Download Molar Heat Capacities of an Ideal Gas

The First Law of Thermodynamics (23)
A container with boiling water placed in a room cools off and finally
comes into thermal equilibrium at the room temperature tells us: Heat is
a form of energy that flows between a system and its surrounding
because of a temperature difference between them.
Abstractly:
Any system participating in any thermodynamic process is able to
transfer certain amount of energy either into or out of the system. The
energy exchanged between the system and its environment appears as
either heat or work or as both of them. This means that a system does
not contain heat or work. Only through a thermodynamic process in
which a system evolves from one equilibrium state to another can we
talk about energy transfer.
Later, we learn (I hope that would be the case), that heat Q and work W
are not intrinsic thermodynamic properties of a system and are not state
functions such as temperature, internal energy, entropy, …
Heat Transfer
The mechanisms by which a system and its environment transfer heat
are found to fall into three methods: conduction, convection and
Radiation. In this course we pay attention to thermal conduction and
state a few facts about thermal convection and thermal radiation
processes.
Thermal Conduction
If we leave a frying pan on a burner for some time, it will get very hot.
However, not all parts of the pan become equally hot in a given time
interval. In fact, the handle of the pan is much cooler than the rest of the
pan. As heat is energy and energy does work, the hot molecules of the
pan will be forced to have larger amplitude of oscillations than when
they were in thermal equilibrium at room temperature. As a result, we
have transfer of heat energy from high temperature region to cooler
temperature region by means of collisions of molecules.
Heat is transferred by means of molecular collisions between hot and
cooler molecules across the dimension of a body without transfer of any
material medium. If material were also transferred as heat, the body
would be permanently deformed.
Conduction, Computational Aspect
Under steady state condition dQ dt  H  constant, the one-dimensional heat flow
equation is:
dQ
dT
 A
,
dt
dx
(1)
where  is thermal conductivity of the material and A is the cross sectional area,
the – sign is necessary and A is the cross sectional area. The “–“ sign is necessary
to make dQ dt  H  0 positive.
H
dx , H  dQ  constant
Integrating Eq(1), we obtain: dT=dt
kA
T
x
H
H
H
T dT   kA 0 dx or T(x)-T1 =  kA x , or T(x)= T1  kA x
1
Heat flow rate can be obtained by: T 2 = T1 -
H
KA
l => H =
(T1-T 2 )
KA
l
The temperature distribution along the length of a slab of cross sectional area A
and length L takes the form:
1
l
T(x)= T1  (T1  T2 ) x,
H=
kA
(T1  T2 )
l
l
In particular T at x = is:
2
1
T ( l 2 )  T1  T2 
2
= average of the two fixed
temperatures T1 and T2 .
T1
The steam pipe of radius r1 is in thermal equilibrium with the steam. As
heat flows radially outward, we integrate heat equation from r1 to r 2 to
obtain:
dQ
dT
 H   kA
, where A  2rD . Accordingly:
dt
dr
r2
TR
2kD(Ts  TR )
dr
H


2

kD
dT

H

r r

Ts
ln( r2 )
1
r1
Now insert H into heat equation and integrate between r1 and r to show
that T(r) has the form:
.
T (r )  Ts 
Ts  TR
r
ln( )
ln( r2 ) r1
r1
After, going through sample problem 23-1, make use of the results given
there to obtain the temperature at the interface as:
T
1
R2THot  R1Tcold 
R1  R2
where R1 and R2 are the R-values of the respective mediums. See Eq(232) of your text.
R
L

