Kinetic Calculation of Pressure
A1
A2
L
L
L
Assume N molecules, each mass = m.
L
A2
A1
Vx
+ ve
Consider one molecule moving in x-direction:
Initial momentum ( before collide on A1 ) = mVx
Final momentum ( after collide on A1 ) = - mVx ( elastic collision )
 change in momentum = - mVx - mVx = - 2 mVx
Suppose the molecule reach A2 without striking any other particle on the way.
2L
 time for successive collision on A1 =
( period )
Vx
V
 number of collision on A1 per unit time = x (  frequency )
2L
V
change of momentum of the molecule per unit time = ( - 2mVx)( x ) = 2L
2
mV X
(
)
L
= force acting on the molecule by A1.
By Newton's 3rd Law:
2
mV X
 force on A1 by the molecule = + (
)
L
2
 pressure on A1 due to this molecule = p =
F
=(
A
m(
For N- molecules, total pressure on A1:
P = p1 + p2 + p3 + p4 +
=
mVX 1
2
L3
+
mVX 2
L3
VX
)
2
L ) = mV X
L2
L3
+ pN
2
+
mVX 3
2
+
L3
+
mVX N
2
L3
Where Vxi = x-component of velocity of ith molecule.
P=
m
( Vx12 + Vx22+ Vx32 +
L3
+ VxN2 )
N
= number of molecule per unit volume = particle density
L3
N
 L3 =
n
2
2
2
2
m(V X 1  V X 2  V X 3                V XN )
P=
N
n
2
2
2
2
(V X 1  V X 2  V X 3                V XN )
=(nm)
=  V X2 (  = density of gas
N
If n =
average value of (VX)2 = V X2
inside container )
 V2 = V 2X VY2  VZ2  V 2  V X2  VY2  VZ2
V2
For random motion : V  V  V 
3
2
X
Hence, P =
1
 V2
3
or
2
Y
P=
2
Z
1 Nm 2
V
3 
or
Pv =
1
NmV 2
3
Further:
nM
v
& Pv = nRT
 =
(1)
M = molar mass of gas
n = no. of moles of gas
(2)
( 2)
P nRT RT
 

(1)
 nM
M
 V2 
3P


3RT
M
root-mean-square velocity ( V RMS )
e.g.
Find the Vrms of oxygen gas at 27 0 C. ( given that the molecular mass of oxygen gas
= 32 )
Ans: ~ 480 ms -1
Kinetic interpretation of temperature
 Pv =
1
1
NmV 2  nM V 2
3
3
 Pv =
2 1

nM V 2 

3 2

total K.E. of the molecules. ( called the thermal energy / internal energy )
But, Pv = nRT
2 1


nM V 2  = nRT

3 2

3
1

  M V 2  = RT
2
2

( = K.E. per mole )
Divide both side by N ( avogadro’s number )
1 M 2  3 R
 
V   ( )T
2 N
 2 N
Boltzman constant, k = 1.38x10-23 J / molecule.K

1
3
mV 2  kT
2
2
Hence, average K.E. per molecule  T ( K )
 different ideal gases have the same average K.E. per molecule at equal
temperature.