Download 1 7.3 Heat capacities: extensive state variables (Hiroshi Matsuoka

1
7.3 Heat capacities: extensive state variables
(Hiroshi Matsuoka)
Specific heats and molar heat capacities
Heat capacity for 1 g of substance is called “specific heat” and is useful for practical
applications.
For example, the specific heat at constant pressure of liquid water at room
temperature and under the atmospheric pressure is roughly 4 J (g " K) while that of solid copper
is roughly 0.4 J (g " K) .
As mentioned above, heat capacities indicate how easily the
temperature of the system can be changed by some heat flowing into the system. Having the
!
larger specific heat at constant pressure, liquid water is therefore harder to warm up or cool down
!
than a piece of solid copper with the same mass as the water.
Molar heat capacity is actually more physical than specific heat because molar heat capacity
corresponds to Avogadro’s number of atoms or molecules and is therefore proportional to heat
capacity per atom or molecule so that by comparing molar heat capacities of various systems, we
can gain some insight into how the total energy of each of these systems is divided among its
constituent atoms and molecules. More specifically, molar heat capacity at constant volume c v
can tell us how the total energy is distributed among different types of kinetic and potential
energies on the microscopic level. This is because c v is directly related to the molar internal
!
energy u by
# "u &
cv = % ( .
$ "T ' v
!
For example, look at the table of c v for various gases below.
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Molar heat capacity at constant volume c v for low-density gases at 15 °C and 1 atm
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Substance
cv R
!
!
Ar
1.50
He
CO
!
!
!
!
1.50
2.49
H2
2.45
N2
2.49
CO 2
3.40
2
This table suggests that there is some connection between the value of c v and the internal structure
of constituent molecules of gases as we find (we will address this issue in the next section).
Monatomic gases:
!
c v ~ ( 3 2) R
Diatomic gases:
c v ~ (5 2) R
Triatomic gases:
c v ~ ( 7 2) R
!
!
The table below also suggests that for elemental solids, c P ~ 3R = 25 J (mol" K) . Where does this
!
constant of 3R comes from? We will discuss this question in the next chapter.
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Specific heats & molar heat capacities for liquids & solids at room temperature (~300 K)
Substance
Specific heat
cP
( J (g " K ) )
( J (mol" K) )
Water
4.186
Ethyl alcohol
!
Copper
2.4
Aluminum
0.9
24.3
Gold
0.126
25.6
Ice(-10 °C)
2.05
36.9
0.386
!
!
75.2
111
24.5
Heat capacity at constant volume, internal energy, entropy, and temperature
!
We have thus far defined temperature as a label for equilibrium states so that two states with
the same value of temperature are in thermal equilibrium with each other. We have also noted
that energy flows as heat from a system with a higher temperature to another with a lower
temperature so that we have effectively ordered equilibrium states according to their
temperatures: if heat flows from a system to another, the we know the temperature of the former
is higher than the latter. What is then temperature anyway, especially on the microscopic level?
It is a difficult question to answer at this point because temperature is a concept closely coupled
with entropy, a concept we have not defined yet. Also, in discussing the equations of state for
gases, we have mentioned that as we increase the temperature of a gas, the average speed as well
as the average kinetic energy of the molecules in the gas increase. How is the temperature of a
system related to its kinetic energy after all?
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To give you some idea, let’s first look at a diatomic gas such as a nitrogen gas in the air at
room temperature. As mentioned above, the heat capacity at constant volume of the nitrogen gas
at room temperature is
5
CV = nR.
2
As the internal energy U is related to CV by
!
# "U &
! % ( = CV ,
$ "T 'V ,n
we obtain
!
5
U = nRT ,
2
where we have set U (T = 0 K,V,n ) = 0 . It turns out that this internal energy is a sum of two
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terms, one of which comes from the kinetic energy for the translational motion of each molecule
as a whole and the other of which comes from the kinetic energy for the rotational motion of
!
each molecule:
3
3
U = U trans + U rot = nRT + nRT = Nk B T + NkB T ,
2
2
where N is the total number of the molecules and satisfies nR = NkB . Clearly, the average
!
kinetic energy per molecule ( 3 2) k B T is proportional to the temperature of the gas. As we will
see in Ses.8.1, the entropy of the gas is given by
!
!
!
(" % 5 2 " %+
T
V
S (T,V,n ) = S (T0 ,V0 ,n ) + nRln*$ ' $ '-,
*)# T0 & # V0 &-,
4
where a reference state is chosen to be at (T0 ,V0 ,n ) . We can then express the entropy as a
function of the internal energy U:
!
(" % 5 2 " %+
U
V
S (U,V,n ) = S (U 0 ,V0 ,n ) + nRln*$ ' $ '- ,
*)# U 0 & # V0 &-,
where U = (5 2) nRT and U 0 = (5 2) nRT0 . We the find
!