Can you now see that if between hot reservoir at TH and cold reservoir at
T c , we have several heat conducting materials with R-vales: R 1 , R 2 , R 3 ,
…, and all with same cross sectional area A, The rate of heat flow
dQ  H is simply:
dt
A(TH  Tc )
Hseries = R  R  R  ........
1
2
3
The warm liquid at the bottom expands and rises because of reduced
density. It is replaced by cooler, denser liquid that then becomes heated
and rises. Upon reaching to the top, it cools and hence sinks back to the
bottom. These convection currents continue to flow through the liquid.
A similar situation occurs on the outer surface of the sun. The granules
observed are rising columns of hydrogen gas heated by solar surface
temperature (6000K). The rising gas eventually cools off and falls back
down toward the surface.
Convection
plays a role in
atmospheric
temperature,
earth’s magnetic
field ,continental
drift(resulting
earthquakes), off
and on shore
breezes.
The
distinguishing
factor: transfer
of heat by
radiation
requires no
material
medium. Details
are quantum
mechanical.
The First Law of Thermodynamics
Just as in mechanics, here, in thermodynamics, we also choose a system
to study. The choice of the system is quite arbitrary. Once the system is
selected then all energy exchanges between the system and its
environment must be identified. The law of conservation of energy for a
thermodynamic system where internal energy is the only type of energy
the system might have, reads:
 E int =  Q + W
In this law Q is the heat energy transferred (as heat) between the system
and its environment. Q does not include the heat exchange between
different parts within a system.
W is the amount of work done on the system (or by the system) by all
forces between the system and its environment. W does not include
work done by internal forces within the system.
We choose by way of convention:
Q>0 If heat is added to the system
W>0 If work is done on the system
It is clear that if both Q and W are positive, then the internal energy
increases and hence  Eint* > 0.
*
It is assumed that the selected system as a whole has no center of mass
motion (Kt or Krot). For systems where K = Kt + Krot #0, we must add 
K to  Eint .
The thermal behavior of a system is described in terms of
thermodynamic variables. Hence for an ideal gas these variables are P,
V, T, and N, the number of molecules of the gas (or, n the number of
moles of the gas). Different types of the systems are generally described
by different sets of variables. For example for a stretched rubber band,
the thermodynamic variables are: tension, length, temperature, and mass.
When a system is in thermodynamic equilibrium with its environment,
its thermodynamic variables are related by an equation of state. We have
seen two of these equations: Ideal gas equation and Van der Waals
equation. If the system is not in thermal equilibrium, there is no equation
of state, and in fact, there may not be a well defined temperature,
pressure, etc., that we can specify for the system.
Although both Q and W are process dependent, their sum Q + W does
not.
When the system evolves from an
initial equilibrium state (i) to some
final equilibrium state (f), the values
Qi (i = 1, 2, 3, 4) and Wi are in
general different depending which
thermodynamic process (or path) has
been selected. However, the sums:
Q i + W i remains the same along
each path. This means:
E1 ,  E2  E3  E4 .
A function of this type depending solely on the equilibrium values of the
system is called a state function (like P.E. due to gravity).
Accordingly, if for example, a gas in a container at well defined
temperature and pressure is made to explode by means of spark, then
some of the gas may condense, different gas molecules may combine to
form new compounds, … But eventually the system will settle down to a
new equilibrium state. It should be clear that the system is not in thermal
equilibrium during its transition; however, its behavior is still governed
by the first law since the process starts and ends with the system in
equilibrium states.
Heat Capacity and Specific Heat
The idea of heat capacity is directly related to temperature change that a
body undergoes as the result of adding or removing heat to the body.
One defines heat capacity C by:
dQ
C=
[C] = J K .
dT ,
As the amount of heat added to the system is process dependent the
resulting temperature change dT is also process dependent.
Consequently, one speaks of heat capacity for a given process.
If m is the mass of the body, then experiment shows that equal amount
of heat dQ for given process added to equal amount of mass m involving
different substances will result different temperature change. Hence, one
defines specific heat c = C/m as heat capacity per unit mass. Specific
heat c just like heat capacity must also be process dependent.
Accordingly, we have:
c=
1  dQ 


