculated from the change in potential
energy of the mass, ΔE
= w = Fh = mgh
potential
where g is the gravitational acceleration and h is the change in the
height of the mass, m.
•The sign convention for work is as follows. If the height of the
mass in the surroundings is lowered, w is positive, and if the height
is raised, w is negative. In short, w > 0 if ΔU > 0. It is common
usage to say that if w is positive, work is done on the system by the
surroundings. If w is negative, work is done by the system on the
surroundings.
2-4
How to calculate work
•Expansion work
G G
w = ∫ F • dl = − ∫∫∫ Pexternal dAdl = − ∫ Pexternal dV
•Electrical work
welectrical = q ⋅ φ
= φ ⋅ I ⋅t
mass
mass
pi,Vi
H2
+
φ Is the electrical
potential, usually in
Volts.
O2
H2O
-
pi,Vf
Initial
State
H2
+
electrical
generator
-
electrical
generator
mass
mass
2-5
H2O
O2
Final
State
EP2.1a) Calculate the work involved in expanding 20.0 L
of an ideal gas to a final volume of 85.0 L against a
constant external pressure of 2.50 bar.
w = − ∫ Pexternal dV = − Pexternal (V f − Vi )
105 Pa
10−6 m3
= −2.50bar ⋅
= −16.3 J
(85.0 L − 20.0 L ) ⋅
bar
L
c) A current of 3.2 A is passed through a heating coil for 30
s. The electrical potential across the resistor is 14.5 V.
Calculate the work done.
w = ∫ φ dq = φ q = φ It = 14.5V ⋅ 3.2 A ⋅ 30 s = 1.39 kJ
1 Volt = 1 Joule/Coulomb
2-6
Heat is defined in thermodynamics as the quantity of
energy that flows across the boundary between the system
and surroundings because of a temperature difference
between the system and the surroundings.
•Heat is transitory, in that it only appears during a change in state of
the system and surroundings. Only energy, and not heat, is associated
with the initial and final states of the system and the surroundings.
•The net effect of heat is to change the internal energy of the system
and surroundings in accordance with the First Law. If the only
change in the surroundings is a change in temperature of a reservoir,
heat has flowed between system and surroundings. The quantity of
heat that has flowed is directly proportional to the change in
temperature of the reservoir.
2-7
•The sign convention for heat is as follows. If the temperature of the
surroundings is lowered, q is positive, and if it is raised, q is negative.
In short, q > 0 if ΔU > 0. It is common usage to say that if q is
positive, heat is withdrawn from the surroundings and deposited in
the system. If q is negative, heat is withdrawn from the system and
deposited in the surroundings.
Rest of universe
Thermometers
outer water bath
Isolated composite
system:
Sealed
reaction
vessel
Provides practical
way to limit size of
surroundings
Heating
coil
inner water bath
Touter = Tf
2-1
Tf
Distinguishing work from heat
electrical
generator
I
Propane
mass
heating coil
I
Bunsen burner
a
b
Two systems, I and II are enclosed in a rigid adiabatic enclosure.
System I consists solely of the liquid in the beaker for each case.
System II consists of everything else in the enclosure. Part a
show a process in which the liquid is heated using a flame. Part b
shows heating using a resistive coil through which an electrical
current flows.
2-9
Surroundings are important
• In both cases ΔU=0 (no work or heat was
transferred in or out of the box.)
• In the first one to do the energy balance,
the change in chemical energy heated the
water. 0 = ΔU = ΔU Chem + qWater
• In the second one the loss of potential (I
hate to call it work) became heat.
0 = ΔU = ΔU Mass + qWater
2-1
Heat capacity
q
dq
/
C = lim
=
ΔT → 0 T − T
dT
f
i
•C has the units J K-1 kg-1 in SI units. It is an extensive quantity.
•Cp and Cv distinguish method of measurement; the first at
constant pressure and the second at constant volume.
w = φ ⋅ I ⋅t
•Cm is an intensive quantity with the units J K-1 mol-1.
