Transcript New_Equations_for_ideal_gases

Equations for ideal gases
pV  nRT
pV  NkT
The ideal gas equation and
its variant for N molecules
1
2
pV  Nmcrms
3
Relates pressure to the density of a
gas and root mean square velocity.
You may have to manipulate this one
3
1 2
3RT
kT  mcrms 
2
2
2N A
Relates absolute temperature to
average molecular energy
Particle in a Box
Deriving
1
2
p  Nmcrms
3
The particle in a box
•
The idea is to derive an expression for to
relate the pressure and volume of a gas
to the movement of the many individual
molecules which make it up
•
We are building a MATHEMATICAL
MODEL of what is happening and to do
this we will need to make several
assumptions
Assumptions
1.
2.
3.
4.
5.
6.
Intermolecular forces are negligible except during collisions.
All collisions are perfectly elastic
The volume of the molecules themselves can be neglected
compared with the volume occupied by the gas.
The time for any collision is negligible compared with the time
spent between collisions.
Between collisions molecules move with uniform velocity.
There is a sufficiently large number of molecules for statistics to
be meaningfully applied.
These conditions imply that all of the internal energy of the gas is
kinetic.
• We also start with the knowledge that :
1.Pressure = Force / Area
2.Force = rate of change of momentum
We start with one molecule and take the simplest case, that I
is moving only in the x direction heading for face A
A
u
m
y
z
x
The molecule has mass m and velocity u so its momentum is mu
The molecule collides with A and bounces back in the
opposite direction. Its momentum velocity has changed from
u to -u
A
u
m
y
z
x
So its momentum has chaged from mu to -mu
This means that its change in momentum is
2mu
A
u
m
y
z
x
So its momentum has chaged from mu to -mu
its rate of change of momentum is given by:
2mu
t
this is equal to the force acting on the particle
The time (∆t) between
successive collisions
is the time it takes to
travel from wall A to
the far wall of the box
and back.
y
A
m
u
The distance it has to
travel is 2x.
x
It does this travelling
at velocity u
remembering that velocity is distance travelled / time
taken
2x
u
t
So the time taken between collisions (∆t) is
2x
t
u
so the rate of change of momentum of the molecule
2mu
2x
( )
u
Is
ie
mu
x
2
Now this is equal to the force of the wall which acts on the particle to change its
momentum and by Newton’s third law this must be equal to the
force of the particle acting on the wall.
So the force on the wall from this one particle is
mu
F
x
2
As pressure is force / area
and the area of wall A is y x z
The pressure on wall A due to
this one molecule is
y
A= y x z
z
mu 2
(
)
F
p  x
A
yz
mu 2 mu 2
p

xyz
V
mu 2
P
V
This is an expression for the pressure due to 1 molecule of the ideal
gas moving in the x direction.
BUT there are N molecules in the gas and they have individual
velocities: u12,+ u22+ u32,……uN2
m
2
P  Nurms
V
m
2
P  Nurms
V
But there are three components of velocity
u in the x direction , say v in the y direction and w in the z direction
v
w
u
Then taking c2rms = u2rms+ v2rms+w2rms
So u2rms = 1/3c2rms
1m 2
p
Ncrms
3V
or
1
2
pV  Nmcrms
3