m  dT 
for a given process.
The specific heat for a given process is the heat required to change the
temperature of a unit mass of a substance by 1 degree using that process.
Or:
dQ  mcdT  Q  m  cdT
For cases in which c is temperature independent, we have:
Q  mcT , which is the familiar calorimetric equation.
Molar Heat Capacity
If we multiply specific heat c by molecular mass M of a substance, we
obtain the molar heat capacity for the substance. Namely:
1 dQ 1 dQ
C Molar = Mc = M
=
, m =nM
m dT
n dT
when this relation is applied to solids, one finds that it is temperature
dependent, and:
i  lim Cmolar  0 as T -> 0
lim Cmolar = constant 25J/mol.K as T  large value for all solids.
This limiting value 25J/mol.K at high temperature is known as the
Dulong-Petit value.
ii 
The complete expression of molar heat capacity as a function of
temperature is given by quantum mechanical analysis (see figure)
Study of heat capacity as a function of temperature yields understanding
of various transition properties of elements. In particular, Tcritical below
which a body shows super conductivity, and at what temperature crystal
structure of a substance is disordered.
Quantum Mechanical Results on Cv (Molar heat capacity of solids at
constant volume)
Work and Ideal Gas
As the internal energy of a system involves both work W and heat Q,
exchanged between a system and its environment, we need to have a
way of computing W and Q. We first look at W and restrict our attention
to an ideal gas situation for which there exists an equation of state:
[PV=nRT]. Consider n moles of an ideal gas in a container equipped
with frictionless movable piston of cross sectional area A. Suppose the
temperature of the gas is increased form T 1 to T 2 .
The work done on the piston by the gas while the temperature of the gas
is increasing is: W on the piston =  F(x) dx
by the gas
By Newton’s third law the work done on the gas
by the piston is then:
W on the gas = -  F(x) dx = -  PAdx = - 
by the piston
v2
v1
PdV.
This work is generally represented as the area
under pressure curve on a PV-diagram between
volume V 1 and volume V 2 .
It is important to keep in mind when calculating W that:
i)
The process used to carry out volume change V  V2  V1 was
slow enough that all intermediate stages of expansion, the gas
considered to be in equilibrium (expansion is made possible
by a series of quasi-static equilibrium states). If the
expansion is rapid or chaotic, there exists no well defined
pressure volume or temperature changes (i.e no well-defined
equation of state to be used) .
ii) As pressure forces are non-conservative, the work W
becomes thermodynamically path (or process) dependant.
For Example:
WAD  WAB  WBD
1.  0  P (V  V )
f
f
i
Since
WAD  0 Work Done on the
gas is negative .
2. WAC  WC D   pi (V f  Vi )
Now W1. AD
 W2. AD with the
difference of :
W  W1. AD  W2. AD  ( p f  Pi )V
Hence computing W, requires a knowledge of the process used. We now
compute W, for several well known processes in thermodynamics.
These are:
Constant Volume Process V  0
Constant Temperature process T  0
Constant Pressure Process P  0
Constant Heat Process Q  0
1. Constant Volume Process,
W=0
a. For W to be zero, it is not
enough for the change in
volume to be zero.
Volume must remain
constant all throughout
the process. (As an example of w=0, one may heat up a gas
in a container while keeping the volume of the container
fixed during heating process.
2. Constant Pressure Process
a. In this care, we may heat up a gas in a container and at the
same time expand its volume in a manner to keep its pressure
unchanged during the heating period. The amount of work
f
is: Wi f   Pdv   P(v f  Vi )   PV
i
If
If
V is
positive (expansion) then the work done on the gas is negative.
V  0 (compression), the work done on the gas is positive.
Work Done on a gas by a process
or any process that begins and ends
with the same pressure, will not
give the same result as compared to
a process in which the pressure is
kept constant all throughout the
process.
Work Done at Constant Temperature (Isothermal process)
The work done on an ideal gas in an isothermal process involving either an
expansion or contraction is given [ using the equation of state: PV=nRT]
b
Wab
b
b
nRT
dV
V
  PdV  
dV  nRT 
 nRT ln( b )
V
V
Va
a
a
a
For expansion where V b ,> V a , W a  b is negative and for the case of
compression in which V b < V a , W a  b > 0.
Work Done Adiabatically (Thermal Isolation)
If we do work on n moles of an ideal gas confined
within a completely insulating container, then its
temperature, pressure, and volume will change. It
turns out that in addition to the usual equation of state
(PV=nRT), we also have an adiabatic equation of
state for an ideal gas. Namely:
PV  = constant、
Where  , which is a dimensionless parameter is
called the ratio of specific heats. One can easily
dP 
 dP 
 isot ;
 = 
 dV 
 dV  adia
show that the slopes 
and since 

Cp
Cv ,
 1,
the slope is
steeper at the point of intersection of the curves PV   constant, and
PV=constant for the adiabatic curve.
dP 
Adiabatic Case: 

 dV 
Adiabatic case


P
V
a
Cp
 1.
Cv
 dP 

 dV 
Isothermal Case: 

P
,
V,
a
b
PV   constant and PV=nRT, we
Using the equations: W ab  a PdV ,
obtain for the work W on the gas in an adiabatic process the expression:
1
Pb Vb  PaVa ).
W ab =
 1