•Convenient to measure q through electrical work
•Numerical value depends on path between initial and final
states. Most common are constant V or P.
2-12
Cp,m (J K-1mol-1)
60
50
40
30
20
10
solid
100
liquid
200
gas
300
400
Temperature (K)
•CP,m for the solid rises rapidly at low T vibrations of the solid
activated.
•CP,m increases discontinuously as the solid melts to form a liquid
because the liquid retains the local vibrational modes of the solid
shifted to lower more accessible frequencies.
•CP,m decreases discontinuously at the vaporization temperature
because local vibrational modes in the liquid are converted to
translations, which can’t take up 2-13
as much energy as vibrations.
Once CP,m is known, q can be determined from ΔT
of reservoir
Tsys , f
qP =
∫
CPsystem ( T ) dT = −
Tsys ,i
Tsurr , f
∫
CPsurroundings (T ) dT
Tsurr ,i
Water is a convenient choice of material for a heat bath in
experiments because CP is nearly constant at the value
4.19 J g-1 K-1 or 75.4 J mol-1 K-1 over the range 0°C-100°C.
From the idea of the equipartition of energy: each mode that can
access the energy gets 1 R added to the heat capacity. For water
2
For Cv: We have 3 rotational and 3 translational degrees of
freedom and 3 vibrational degrees of freedom. This gives at most
3+3+2*3=12 modes, or 48 J mol-12-14K-1
CP,m > CV,m for gases
w
For ideal gas
mass
CP − CV = nR or
piston
mass
piston
q
pi,Ti
pi,Tf
cylinder
cylinder
Initial State
Constant
Pressure
Heating
Final State
Vi,Ti
Vi,Tf
Initial State
Final State
Constant
Volume
Heating
2-15
CP ,m − CV ,m = R
State functions and path functions
Change particle speed following the sequence v1 → v2 → v3 → v4.
1
1
1
1
1
1
ΔEkinetic = ⎛⎜ mv 22 − mv12 ⎞⎟ + ⎛⎜ mv 32 − mv 22 ⎞⎟ + ⎛⎜ mv 42 − mv 32 ⎞⎟
2
2
2
⎝2
⎠ ⎝2
⎠ ⎝2
⎠
1 2 1 2⎞
⎛
= ⎜ mv 4 − mv 1 ⎟
2
⎝2
⎠
f
ΔU = ∫ dU = U f − U i
i
v∫ dU = U
f
−U f = 0
Condition defines state
function
U (or any state function) can be expressed as an infinitesimal
quantity, dU, that when integrated, depends only on the initial and
final states. dU is called an exact differential.
Is w a state function?
mass
piston
mechanical
stop
p1,V1,T1
mass
mass
piston
piston
p2,V2,T2
p ,V2,T3
3
T3
T3
cylinder
T3
Initial
State
w = − Pexternal ΔV
Intermediate
State
Final
State
T1 < T2 < T3
Repeat process with
different masses. For
each path
P1,T1 → P3,T3
therefore ΔU same
Work different for each path. Therefore,
work is a path function, not a state
function.
Assume in all cases that Pext > Pint so compression
happens. For different masses2-17
different amount of work
but system change is the same.
Is q a state function?
If w is a path function and U is a state function,
q = ΔU-w must be a path function.
mass
piston
mechanical
stop
p1,V1,T1
mass
mass
piston
piston
p2,V2,T2
p ,V2,T3
3
f
∫
i
Δq ≠ dq
/ ≠ q f − qi
cylinder
T3
Initial
State
T3
T3
Intermediate
State
2-18
Final
State
f
∫
i
Δw ≠ dw
/ ≠ w f − wi
Thermodynamics can only
Quasi-static process
be applied to systems in
internal equilibrium, and a
1000
requirement for
800
equilibrium is that the
600
T(K)
400
overall rate of change of
200
all processes such as
6
3×10
diffusion or chemical
reaction is zero. How do
6
2×10
P(Pa)
i
we reconcile these
statements with our
6
1×10
f
calculations of q, w, and
ΔU associated with
0
0.0125 0.01 0.00750.0050.0025
processes in which there is
a macroscopic change in
the system?
