# Specific heat of gases

Μηχανική

29 Οκτ 2013 (πριν από 4 χρόνια και 6 μήνες)

111 εμφανίσεις

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:

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.

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.

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

axis is “frozen out”

axes imparts energy

Rotational states

Quantum mechanics: rotational and
vibrational energies not continuous but
quantised

Rotation: spacing ~1/
I

small steps, near
-

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

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