Page 17 - Hsslive-Plus-One-Physics-Class-Notes-Part-III (1)

Heat, Thermodynamics & Kinetic Theory {16} Vinodkumar M, St. Aloysius H.S.S, Elthuruth, Thrissur.
3 R R
KE of one molecule of gas = T ; N = Avagadro number. Here = k (Boltzmann’s constant)
2 N N

3
Average KE of one molecule of gas = k T..
2
 KE  T..

Average KE of gas molecules is directly proportional to absolute temperature.
Mean, rms and most probable speeds.

Mean speed( ) : It is the arithmetic mean of the speeds of gas molecules at a given temperature.
v
k T 8 R T
v   ; M = mass of one mole of gas.
 M  M

3 k T 3 R T
rms speed (c or ) c =  ( M is the mass of one mole of gas, m = mass of one
c
rms m M
molecule of gas.)
Most probable speed (v ): The most probable speed is the speed at which the maximum number of molecules
P
moves in a given gas at a given temperature.

Mean free path.
Mean free path of a molecule in a gas is the average distance treavelled by the molecule between
two successive collissions.
Molecules of a gas have finite size and behave like rigid spheres. The molecular motion is random and
hence they collide against one another frequently. Between two successive collissions, a molecule move
alo0ng a straight path with uniform velocity. This path is known as free path.
Let  ,  ,  ..........  are the free paths travelled by the molecules in n successive collissions, then
1 2 3 n
total distan ce travelled      .............  
mean free path  i.e.   1 2 3 n
total no of collissions. n
Derivation of expression for mean free path.
Assume that only one molecule is in motion and all other molecules
are at rest. Let d be the diameter of each molecule. The moving molecule d
will collide against all those molecules whose centres lie within a distance
d from the centre of the moving molecule. Suppose is the distance

travelled by the moving molecule. The moving molecule will make a d
collission with all those molecules whose centres lie inside
a volume d 2  
Suppose n is the no. of molecules per unit volume in the gas, then no. of collissions = no. of molecules in

dis tan ce travelled  1
2 2    
the volume d  = n d  . Now no. of collissions n d 2  n d 2 .
In the above derivation, we have assumed that all the molecules are stationary. But this is not correct. So the
chances of collission by a molecule ios greater. Taking this into account, the mean free path can be shown to
1 m
be 2 times less than that shown above. i e. .   2 i e. .   2
2 n d 2  d n m

total no of molecules. m
Here n m   mass of one molecule    , density of gas
volume V