chapter 16 kinetic theory of gases. ideal gas model 2 1. large number of molecules moving in random...

Chapter 16

Kinetic Theory of Gases

Ideal gas model

2

1. Large number of molecules moving in random directions with random speeds.

2. The average separation is much greater than the diameter of each molecule.

3. Obey the laws of classical mechanics; interact only when they collide. (ignore potential energy)

4. All collisions are perfectly elastic.

“Many elastic particles move randomly, no interaction”

Pressure in a gas (1)

3

like an umbrella in rain

Consider a gas in cube

pressure → collisions of the molecules

For one collision: 2 xp mv

.

. x

y

zl

Until next collision: 2

x

lt

vAverage force:

2

xmvp

Ft l

Pressure in a gas (2)

4

Average force:2

xmvpF

t l

Other kinds of collision?

For all molecules in the cube:2

ix

i

mvF

l

2

2( ) ix

ix

vv

N

2 x

mNv

l

2 2 2 2+ x y zv v v v

2 2 2= x y zv v v

22

3

x

vv random velocities

2

3

vmN

l.

. x

y

zl

. .

.

..

.

Pressure in a gas (3)

5

2

3

vmNF

lNet force:

Pressure on the wall:

/N N V

2

3

mvF NP

S Sl

Number density:

Average (translational) kinetic energy: 21

2kE mv

Pressure in a gas:2

3

N kP E

.

. x

y

zl

. .

.

..

.

S

Pressure in explosion

6

Example1: Gunpowder explodes in 10cm3 space. The explosion produces 0.1mol gas and 2×105J energy. Estimate the instantaneous pressure.

Solution: Explosion → extra pressure

2

3 N kP E 2

3

k

NE

V

2

3 totalE

V5

5 3

2 2 10

3 10

J

m

10 51.3 10 1.3 10 Pa atm

Molecular interpretation of temperature

7

Pressure in a gas:2

3

N kP E

Compare with the ideal gas law: PV NkT

3

2 kE kT

The average translational kinetic energy of molecules in an ideal gas is directly proportional to the absolute temperature.

Molecular interpretation of temperature:

Molecule kinetic energy

8

Example2: (a) What is the average kinetic energy of molecules in an ideal gas at 37℃? (b) If H2 and

O2 are both at 37℃, which kind of molecules

moves faster on average?Solution: (a) Average kinetic energy:

3

2kE kT 23 213

1.38 10 310 6.42 102

J

(b) Same T → same average kinetic energy

H Om m H Ov v 2 ?kE

vm

Mean speed and rms speed

9

Example3: 5 particles have the following speeds, given in m/s: 1, 2, 3, 4, 5. Calculate (a) the mean speed and (b) the root-mean-square (rms) speed.

Solution: (a) Mean speed: 3 /v m s

(b) rms speed : 2 2 2 2 2

2 1 2 3 4 5

5

rmsv v 11 3.32 / m s

! rmsv v

vrms for an ideal gas:

for an ideal gas? v

2 3 =k

rms

E kTv

m m

Distribution of molecular speeds

10

How many molecules move faster than vrms?

Distribution

Gauss distribution

Maxwell distribution of speeds:3

22( ) 4

2

mf v N v

kT

21

2

mv

kTe

▲ Ideal gas in equilibrium state at temperature T

f (v)

v

Maxwell distribution of speeds

11

3

22( ) 4

2

mf v N v

kT

21

2

mv

kTe

constants: N, m, k, T

f (v)

v

f (v) dv: number of molecules with speed between v and v+dv

dv

f (v)

v1 v2

Number of molecules with speed v1< v <v2 :2

1

( ) v

vN f v dv

Sum all molecules:0

( )

f v dv N

Shape of the curve

12

3

22( ) 4

2

mf v N v

kT

21

2

mv

kTe

Most probable speed vp

f (v)

v

( )0

df v

dv

vp

2 p

kTv

m

Distribution for different temperature:

f (v)

v300K

600K

Chemical reaction

& temperature

Average speed

13

Solution: How to calculate average values?