2-19
Reversible process: an infinitesimal opposing change in the
variable that drives the process causes a reversal in the direction of
the process.
Infinitesimal temperature changes
in a 2 phase gas – liquid system
at the boiling temperature
pulley
1 kg
1 kg
2-20
Irreversible process: a large opposing change in the variable
that drives the process is needed to cause a reversal in the direction
of the process.
A mechanical stop holding a piston is removed. As a result, a gas
is compressed.
mass
piston
mass
mechanical
stops
pi,Vi
Initial
State
cylinder
piston
pf,Vf
cylinder
2-21
Final
State
Calculating work in
an irreversible
process at constant
Pexternal
T, P2, V1 → T, P2, V2 ,
P2 < P1 q1 > 0
Initially, Pexternal = P1 , reduce to
Pexternal = P2, increase again to
Pexternal = P1 Keep the
temperature constant throughout
the process.
Indicator diagram
w > 0, q 0
w > 0, q < 0
T, P1, V2 → T, P1,V1
q2 < 0
ΔU = 0 because cyclic
process
wtotal = − P2 (V2 − V1 ) − P1 (V1 − V2 ) = − ( P2 − P1 )(V2 − V1 ) > 0
Calculating reversible work (IG)
wexp ansion
dV
V2
= − ∫ Pexternal dV = − ∫ PdV = − nRT ∫
= − nRT ln
V
V1
For
reversible
cyclic
process
ΔU=w=q= 0
2-23
P
P
V
P
P
V
P
V
V
P
V
V
For compression, wreversible < wirreversible
For expansion, wreversible < wirreversible
The maximum work that can be extracted (w<0) from a
process between the same initial and final states is
obtained under reversible conditions. The meaning of
2-24
revesible vs irreversible path.
Determining ΔU
The First law states that ΔU = q + w . Imagine that the
process is carried out under constant volume conditions and
that non-expansion work is not possible. Because under
these conditions w = − P
dV = 0
∫
external
ΔU = qV
Simple example: Heat a gas in a steel cylinder. The heat is
directly related to the increase in internal energy of the gas.
2-25
Introducing Enthalpy, a new State Function
What does the First Law look like under reversible and
constant pressure conditions? ΔP = 0
By constant pressure we mean that the external pressure is kept
constant and the internal pressure always equals it; the volume
of the box may readjust during a transformation. Pex = P
/ P − Pex dV = dq
/ P − PdV
dU = dq
f
∫ dU = U
i
(U
f
f
/ P − ∫ PdV = qP − ( Pf V f − Pf Vi )
− U i = ∫ dq
+ Pf V f ) − (U i + Pf Vi ) = qP
If we define a new state function:
H ≡ U + PV
2-26
Then it follows for
processes carried out at
constant P:
ΔH = q P
Calculating q, w, ΔU, and ΔH for Processes
Involving Ideal Gases
In order to describe a fixed amount of an ideal gas (i.e. n is
constant), the values of two of the variables P, V, and T must be
known. Is this also true for ΔU for processes involving ideal
gases?
ΔU = qV = CV ( T f − Ti )
ΔH = qP = CP ( T f − Ti )
Because U(H) is a function of T only for an ideal gas, this
equation is also valid for processes involving ideal gases
in which V is not constant. Therefore, if we know CV(CP),
T1 and T2, we can calculate ΔU(ΔH), regardless of the path
between the initial and final states.