The Internal Energy of an Ideal Gas, Equipartition of Energy
1. An ideal monatomic gas is viewed as a collection of point-like
masses. There is no energy other than translational kinetic energy that
we can associate with each of these particles. Hence, the internal energy
of a monatomic gas is simply equal to its average translational Kinetic
energy. To each gas molecule that is free to move in space we associate
3
kT as its kinetic energy. For N such molecules, we have:
2
3
3  3
Eint  NK t  N  kT   NkT  nRT
2
2  2
Recall: N  nN A , and R  kN A
Accordingly, the internal energy of an ideal monatomic gas depends
only on its equilibrium temperature T. Hence, for a fixed number of
moles, the change in E int is caused solely by a change in its temperature:
Eint 
3
nRT
2
2. A diatomic ideal gas molecule, such as: H 2 , N
2 , O 2 , CO, … not only it has the usual
translational kinetic energy, but now there are two
possible rotation axes through its center of mass.
1
1
1
1
1
2
2
K per molecule  Kt  K rot  mvx2  mv y2  mvz2  I x x  I y y
2
2
2
2
2
There are five independent terms for Ktot of a single diatomic molecule.
According to the equipartition of energy theorem, to each term (also
known as degree of freedom), one associates 1 kT of energy. Hence, E
2
for N ideal diatomic gas molecules is:
5
5  5
Eint .  N  kT   NkT  nRT .
2
2  2
3. An ideal polyatomic gas such as: CO 2 , NH 3 , CH 4 , … where each
molecules has more than two atoms, there will be one additional axis of
rotation as compared to a diatomic gas molecule. Therefore, the total
kinetic energy per polyatomic molecule having six degrees of freedom
is:
K PerMolecule  K t  K rot 
1 2 1 2 1 2 1
1
1
2
2
2
mvx  mvy  mvz  I x  x  I y  y  I z  z
2
2
2
2
2
2
This kinetic energy expression includes the special cases in which the
molecules may lay a long straight line, such as the three atoms in CO 2 .
By equipartition of energy theorem, we associate
1
kT to
2
each degree of
freedom, resulting an internal energy for N such molecules the
6
2
expression: E Int=NK total of a single molecule  N ( KT )  3NKT  3nRT .
We note that in all cases discussed so far, the internal energy depends
only on temperature. We have not so far included the vibration kinetic
energy, which could be significant for diatomic and polyatomic gas
molecules at high temperatures.
The equipartition of energy theorem can also be applied to solids. In its
simplest form the energy of an atom in a solid is composed of its
oscillations in three dimensions about its equilibrium lattice site and its
interaction with neighboring atoms through its potential energy. With
only these two forms of energy, an atom in a solid has six degrees of
freedom which give rise to an internal energy of: 6 kT per atom. Hence for
2
N such atoms we have:
6 
Eint  N  kT   3
2 
NkT  3nRT
Adding an amount of heat energy Q to a
sample of solid to raising its temperature will
not change the volume of the sample.
Therefore, no work W is done on the sample.
Assuming all added heat Q has been absorbed
by the sample, then the change in the internal
energy is:
Eint  Q .
Or:
Q  3nRT .
The molar heat capacity in this constant volume process is: Cv=
1 Q
 3R. Using the value of R= 8.314 J
the value of C v  25J mol.K .
mol.K
n T
As we have seen, this is the limiting value of molar heat capacity at high
temperature. As Cv T  is a function of temperature, other values of C v for
different temperature are predicted remarkably well based on the
quantum mechanical description forwarded by Debye’s generalization of
Einstein’s model of molar heat capacity of solids.
Molar Heat Capacities of an Ideal Gas
We have already seen that the internal energy of an ideal gas depends on
temperature T. Moreover,
molar heat capacity
1 dQ
, Since
CV 
n dT
v  cons tan
Va=Vb=V=constant, Work
W=0 and
dQ=dEint hence:
CV 
1 dEint
,
n dT
3
2
5
2
We have seen Eint (monatomic)  nRT , Ediatomic  nRT
,
and Epolyatomic=3nRT. Accordingly:
CV 
3
5
R(monatomic), CV  R(diatomic), CV  3R( polyatomic).
2
2
To calculate CP, we first calculate dWa c   Pdv  d ( pv)  nRdT .
By energy equation we have: dEint  dQ  dW  dQ  nRdT . or
dQ  dEint  nRdT .
Now,
1 dQ 1 dEint  nRdT

n dT n
dT
1 dE
C P  R  ( int )  R  CV
n dT

CP 
CV
The ratio of molar heat capacities can now be evaluated according to:
Cp
R
R
2 5
 1
 1    Monoatomic
3
CV
Cv
3 3
R
2
C
R
R
2 7
  p  1
 1
 1    Diatomic
5
CV
Cv
5 5
R
2
4
   Polyatomic
3
 
 1