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7/30/2019 7 the First Law of Thermodynamics (Adv)
1/21
Lecture 2 The First Law of Thermodynamics (Ch.1)
Lecture 1 - we introduced macroscopic parameters that describethe state of a thermodynamic system (including temperature), the
equation of state f(P,V,T) = 0, and linked the internal energy ofthe ideal gas to its temperature.
Outline:
1. Internal Energy, Work, Heating
2. Energy Conservation the First Law
3. Enthalpy
4. Heat Capacity
7/30/2019 7 the First Law of Thermodynamics (Adv)
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Internal Energy
systemU = kinetic + potential
system
boundary
environment
The internal energy of a system of particles, U, is the sum of the kineticenergy in the reference frame in which the center of mass is at rest and the
potential energy arising from the forces of the particles on each other.
Difference between the total energy and the internal energy?
The internal energy is a state function it depends only on
the values of macroparameters (the state of a system), not
on the method of preparation of this state (the path in the
macroparameter space is irrelevant).
U = U (V, T)In equilibrium [ f(P,V,T)=0 ] :
U depends on the kinetic energy of particles in a system and an averageinter-particle distance (~ V-1/3) interactions.
P
VT A
B
For an ideal gas (no interactions) :
U = U (T) - pure kinetic
7/30/2019 7 the First Law of Thermodynamics (Adv)
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7/30/2019 7 the First Law of Thermodynamics (Adv)
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Work and Heating (Heat)
We are often interested in U , not U. U isdue to:Q - energy flow between a system and itsenvironment due to T across a boundary and a finite
thermal conductivity of the boundary
heating (Q > 0) /cooling (Q < 0)(there is no such physical quantity as heat; to
emphasize this fact, it is better to use the term
heating rather than heat)
W - any other kind of energy transfer acrossboundary - work
Heating/cooling processes:
conduction: the energy transfer by molecular contact fast-movingmolecules transfer energy to slow-moving molecules by collisions;
convection: by macroscopic motion of gas or liquid
radiation: by emission/absorption of electromagnetic radiation.
HEATING
WORK
Work and Heating are both defined to describe energy transfer
across a system boundary.
7/30/2019 7 the First Law of Thermodynamics (Adv)
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The First Law
For a cyclic process (Ui = Uf) Q= - W.If, in addition, Q = 0 then W =0
Perpetual motion machines come in two types: type 1 violates
the 1st Law (energy would be created from nothing), type 2 violates
the 2nd Law (the energy is extracted from a reservoir in a way that
causes the net entropy of the machine+reservoir to decrease).
The first law of thermodynamics: the internal energy of a system can be
changed by doing work on it or by heating/cooling it.
U = Q + WFrom the microscopic point of view, this statement is equivalent to
a statement ofconservation of energy.
P
VT
An equivalent formulation:
Perpetual motion machines of the first type do not exist.
Sign convention: we considerQ and W to be positive if energy flows
into the system.
7/30/2019 7 the First Law of Thermodynamics (Adv)
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Quasi-Static Processes
Quasi-static (quasi-equilibrium) processes sufficiently
slow processes, any intermediate state can be consideredas an equilibrium state (the macroparamers are well-
defined for all intermediate states).
Examples ofquasi-
equilibrium processes:
isochoric: V = const
isobaric: P = const
isothermal: T = const
adiabatic: Q = 0
For quasi-equilibrium processes, P, V, T are
well-defined the path between two statesis a continuouslines in the P, V,T space.
P
V
T
1
2
Advantage: the state of a system that participates in a quasi-equilibriumprocess can be described with the same (small) number of macro
parameters as for a system in equilibrium (e.g., for an ideal gas in quasi-equilibrium processes, this could be T and P). By contrast, for non-equilibrium processes (e.g. turbulent flow of gas), we need a huge numberof macro parameters.
7/30/2019 7 the First Law of Thermodynamics (Adv)
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Work
The sign: if the volume is decreased, W is positive(bycompressing gas, we increase its internal energy); if the
volume is increased, W is negative (the gas decreasesits internal energy by doing some work on the
environment).
2
1
),(21V
V
dVVTPW
The workdone by an external forceon a gasenclosed within a cylinder fitted with a piston:
W =(PA) dx = P (Adx) = - PdV
x
P
W = - PdV - applies to anyshape of system boundary
The work is not necessarily associated with the volume changes e.g.,
in the Joules experiments on determining the mechanical equivalent of
heat, the system (water) was heated by stirring.
dU = Q PdV
A the
pistonarea
force
7/30/2019 7 the First Law of Thermodynamics (Adv)
8/21
W and Qare not State Functions
P
V
P2
P1
V1 V2
A B
CD
- the work is negative for the clockwise cycle; if
the cyclic process were carried out in the reverse
order (counterclockwise), the net work done on
the gas would be positive.
