Lecture6.pdf

CHAPTER 2 (Internal Energy and Kinetic Energy)

Lecture 6

Connection between the translation kinetic

energy

K

tr

and

T

for Ideal Gases

Pressure – the result of collisions between the molecules

and walls of the container.

v

_x Piston _{area A}

L

Volume = LA

For each (elastic) collision:



p

_x

= 2

m v

_x

Intervals between collisions:



t = 2 L/v

_x

Strategy: Pressure = Force/Area = [



p

/



t

]/Area

2

2

1

1

2 /

x

i x

x

mv

P

mv

L

v A

V





2 2

x N

x

N

m

v

mv

PV







-Connection between

K

_tr

and

T

for Ideal Gases (cont.)

2 2 x N i

x

N

m

v

mv

PV







T

Nk

PV



_B

2 2 2 2 2

2

3

2

1

2

1

x z y x

tr

m

v

m

v

v

v

m

v

K











T

k

v

m

x

2



B

T

k

K

_{tr B}

2

3



The temperature of a gas is a direct measure of the average

translational kinetic energy of its molecules

! Average kinetic energy of the

translational motion of molecules:

tr B

3

3

U = K =

Nk T

PV

2



2

The internal energy U of a monatomic ideal gas

is independent of its volume,

and depends only on

T

(



U

=0 for an isothermal process,

T

=const

).

Units for Energy, Temperature

23

3

,

1.38 10

/

2

tr B B

K



k T

k







J K

The kinetic energy is proportional to the temperature, and the

Boltzmann constant

k

_B

is the coefficient of proportionality that provides

one-to-one correspondence between the units of energy and

temperature.

A convenient unit for energy is an

electron-Volt

defined as:

The kinetic

energy acquired by an electron accelerated by electrostatic potential

difference of 1V).

23

19

1.38 10

/

100

100 ,

8.625

1.6 10

/

B

J K

K

At T

K

k T

meV

J eV

















19

1.6 10

Comparison with Experiment

T

Nk

U

_B

2

3



- for a point mass with three degrees of freedom dU/dT(300K) (J/K·mole) Monatomic Helium 12.5 Argon 12.5 Neon 12.7 Krypton 12.3 Diatomic

H₂ 20.4 N₂ 20.8 O₂ 21.1

CO 21

Polyatomic

H₂0 27.0 CO₂ 28.5

Testable prediction: if we put a known dU into a sample of gas, and measure the resulting change

dT, we expect to get







mole

J/K

5

.

12

J/K

10

38

.

1

mole

10

6

2

3

2

3

23 1 -23













 B

Nk

dT

dU

Conclusion: diatomic and polyatomic gases can store thermal energy in forms other than the translational kinetic energy of the molecules.

Degrees of Freedom

Indeed, the polyatomic molecules have more than just three degrees of freedom – they rotate and vibrate.



2 2 2

 

2 2 2



2

1

2

1

z z y

y x

x

I

I

I

z

y

x

M

K































Thus, for instance, the degrees of freedom of an ideal gas, in a thermodynamic description, are

P and V(or a pair of other variables).

The independent coordinates that describe the position of a mechanical system in space - e.g., for a rigid body, it is sufficient to specify the coordinates of its center of mass (x,y,z) and three angles (



_x

,



_y

,



_z).

I_x –the moment of inertia for rotations around the x-axis, etc.

The degrees of freedom of a system are a collection of independent variables required to characterize the system.

Diatomic molecules: 3 + 2 = 5 transl.+rotat. degrees of freedom (QM: degrees of freedom corresponding to rotations that leave the molecule completely unchanged do not count )

Problem 1.16 The “exponential” atmosphere

Consider a horizontal slab of air whose thickness is dz. If this slab is at rest, the pressure holding it up from below must balance both the pressure from above and the weight of the slab. Use this fact to find an expression for the variation of pressure with altitude, in terms of the density of air,



.

Assume that the temperature of atmosphere is independent of height, the average molecular mass m.

z

z+dz

P(z+dz)



A

P(z)



A

Mg

area A

P

(

z



dz

)



A



Mg



P

(

z

)



A

A

Mg

z

P

dz

z

P

(



)



(

)





Adz

M





g

dz

dP







the density of air:

T

k

Pm

T

k

PV

N

V

Nm

V

M

B B



























P

kT

mg

dz

dP

















kT

mgz

P

z

P

(

)

(

0

)

exp

Assuming T is independent of z

Van Der Waals equation of state

Next account for intermolecular attraction which

will reduce pressure as molecules are forced closer

together. This term is proportional to

v

-2

b

v

RT

P





Ideal gas P=RT/v

van der Waals equation of state



v

b



RT

v

a

P

v

a

b

v

RT

P















 







2

2

,

or

Then,

This equation has a critical value of

T

which suggests a

phase change. The next slide shows graphs for several

values of

T

.

Critical Values

2

27

27

8

3

b

a

P

Rb

a

T

b

v

C

C

C







van der Waals equation of state

2

',

',

'

3

1

8

'

'

'

'

3

3

C

C

C

v

v v

P

P P

T

T T

P

v

T

v































This can be expressed in term of dimensionless

coordinates,

P

'

,

v

'

, and

T

'

with the following

Substitutions:

2

'

3

1

'

3

'

8

'

v

v

T

P



Determination of critical values

2 3 2 3 2 3 4

2 3 4

2

2 (

)

0 for

...(1)

(

)

2

6

3 (

)

0 for

...(2)

(

)

T

T

P

RT

a

a V

b

V

V

V

b

V

RT

P

RT

a

a V

b

V

V

V

b

V

RT







  















































2

8

3 ,

,

27

27

c

c

c

a

a

V

b

T

P

Rb

b







3

2

1

(

)

3

2

Divide eq on eq

 

V

V

  

b

V

b