Specific heat of gases

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29 Οκτ 2013 (πριν από 4 χρόνια και 2 μήνες)

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Specific heat of gases

Specific heat of monatomic
gas I


Add heat keeping the volume constant. Energy
can only go into kinetic energy of atoms:





Conclusion:

Specific heat of a monatomic
gas II


Add heat keeping the pressure constant.
Energy can go into kinetic energy of atoms, or
into work when the gas expands:





Some mathematical gymnastics:

Specific heat of a monatomic gas
III


Substitute:





Conclusion:



Valid for all ideal gases


Equipartition of energy


According to classical mechanics, molecules
in thermal equilibrium have an average energy
of associated with each independent
degree of freedom of their motion
provided
that the expression for energy is quadratic.



monatomic gas:

x
,
y
,
z
motion:
since the terms are quadratic

Degrees of freedom for
molecules


We take
m

to be the mass of the molecule,
then as for monatomic gases



However, there is now internal energy due to
rotation and vibration:



Diatomic molecule: rotation


Think of molecule as dumbbell:



Centre
-
of
-
mass (CM) moves with K.E.



Rotation about CM with K.E.



All quadratic:
f
= 6

Diatomic molecule: vibration


Atoms attract each other
via Lennard
-
Jones potential



Atoms stay close to
equilibrium



Can be approximated by a
harmonic oscillator
potential

Harmonic oscillator


Average P.E. = average K.E.



P.E. is quadratic so
associated with it



K.E. is the same so also



Total:
kT

or
f

= 2.

Diatomic molecules:
conclusion?


According to classical physics there should be
8 degrees of freedom for each diatomic
molecule…



From
C
p

and
C
V

measurements for H
2
:


below 70 K:
f

= 3


at room temperature:
f

= 5


above 5000 K:
f

= 7

Question


If e.g. vibration of H
2

does not contribute to
the specific heat, then that is because


a) The vibration is independent of temperature

b) The vibration doesn’t change in collisions

c) The vibration is like a very stiff spring

Quantum mechanical effects


Translation: 3 degrees



Rotation: 2 degrees




Vibration:


at room temperature often also “frozen out”


at high temperatures K.E. + P.E. = 2 degrees

Rotation about this

axis is “frozen out”

Rotation about these

axes imparts energy

Rotational states


Quantum mechanics: rotational and
vibrational energies not continuous but
quantised



Rotation: spacing ~1/
I


small steps, near
-
continuous about two axes


big steps about axis through atoms


If step >>
kT

: not excited, no rotation


Even at
T
=10
4

K no rotation about third axis

Vibrational states


Vibration: harmonic oscillator


Spacing ~1/

m


step >>
kT

for light molecules (O
2
, N
2
)
but excitations possible for heavy
molecules such as I
2


Potential energy: on average equal to
kinetic energy


vibration contributes
kT

when it does





Solids


Model for solid:






Elemental (“monatomic”) solids have
K.E. and P.E. to give
C

= 3
R

(Rule of
Dulong and Petit)

Freezing out
-

again


At lower temperatures, some vibrational levels
cannot be reached with
kT
energy. At very low
temperatures
C



T
3
.



For metals the electrons (which can move
freely through the metal) contribute and
C

=
aT

+
bT
3
(Einstein
-
Debye model)

Specific molar heat


Dulong & Petit: elemental solids C


25 J mol
-
1

K
-
1

Question 2


A room measures 4.00

2.40

2.40 m
3
. Assume
the “air molecules” all have a velocity of 360
ms
-
1

and a mass of 5

10
-
26

kg. The density of
air is about 1 kg m
-
3
. Calculate the pressure in
this room.

Isothermal expansion of an ideal
gas


Isothermal expansion: pressure drops as
volume increases since
pV
=
nRT
= constant



The internal energy only depends on the
temperature so it doesn’t change



Two equations hold at the same time:





pV
=
nRT

and
T
2

=
T
1

Question


An identical volume of the same gas expands
adiabatically to the same volume. The pressure
drops



a) more than in the isothermal process


b) by the same amount


c) less than in the isothermal process


Adiabatic expansion of an ideal
gas I


Q

= 0 so

U

+
W

= 0




Remember:

U

=
nC
V

T



Use ideal gas law:



Substitute:

Adiabatic expansion of an ideal gas
II


Divide by
nC
V
T
:




Prepare for integration:




Define so that

Adiabatic expansion of an ideal gas
III


Integrate:




Play around with it:




Use



Question


Consider isothermal and adiabatic expansion of
a Van der Waals gas. Do
T
2

=
T
1
and
pV

= C

hold for this gas?



a) yes; yes


b) yes; no


c) no; yes


d) no; no

Question


A gas is compressed adiabatically. The
temperature



a) rises because work is done on the gas


b) rises depending on how much heat is added


c) drops because work is done by the gas


d) drops depending on how much heat is added

PS225


Thermal Physics
topics


The atomic hypothesis


Heat and heat transfer


Kinetic theory


The Boltzmann factor


The First Law of Thermodynamics


Specific Heat


Entropy


Heat engines


Phase transitions