01212
211122
VVPP
VVPVVPWWW CDABnet
2
1
),(21V
VdVVTPW
- we can bring the system from state 1 to
state 2 along infinite # of paths, and for each
path P(T,V) will be different.
U is a state function, W - is not thus, Qisnot a state function either.
U = Q + W
Since the work done on a system depends not
only on the initial and final states, but also on theintermediate states, it is not a state function.
PV diagram
P
V
T1
2
7/30/2019 7 the First Law of Thermodynamics (Adv)
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Comment on State Functions
U, P, T, and V are the state functions, Q and W are not. Specifying an initial and finalstates of a system does not fix the values of Q and W, we need to know the wholeprocess (the intermediate states). Analogy: in classical mechanics, if a force is not
conservative (e.g., friction), the initial and final positions do not determine the work, the
entire path must be specified.
x
y z(x1,y1)
z(x2,y2)
dyy
zdx
x
zzd
xy
dyyxAdxyxAad yx ,, - it is an exact differential if it is
x
yxA
y
yxA yx
,,
yxzdyydxxzad ,, the difference between the values of some (state) function
z(x,y) at these points:
A necessary and sufficient condition for this:
If this condition
holds:
y
yxzyxA
x
yxzyxA yx
,,
,,
e.g., for an ideal gas:
dV
V
TdT
fNkPdVdUQ
B 2 - cross derivatives
are not equal
dVPSdTUd
U
VS- an exact differential
In math terms, Q and W are not exact differentials of some functionsof macroparameters. To emphasize that W and Q are NOT the statefunctions, we will use sometimes the curled symbols (instead ofd)for their increments (Q and W).
7/30/2019 7 the First Law of Thermodynamics (Adv)
10/21
Problem
Imagine that an ideal monatomic gas is taken from its initial state A to stateB by an isothermalprocess, from B to C by an isobaricprocess, and fromC back to its initial state A by an isochoricprocess. Fill in the signs ofQ,W, and U for each step.
V, m3
P,105Pa
A
BC
Step Q W UA B
B C
C A
2
1
1 2
+ -- 0
-- + --
+ 0 +
T=const
7/30/2019 7 the First Law of Thermodynamics (Adv)
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Another Problem
During the ascent of a meteorological helium-gas filled balloon,
its volume increases from Vi = 1 m3 to Vf = 1.8 m
3, and the
pressure inside the balloon decreases from 1 bar (=105 N/m2) to
0.5 bar. Assume that the pressure changes linearly with volume
between Vi and Vf.(a) If the initialT is 300K, what is the finalT?(b) How much work is done by the gas in the balloon?(c) How much heat does the gas absorb, if any?
P
V
Pf
P i
Vi Vf
K2701mbar1
1.8mbar5.0K300
3
3
ii
ff
if
B
BVP
VPTT
Nk
PVTTNkPV
(a)
(b) f
i
V
V
ON dVVPW )(
bar625.1bar/m625.0 3 VVP
(c) ONWQU
Jmbarmbarmbar 333 41066.04.05.08.05.0)( f
i
V
V
BY dVVPW
- work done on a system f
i
V
V
BY dVVPW )( - work done by a system
BYON WW
JJJ 445 105.41061.0105.112
3
2
3
BY
i
f
iiONifBON WT
TVPWTTNkWUQ
7/30/2019 7 the First Law of Thermodynamics (Adv)
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Quasistatic Processes in an Ideal Gas
isochoric ( V = const )
isobaric ( P = const )
021 W
TCTTNkQ VB 02
31221
0),( 122
1
21 VVPdVTVPW
TCTTNkQ PB 02
51221
21QdU
2121 QWdU
V
P
V1,2
PV= NkBT1
PV= NkBT21
2
V
P
V1
PV= NkBT1
PV= NkBT21
2
V2
(see the last slide)
7/30/2019 7 the First Law of Thermodynamics (Adv)
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Isothermal Process in an Ideal Gas
1
221 ln),(
2
1
2
1V
VTNk
V
dVTNkdVTVPW B
V
V
B
V
V
f
iBfi
V
VTNkW ln
Wi-f> 0 ifVi>Vf (compression)
Wi-f< 0 ifVi
7/30/2019 7 the First Law of Thermodynamics (Adv)
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Adiabatic Process in an Ideal Gas
adiabatic(thermallyisolatedsystem)
PdVdTNkf
dUTNkf
U BB 22
( f the # of unfrozen degrees of freedom )
dTNkVdPPdVTNkPV BB PVPdVfVdPPdV 2
fP
dP
fV
dV 21,0
21
constVPPVP
P
V
V
111
1lnln
The amount of work needed to change the state of a thermally isolated systemdepends only on the initialand finalstates and not on the intermediate states.