!
!
# "S &
5
1 1
% ( = nR = ,
$ "U 'V ,n 2 U T
which is also a general relation between entropy and temperature: a rate of change of entropy
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with respect to a change in internal energy is always the inverse temperature, 1 T . As we will
see later that the entropy of a system is, on the microscopic level, related to the number W of
energy eigenstates whose energy eigenvalues are near U or in an interval [U,U + dU ] , where
!
dU << U , by
S (U,V,n ) = k B lnW (U,V,n ) .
!
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W basically counts the number of ways of dividing the energy U among the constituent
!
molecules. The above relation between S and T then becomes a relation between W and T:
# " lnW &
1 # "W &
1 # "S &
1
,
%
( =%
( = % ( =
W $ "U 'V ,n $ "U 'V ,n k B $ "U 'V ,n k B T
which means that the fractional increase of W is proportional to 1 T : the higher the temperature
!
is, the smaller the fractional increase of the number of eigenstates degenerate in energy
becomes. W is analogous to the surface area A = 4 "r 2 of a sphere while U is analogous to its
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diameter D and kB T is analogous to its radius r as
!
& 1
1 dA
1 # d
=
4 "r 2 ( = .
2%
' r
A dD 4 "r $ 2dr
!
!
5
As another example, consider an insulator solid whose heat capacity at constant volume of
the nitrogen gas at low temperatures much lower than its Debye temperature " is found (see
Sec7.4.2) to be
!
3
#T&
CV = anR% ( ,
$" '
where a is a positive constant. The internal energy U is then
!
# T &4
1
U = anR"% ( ,
$" '
4
where we have set U (T = 0 K,V,n ) = 0 and " is a function of the molar volume v of the solid. It
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turns out that this internal energy is a sum of two terms, one of which comes from the vibrational
kinetic energy of each atom and the
! other of which comes from the potential energy among the
!
atoms due to inter-atomic forces:
U = U kin + U pot
!
# T &4 1
# T &4 1
# T &4 1
# T &4
1
= anR"% ( + anR"% ( = aNk B"% ( + aNk B"% ( ,
$" ' 8
$" ' 8
$" ' 8
$" '
8
where N is the total number of the atoms and satisfies nR = NkB . In this case, the average kinetic
energy per atom is not simply proportional to the temperature of the solid. The entropy of the
solid is then given by
!
# T &3
1
S (T,V,n ) = anR% ( ,
$" '
3
where we have used the third law of thermodynamic: S (T = 0 K,V,n ) = 0 . We can then express
!
the entropy as a function of the internal energy U:
!
# 4U & 3 4
1
S (U,V,n ) = anR%
(
$ anR" '
3
!
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so that we obtain the general relation between S and T as we have found for the nitrogen gas at
room temperature:
# "S &
1 # anR) & 1
% ( = %
(= .
$ "U 'V ,n ) $ 4U ' T
At temperatures higher than the Debye temperature " , CV is found (see Sec7.4.2) to be
!
CV " 3nR,
! !
so that the internal energy U is roughly
!
U " 3nRT ,
where we have set U (T0 ,V,n ) = 3nRT0 and T0 is a reference temperature that satisfies T0 >> " .
!
This internal energy is also a sum of the total vibrational kinetic energy of the atoms and the total
potential energy among the atoms due to inter-atomic forces:
!
!
!
3
3
3
3
U = U kin + U pot = nRT + nRT = Nk B T + Nk B T ,
2
2
2
2
Note that the average kinetic energy per atom is proportional to the temperature of the solid. The
entropy of the solid is then given by
!
"T%
S (T,V,n ) = S (T0 ,V,n ) + 3nRln$ ' ,
# T0 &
so that we can express the entropy as a function of the internal energy U:
!
"U%
S (U,V,n ) = S (U 0 ,V,n ) + 3nRln$ '
# U0 &
and
!
!
# "S &
3nR 1
= .
% ( =
$ "U 'V ,n
U
T
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Appendix: A more careful derivation of CP = CV + TV" 2 # T
To derive this equation, we use the first law of thermodynamics or the law of conservation of
!
energy. Suppose that our system is initially at an equilibrium state (T,V,n ) , where the pressure
is determined through the equation of state P = P (T,V,n ) . If we slowly change the temperature
and volume by infinitesimal amounts dT and dV while keeping the pressure constant, the system
!
ends up in an equilibrium state (T + dT,V + dV,n ) for which the pressure remains at P. The
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volume change dV is related by
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# "V &
1 # "V &
dV = % ( dT = V % ( dT = V)dT .
$ "T ' P.n
V $ "T ' P.n
According to the first law, in this infinitesimal quasi-static isobaric (i.e., constant-pressure)
!
process, the internal energy changes by
qs
dU P,n = "QP qs
,n +"W P,n= C P dT # PdV .