Example4: Determine the average speed of molecules in an ideal gas at temperature T.

v

( )v f v dv

f (v) dv: dN with speed between v and v+dv

Sum of speeds of dN molecules:

0( )

v f v dv Sum of speeds of all molecules:

Average speed:0

( )

v f v dv

vN

8

kT

m

Three statistical speeds

14

Most probable speed vp :

2p

kTv

m1.41

kT

m

8

kTv

m1.60

kT

m

Average speed :v

Root-mean-square speed vrms :

3rms

kTv

m1.73

kT

m

f (v)

vvp

v

vrms

p rmsv v v

Using f (v)

15

Solution: (a) Most probable speed:

Example5: 1mol H2 gas at 300K. (a) What is vp? (b)

How many molecules have speed vp < v < vp +20m/s?

(b) Speed between vp and vp +dv : dN=f (vp) dv

2p

kTv

m

2

mol

RT

m31.58 10 / m s

231

22 24

2

mv

kTmN N v e v

kT

216.33 10 ( 1%)

2vp < v < 2vp +20m/s: / 0.2% N N

3vp < v < 3vp +20m/s: / 0.003% N N

Homework

16

If the distribution of speeds in a N-particles system

is (instead of Maxwell distribution)

00

0

sin( ), 0 constant

0.

vC v v

vf v

v v

(a) C=? (b) number of particles with f(v) > C /2.

v00 v

f(v)

Real gases

17

High T & low P → ideal gas law PV=nRT

Real gas behavior? Shown in a PV diagram:

Higher pressure

molecules be closer

Ep (attractive force) can’t be ignored

molecules get even closer

Changes of phase

18

Curve D → liquefaction

Curve C → critical point c

Critical temperature

vapor & gas

PT diagram: phase diagram

boiling / freezing / sublimation

Using phase diagram

19

Solution: (a) P > 1atm → liquid; P < 1atm → gas

Example6: Describe the phase of water in different

pressure at (a) 100℃; (b) 0℃.

100℃P = 1atm → boiling point

(b) P > 1atm → liquid;

P < P0 → gas;

P0 < P < 1atm → solid

0℃

P0 .P = 1atm → freezing point

P = P0 → sublimation point

Mean free path

20

Collisions between molecules

Mean free path: average distance

traveling between collisions

P396: Molecules →

hard spheres of radius r

2

1

4 2M

N

lr

Collision frequency

21

Solution: (a) Number density at STP:

Example7: Estimate (a) the mean free path of O2

molecules at STP and (b) the average collision frequency. (r≈1.5×10-10m)

2325 3

3

6.02 102.7 10

22.4 10

N

Nm

V

Mean free path:2

1

4 2M

N

lr

89 10 600 m r

(b) Average speed 8 / 420 / v kT m m s

average collision frequency 9 1/ 4.7 10 Mf v l s

*Diffusion

22

Diffusion: substance moves from high concentrated region to

low concentrated region

Random motion of molecules:

1 2N N more molecules moves from 1 to 2 than from 2 to 1

concentrations become equal everywhere

Diffusion equation (Fick’s law): dC

J Ddx

1 2 3

*Distribution of energy

23

All molecules of atmosphere diffuse to the space?

Boltzmann’s distribution of potential energy:

0 exp( ) pN N

E

kT

number density at position Ep =0 0 :N

pE mghIn an gravity field:

NP kT 0 exp( ) mgh

P PkTconstant T

*Height of aircraft

24

Solution: By using the equation of pressure:

Example8: Air pressure gauge on an aircraft reads 0.3atm, 0 outside. What is the height? ℃

0 exp( ) mgh

P PkT

0 0ln ln mol

P PkT RTh

mg P m g P

3

8.31 273 10ln

29 10 9.8 3

9600 m