021 Q 21WdU
to calculate W1-2 , we need to know P(V,T)
for an adiabatic process
2
1
),(21
V
V
dVTVPW
0
11
P
P
V
VP
dP
V
dV
V
P
V1
PV= NkBT1
PV= NkBT21
2
V2
7/30/2019 7 the First Law of Thermodynamics (Adv)
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Adiabatic Process in an Ideal Gas (cont.)
V
P
V1
PV= NkBT1
PV= NkBT21
2
1
1
1
2
11
1121
11
1
1
),(2
1
2
1
VVVP
dVV
VPdVTVPW
V
V
V
V
1+2/31.67 (monatomic), 1+2/5 =1.4 (diatomic), 1+2/6 1.33 (polyatomic)(again, neglecting the vibrational degrees of freedom)
constVPPV
11
An adiabata is steeper than an isotherma:
in an adiabatic process, the work flowing
out of the gas comes at the expense of its
thermal energy its temperature will
decrease.V2
7/30/2019 7 the First Law of Thermodynamics (Adv)
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Problem
Imagine that we rapidly compress a sample of air whose initial pressure is
105 Pa and temperature is 220C (= 295 K) to a volume that is a quarter of
its original volume (e.g., pumping bikes tire). What is its final temperature?
2211
222
111
VPVP
TNkVP
TNkVP
B
B
1
2
1
2
12
1
1121
2
11
2
112T
T
V
VTT
VPTNkV
VP
V
VPP B
constVTVT 1
22
1
11
KKKV
VTT 51474.12954295 4.0
1
2
112
For adiabatic processes:
Rapid compression approx. adiabatic, no time for the energyexchange with the environment due to thermal conductivity
constTP
/1
also
- poor approx. for a bike pump, works better for diesel engines
7/30/2019 7 the First Law of Thermodynamics (Adv)
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Non-equilibrium Adiabatic Processes
- applies only to quasi-equilibrium processes !!!constTV 1
2. On the other hand, U = Q + W = 0U ~ T T unchanged
(agrees with experimental finding)
Contradiction because approach
#1 cannot be justified violent
expansion of gas is not a quasi-
static process. T must remain the
same.
(Joules Free-Expansion Experiment)
constTV 11. V increases T decreases (cooling)
7/30/2019 7 the First Law of Thermodynamics (Adv)
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The Enthalpy
Isobaric processes (P = const):
dU = Q - PV = Q -(PV) Q = U +(PV)
The enthalpy is a state function, because U, P,
and V are state functions. In isobaric processes,the energy received by a system by heating equalsto the change in enthalpy.
Q = Hisochoric:
isobaric:
in both cases, Qdoes not
depend on the
path from 1 to 2.
Consequence: the energy released (absorbed) in chemical reactions at constantvolume (pressure) depends only on the initial and final states of a system.
H U + PV - the enthalpy
The enthalpy of an ideal gas:
(depends on T only)
TNkf
TNkTNkf
PVUH BBB
1
22
Q = U
7/30/2019 7 the First Law of Thermodynamics (Adv)
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Heat Capacity
T
QC
The heat capacity of a system - the amount of energytransfer due to heating required to produce a unit
temperature rise in that system
Cis NOT a state function (since Q is not astate function) it depends on the path
between two states of a system
T
V
T1
T1+dT
i
f1 f2 f3
The specific heat capacitym
Cc
( isothermic C =, adiabatic C = 0 )
7/30/2019 7 the First Law of Thermodynamics (Adv)
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CV and CP
dT
PdVdU
dT
QC
VV
T
U
C
the heat capacity atconstant volume
the heat capacity atconstant pressure
P
PT
HC
To find CP and CV, we need f(P,V,T) = 0 and U = U (V,T)
dT
dVP
V
U
T
U
dVV
UdT
T
UdU
dT
PdVdUC
TV
TV
PT
VPdT
dVP
V
UCC
7/30/2019 7 the First Law of Thermodynamics (Adv)
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CV and CPforan Ideal Gas
nRf
Nkf
C BV22
CV = dU/dT
3/2NkB
5/2NkB
7/2NkB
10 100 1000 T, K
Translation
Rotation
Vibration
CV
of one mole of H2
R
NkP
Nk
PdT
dV
PV
U
CC BB
PTVP
( for one mole )0 nR
f
Nkf
NkCC BBVP
12
1
2
For an ideal gas
TNkf
U B2
# of moles
For one mole of a monatomic ideal gas: RCRCPV 2
5
2
3