We can also start from the same initial state and end with the same final state through a different
!
route. We first slowly change the temperature by dT while keeping the volume constant and then
slowly change the volume from V to V + dV while keeping the temperature at T + dT . In the
first leg of this roundabout process, the internal energy changes by
!
qs
dUV ,n = "QV qs
,n +"W V ,n= CV dT ,
!
where "W V qs
,n= #PdV = 0 as dV = 0. In the second leg of this roundabout process, the internal
!
energy changes by
!
!
qs
dUT +dT ,n = "QT +dT qs
,n +"W T +dT ,n= (T + dT )
&T
!
# (T + dT )
dV % P (T + dT,V,n ) dV
$ T (T + dT )
#
#
#2
dV % PdV = T (V#dT ) % PdV = TV
dV % PdV
$T
$T
$T
8
where
+ 1 % $" (
.
" ,1+ ' * dT /
" (T + dT )
- " & $T )V ,n 0
dV = (T + dT )
dV
(T + dT )
+
.
# T (T + dT )
1 % $#T (
# T ,1+ '
* dT /
- # T & $T )V ,n 0
1T
"
dV + O( dTdV )
#T
and
!
P (T + dT,V,n ) dV = P (T,V,n ) + O( dTdV )
so that we have neglected terms on the order of or smaller than dTdV . Finally, as the internal
!
energy is a state variable, its change dU P,n must be equal to the difference between its final value
U (T + dT,V + dV,n ) and initial value U (T,V,n ) so that !
!
dU P,n = U (T + dT,V + dV,n ) " U (T,V,n )
!
= {U (T + dT,V + dV,n ) " U (T + dT,V,n )} + {U (T + dT,V,n ) " U (T,V,n )}
!
= dUT +dT ,n + dUV ,n
!
from which we get
% #2
(
%
#2 (
CP dT " PdV = ' TV
dT " PdV * + CV dT = 'CV + TV * dT " PdV
$T )
& $T
)
&
and
!
"2
CP = CV + TV
.
#T
This derivation shows that the second term TV" 2 # T comes from "QT +dT qs
,n and that C P " CV
!
because "QT +dT qs
,n is positive as long as dT is positive.
!
!
!
!
!
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SUMMARY FOR SEC.73
1. The heat capacity at constant volume CV of a system is an extensive state variable related to
the internal energy U of the system by
# "U &
CV!= % ( = nc v ,
$ "T 'V ,n
where c v is the molar heat capacity at constant volume and is related to the molar internal
energy u by
!
# "u &
cv = % ( .
$ "T ' v
!
2.
# "U &
! % ( >0
$ "T 'V ,n
and
# "u &
% ( >0
$ "T ' v
and
c v > 0.
imply
CV > 0
!
3. The heat!capacity at constant pressure CP of a system is an extensive state variable related to
the enthalpy H " U + PV of the system by
!
!
# "H &
CP!= % ( = nc P ,
$ "T ' P,n
!
!
where c P is the molar heat capacity at constant pressure and is related to the molar enthalpy h
by
!
# "h &
cP = % ( .
$ "T ' P
4. The enthalpy H of a system is an extensive state variable defined by
! H " U + PV = n ( u + Pv ) = nh (T,P ) .
5. CP and CV are related with each other by
!
CP = CV + TV
!
!
6.
c P and c v are related with each other by
!
c P = c v + Tv
!
"2
$ CV .
#T
!
!
"2
$ cv .
#T
10
For a low-density gas, for which the ideal gas law, Pv = RT , is a good approximation for its
equation of state, we find
c P = c v + R.
!
7. As CP " CV > 0 and c P " c v > 0,
!
!
!# &
"H
% ( >0
$ "T ' P,n
# "h &
% ( > 0,
$ "T ' P
and
which imply that H as a function of T, P, and n is an increasing function of T and that h as a
function of T and P is an increasing function of T.
!
!
8. We can calculate the molar enthalpy h of a system by
T
h (T,P ) = h (T0 ,P ) +
# c (T ",P )dT ".
P
T0
9. We can calculate the molar internal energy U of a system by
!
u(T,P ) = h (T,P ) " Pv (T,P ) .
10. At room temperature and under the atmospheric pressure
!
Monatomic
gases:
Diatomic gases:
Triatomic gases:
Elemental solids:
c v ~ ( 3 2) R
c v ~ (5 2) R
c v ~ ( 7 2) R
!
!
!
c P ~ 3R
! the homework question in Sec.7.3
Answer for
HW#7.3.1
For low-density gases: Pv = RT , which leads to
!=
1
T
and
!T =
1
P
so that
2
"2
Pv
(1 T )
cv = c P ! Tv
= c P ! Tv
= cP !
= cP ! R.
#T
(1 P )
T