jas gas kinetics rev1 0

Drs. Jaslin Ikhsan, M.App.Sc., Ph.D.
Chemistry Ed. Department
State University of Yogyakarta

Kinetic Theory of Gases I:
 Gas Pressure
 Translational Kinetic Energy
 Root Mean Square Speed

GASES
 Gases are one of the most pervasive aspects of our
environment on the Earth. We continually exist with
constant exposure to gases of all forms.
 The steam formed in the air during a hot shower is
a gas.
 The Helium used to fill a birthday balloon is a gas.
 The oxygen in the air is an essential gas for life.

GASES

A windy day or a still day is a result of the difference in

pressure of gases in two different locations. A fresh breeze on
a mountain peak is a study in basic gas laws.

Mixture of gases

The Kinetic Molecular Model for Gases








Gas consists of large number of small
individual particles with negligible size
Particles in constant random motion and
collisions
No forces exerted among each other
Kinetic energy directly proportional to

temperature in Kelvin
3
KE 

2

 R T

The Ideal Gas Law

PV  nRT

in K

n: the number of moles in the ideal gas
N
n
NA

total number

of molecules

Avogadro’s number: the number of
atoms, molecules, etc, in a mole of
a substance: NA=6.02 x 1023/mol.

R: the Gas Constant: R = 8.31 J/mol · K

Pressure and Temperature
Pressure: Results from collisions of molecules
on the surface
Pressure:

F
P
A

Force:

dp

F
dt

Force
Area

Rate of momentum
given to the surface

Momentum: momentum given by each collision

times the number of collisions in time dt

Only molecules moving toward the
surface hit the surface. Assuming the
surface is normal to the x axis, half the
molecules of speed vx move toward
the surface.

Only those close enough to the

surface hit it in time dt, those within
the distance vxdt

The number of collisions hitting an area A in
time dt is
1 N 
 A vx  dt
2 V 
Average density

The momentum given
by each collision to
the surface 2mvx

Momentum in time dt:
1 N 
 Av x dt
dp  2mv x  
2 V 

Force:

1 N 
dp
 A v x
F
 2mvx  


2 V
dt

Pressure:

F N 2
P   mv x
A V

Not all molecules have the same v x  average v2x
N 2

P  mv x
V



2
vx

1 2 1 2 2 2
= v  v x  v y  vz
3
3

2
vx

1 2 1 2
= v  vrms
3
3



vrms is the root-mean-square speed
2

vrms  v 

2
vx

2
+ vy

2
+ vz

3

1 N 2 2 N 1 2 

Pressure: P 
mv 
mv




3V
3 V 2

Average Translational Kinetic Energy:
1 2 1 2
K  mv  mvrms
2
2

Pressure:

2 N
P   K

3 V

2
From PV   N  K and PV  nRT
3

Temperature:

3 nRT 3
K 
  k BT
2 N
2

R
23
Boltzmann constant: k B 
 1.38  10
J/K
NA

1
2
From PV   N  mvrms
3
N
and PV  nRT  N RT
A

vrms

3RT

M

Avogadro’s number

N  nN A

Molar mass

M  mN A

Pressure  Density x Kinetic Energy

Temperature  Kinetic Energy

(a) Compute the root-mean-square speed of a nitrogen
molecule at 20.0 ˚C. At what temperatures will the root-meansquare speed be (b) half that value and (c) twice that value?

vrms

(a)

vrms

3RT

M

3RT
38.31 J/mol  K  293 K


 511 m/s
-3
M
28.0 10 kg/mol
2
vrms

2
vrms

(b) Since vrms  T
for 0.5 vrms

for 2 vrms

T 

T
฀

2

T  0.5 T  73.3K = -200 C
฀฀
2

3

฀

T  2 T 1.17 10 K = 899 C
฀฀

Please estimate the root mean square
mean velocity of Hydrogen gas.
A. 2000

B. 1000

E. 250

D. 100

C. 500
m/s

What is the average translational kinetic energy of nitrogen
molecules at 1600K, (a) in joules and (b) in electron-volts?

(a)





3
3
K  k BT  1.38 10 23 J/K 1600K 
2
2
 3.3110 20 J

(b) 1 eV = 1.60 x 10-19 J

3.31 10 20 J
K
 0.21 eV
19
1.60 10 J/eV

3
K   kB T
2