2-27
Further important relations for ideal gas
CP − CV = nR
w = − Pexternal (V f − Vi )
Irreversible P-V work at
constant P
Vf
dV
w = − nRT ∫
= − nRT ln
V
Vi
2-28
Reversible P-V
work
Example Problem 2.5
A system containing 2.50 mol of an ideal gas for which CV,m =
20.79 J mol-1K-1 is taken through the cycle indicated below in the
direction indicated by the arrows. The curved path corresponds to
PV = nRT , where T = T1 = T3
16.6 bar, 1.00 L
15.0
16.6 bar, 25.0 L
1
2
12.5
10.0
P(bar) 7.5
5.0
3
2.5
0
0.665 bar, 25.0 L
5
10
15
20
V(L)
2-29
25
a) Calculate q, w,
and ΔU and ΔH for
each segment, and
for the cycle.
b) Calculate q, w,
and ΔU and ΔH for
each segment, and
for the cycle in
which the direction
of each process is
reversed.
Segment 1→ 2
CV
( PV
2 2 − PV
1 1)
nR
2.50 mol x 20.79 J mol −1 K −1
=
(16.6 bar x 25.0 L − 16.6 bar x 1.00 L )
−1
−1
2.50 mol x .08314 Lbar K mol
= 99.6 kJ
105 N m −2
w = − Pexternal (V2 − V1 ) = 16.6 bar x
x ( 25.0 x 10−3 m 3 − 1.00 x 10−3 m 3 ) = −39.8 kJ
bar
ΔU1→2 = CV (T2 − T1 ) =
q = ΔU − w = 99.6kJ + 39.8kJ = 139.4kJ
PV
16.6 bar x 25.0 L
3
2 2
T2 =
2.00
x
10
K
=
=
−1
−1
nR 2.50 mol x .08314 Lbar K mol
ΔH1→2 = ΔU1→2 + Δ ( PV ) = ΔU1→2 + nR (T2 − T1 )
= 99.6 x 103 J + 2.5 mol x 8.314 J mol −1K −1 x ( 2000 K − 79.9 K ) = 139 kJ
2-30
Segment 2→ 3
w = 0 because ΔV = 0
CV
( PV
3 3 − PV
2 2)
nR
2.50 mol x 20.79 J mol −1 K −1
=
( 0.665 bar x 25.0 L − 16.6 bar x 25.0 L ) = −99.6kJ
−1
−1
2.50 x .08314 Lbar K mol
ΔU 2→3 = q2→3 = CV (T3 − T2 ) =
ΔH 2→3 = −ΔH 1→2 because T3 = T1
2-31
Segment 3 → 1
ΔU 3→1 = 0 and ΔH 3→1 = 0 because ΔT = 0
PV
16.6 bar x 1.00 L
3 3
=
= 79.9 K
T3 =
−1 −1
nR 2.50 mol x 0.08314 Lbar mol K
V1
w3→1 = −nRT ln
V3
1.00
= −2.50mol x 8.314 J mol K x 79.9 K ln
= 5.35kJ
25.0
−1
−1
2-32
Path
q
ΔU
w
ΔH
1→2 139.4 kJ -39.8 kJ 99.6 kJ 139.4 kJ
2→3 -99.6 kJ 0
-99.6 kJ -139.4
kJ
3→1 -5.35 kJ 5.35 kJ 0
0
cycle 34.5 kJ -34.5 kJ 0
0
2-33
Reversible Adiabatic Expansion and Compression
of an Ideal Gas
ΔU = w or CV dT = − Pexternal dV
dV
dT
dV
or equivalently, CV
CV dT = − nRT
= − nR
V
T
V
Tf
Vf
dT
dV
Tf
Vf
∫T CV T = −nR V∫ V CV ln T = −nR ln V
i
i
i
i
1−γ
Tf ⎛ Vf ⎞
⎛ Tf ⎞
⎛Vf ⎞
=⎜ ⎟
ln ⎜ ⎟ = − (γ − 1) ln ⎜ ⎟ or equivalently,
Ti ⎝ Vi ⎠
⎝ Ti ⎠
⎝ Vi ⎠
2-34
CP ,m
γ=
CV ,m
Tf
Ti
=
Pf V f
PV
i i
γ
γ
PV
=
P
V
i i
f f Adiabatic
PV = R ⋅ 300 K Isothermal
2-35