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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 of
the ideal gas to its temperature.
Outl ine:
1. Internal Energy, Work, Heating
2. Energy Conservationthe First Law
3. Enthalpy
4. Heat Capacity
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Internal Energy
systemU = kinet ic + po tential
system
boundary
environment
The internal energy of a system of particles, U, is the sum of the kinetic
energy in the reference frame in which the center of mass is at restand 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 astate functionit 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 average
inter-particle distance (~ V-1/3)interactions.
P
VT A
B
For an ideal gas (no interactions) : U = U(T
) - pure kinetic
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Internal Energy of an Ideal Gas
The internal energy of an ideal gas
with fdegrees of freedom: TNkf
U B2
f 3 (monatomic), 5 (diatomic), 6 (polyatomic)
How does the internal energy of air in this (not-air-tight) room change
with Tif the external P = const?
PVf
Tk
PVNTkN
fU
B
roominBroomin22
(here we consider only trans.+rotat. degrees of freedom, and neglectthe vibrational ones that can be excited at very high temperatures)
- does not change at all, an increase of the kinetic energy of individual
molecules with Tis compensated by a decrease of their number.
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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 Tacross a boundary and a finite
thermal conductivity of the boundary
heat ing (Q > 0)/c oo l ing (Q < 0)
(there is no such physical quantity as heat; to
emphasize this fact, it is better to use the term
heatingrather than heat)
W - any other kind of energy transfer across
boundary - work
Heating/cool in g pro cesses:
conduct ion: the energy transfer by molecular contact fast-moving
molecules transfer energy to slow-moving molecules by collisions;
convect ion: by macroscopic motion of gas or liquid
radiat ion: by emission/absorption of electromagnetic radiation.
HEATING
WORK
Work and Heating are both defined to describe energy transfer
across a system boundary.
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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 1stLaw (energy would be created from nothing), type 2 violates
the 2ndLaw (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 + W
From the microscopic point of view, this statement is equivalent to
a statement of conservation of energy.
P
VT
An equivalent formulation:
Perpetual motion machines of the first type do not exist.
Sign convent ion:we consider Qand Wto be posi t iveif energy flows
in tothe system.
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Quasi-Static Processes
Quasi-stat ic (quasi-equi l ibr ium ) processessufficiently
slow processes, anyintermediate state can be consideredas an equilibrium state (the macroparamers are well-
defined for all intermediate states).
Examples ofquasi-
equi l ibr ium processes:
isochoric: V= const
isobaric: P= const
isothermal: T= const
adiabatic: Q= 0
For quasi-equilibrium processes, P, V, T are
well-def ined the path between two statesis a continuouslinesin the P, V, Tspace.
P
V
T
1
2
Advantage: the state of a system that participates in a quasi-equilibrium
process can be described with the same (smal l ) 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-
equi l ibr ium processes(e.g. turbulent flow of gas), we need a huge number
of macro parameters.
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Work
The sign:if the volume is decreased, Wis posi t ive(by
compressing gas, we increase its internal energy); if the
volume is increased, W is negat ive(the gas decreases
its internal energy by doing some work on the
environment).
2
1
),(21V
V
dVVTPW
The workdon e by an external forceon a gas
enclosed within a cylinder fitted with a piston:
W = (PA ) dx = P (Adx)= - PdV
x
P
W = - PdV - applies to any
shape of system boundary
The work is not necessarily associated with the volume changese.g.,
in the Joulesexperiments on determining the mechanicalequivalent of
heat,the system (water) was heated by stirring.
dU = Q PdV
A the
pistonarea
force
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Wand Qare no tState Functions
P
V
P2
P1
V1 V2
A B
CD
- the work is negative for the clockwisecycle; 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 in i t ia land f inalstates, but also on the
intermediatestates, itis not a state function.
PV diagram
P
V
T1
2
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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 final
states of a system does not fix the values of Q and W, we need to know the whole
process (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, Qand Ware not exact differentials of some functions
of macroparameters. To emphasize that Wand Qare NOT the state
functions, we will use sometimes the curled symbols (instead of d)
for their increments (Qand W).
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Problem
Imagine that an ideal monatomic gas is taken from its initial stateA
to stateBby an isothermalprocess, from Bto Cby an isobaricprocess, and from
Cback to its initial state A by an isochoricprocess. Fill in the signs of Q,
W, and U for each step.
V, m3
P,
105PaA
B
C
Step Q W U
AB
B C
C A
2
1
1 2
+ -- 0
-- + --
+ 0 +
T=const
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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 (=105N/m2) to
0.5 bar. Assume that the pressure changes linearly with volume
between Viand Vf.
(a) If the initial Tis 300K, what is the final T?
(b) How much work is done bythe gas in the balloon?
(c) How much heatdoes the gas absorb, if any?
P
V
Pf
Pi
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 ona system f
i
V
V
BY dVVPW )( - work done bya system
BYON WW
JJJ 445 105.41061.0105.112
3
2
3
BY
i
f
iiONifBON WT
TVPWTTNkWUQ
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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)
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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 if Vi>Vf (compression)
Wi-f < 0 if Vi
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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 onlyon the initialand finalstates and not on the intermediatestates.
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
V P
dP
V
dV
V
P
V1
PV= NkBT1
PV= NkBT21
2
V2
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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 steeperthan 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
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Problem
Imagine that we rapidly compress a sample of air whose initial pressure is
105Pa and temperature is 220C (= 295 K) to a volume that is a quarter of
its original volume (e.g., pumping bikestire). 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 compressionapprox. adiabatic, no time for the energy
exchange with the environment due to thermal conductivity
constTP
/1
also
- poor approx. for a bike pump, works better for diesel engines
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Non-equilibrium Adiabatic Processes
- applies only to quasi-equi l ibr iumprocesses !!!constTV 1
2. On the other hand, U = Q + W = 0
U ~ T Tunchanged
(agrees with experimental finding)
Contradiction because approach
#1 cannot be justified violent
expansion of gas i s not a quasi-
static process. T must remain the
same.
(Joules Free-Expansion Experiment)
constTV 11. Vincreases Tdecreases (cooling)
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The Enthalpy
Isobaric processes (P = const):
dU = Q - PV = Q - (PV) Q = U + (PV)
The enthalpy is a state fun ct ion, because U, P,
and V are state functions. In isobaric processes,the energy received by a system by heating equals
to the change in enthalpy.
Q= H
isochoric:
isobaric:
in both cases, Q
does not
depend on the
path from 1 to 2.
Consequence: the energy released (absorbed) in chemical reactions at constant
volume (pressure) depends only on the initial and final states of a system.
H U + PV - the enthalpy
The enth alpy of an id eal gas:
(depends onT only)
TNkf
TNkTNkf
PVUH BBB
1
22
Q= U
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Heat Capacity
T
QC
The heat capacity o f a system- the amount of energy
transfer due to heating required to produce a unit
temperature rise in that system
Cis NOT a state function(since Qis not a
state function)it depends on the path
between two states of a system
T
V
T1
T1+dT
i
f1 f2 f3
The specif ic heat capacitym
Cc
( isothermic C =, adiabatic C = 0 )
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CVand CP
dT
PdVdU
dT
QC
VV
T
U
C
the heat capacity atconstant vo lume
the heat capacity at
con stant pressureP
PT
HC
To find CPand CV, we need f (P,V,T) = 0and U = U (V,T)
dT
dVP
V
U
T
U
dVV
UdT
T
UdU
dT
PdVdUC
TV
TV
PT
VPdT
dVP
V
UCC
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CVand 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