Kinetic theory of gases

SMART 1 PRESSURE OF AN IDEAL GAS

How much pressure an ideal gas

INTRO
exerts on the walls of the cyllinder ?

2
MAXWELLIAN DISTRIBUTION
With the help of maxwell's distribution curve
we find the different kind of velocities.

KINETIC
THEORY OF
GASES
Collective study of the
3 HEAT CAPACITIES AND DOF
The degrees of freedom of any molecule
and respective Heat capacities.
system of particles

4 VAN-DER WAAL'S EQUATION

How a real gas is different from an
ideal gas ?

5
MEAN FREE PATH
All about the collision of gas
molecules.

☞ Assumptions of kinetic theory of Ideal gases :-

(1). A gas consists of very large number of molecules. These molecules are identical,
perfectly elastic and hard spheres.
(2). They are so small that the volume of the molecule is negligible.
(3). Their motion is completely random.
(4). Time of collision between two molecules is very small.
(5). Collision between two molecules is perfectly elastic.
(6). The effect of gravity is completely negligible.
☞ Expression for the pressure of an ideal gas :-

Let’s suppose we have a cubical container of length l filled with an ideal gas. In which the
molecules are moving with velocities vx , vy and vz in x, y and z directions respectively.
y

vy

vz vx x

F average (total) Δp
P= = , Δp is the change in momentum in one
l 2
l2 second.

Change in momentum of one particle in one collision = 2mvx

2l
Time between two collisions with wall =
vx
2
mvx
Change in momentum of one particle in one second = =F average
l
2 2 2 2
m(v x +v x + v x +.....+v x )
total change in momentum = F average (Total)= 1 2 3 N

The gas is in equilibrium, the molecules move entirely randomly and all directions of motion
are equally probable. Hence,

Σv 2x = Σv 2y= Σv2z
speed of one particle, v 21 =v 2x + v2y +v 2z

or, 2
Σv =3 Σv x
2

Now the total average force can be written as,

2
m Σv
F average (total)=
l 3
m Σv m N Σv2
2
1
and pressure,
2
P= 3 = P= ρ vrms
l 3 3V N 3
☞ Some Important laws :

(a). Boyle’s law : At constant temperature, pressure is inversely proportional to volume.

1
P∝ or P 1 V 1=P 2 V 2
V

(b). Charles’s law : At constant pressure, volume is directly proportional to temperature.

V1 V 2
V∝T or =
T1 T 2

(c). Charles’s law for pressure : At constant volume, pressure is directly proportional to
temperature.
P1 P2
P ∝ T or =
T1 T 2

Combining these laws together, we get PV ∝ T P V =n R T

This is Ideal gas equation of state.

☞ Maxwell’s distribution curve :- The number of molecules in an ideal gas having speeds
in the range v to v+dv as,

3 −mv2
m 2k T
dN =4 π N ( )2 v2 e B
dv
2 π kB T

3 −mv 2
dN m
or, ----- (1)
2k BT
f (v )= =4 π N ( )2 v 2 e
dv 2π kB T

We are not deriving this expression but we need to understand it very carefully. Therefore
dN
we plot a graph between and v ,
dv
f(v)= dN/dv This graph basically gives us the data that how
many particles are having a particular velocity.

v v + dv vmp = This is the speed which have the most

particles.
We can find the expressions of root mean square speed, average speed and most probable
speed from the equation (1) .

Root mean square speed :- To find the rms speed first we’ll calculate the mean square
speed,
∞

∫ v 2 f (v )dv
0
v ms = ∞

∫ f ( v) dv
0
2 2
∞ −mv ∞ −mv
m
)∫ v 4 e ∫v
2k T 4 2 kB T
4π N( dv B
e dv
2π kB T 0 0
v ms = ∞ −mv
2 v ms = ∞ −mv
2

m
)∫ v e ∫ v2 e
2 2k T 2k B T
4π N( dv B
dv
2 π kB T 0 0

m dx dx
Do some substitutions, =B and Bv =x
2
dv= =
2k BT 2B v 1 1
( ) ( )
2B 2 x 2
∞
1
5/ 2 ∫ x
3/2 − x
e dx 3/2
B 0 2 B Γ(5 /2) 3 3 k B T
v ms = = = =
2 B Γ(3 /2) 2 B m
∞ 5/2
1
3/ 2 ∫ x
1/2 − x
e dx
B 0

Now rms speed can be written as,

v rms = √ v ms v rms =
√ 3 kBT
m

Average speed :- The average speed can be written as,

∞

∫ v f ( v) dv 1
∞

= ∫ v f (v )dv
0
v averge = ∞
N 0
∫ f ( v) dv
0

Solving it again with the help of gamma function we get,

v rms =
√ 8k B T
πm

Most probable speed :- It is the speed whose particles are the most, that means to find the
most probable speed we need to find the maxima of the curve given in figure-2 .
d f (v )
Therefore we should first calculate the and equate it to zero,
dv
2
3 −mv
d f (v ) m 2 d 2 2k T
=4 π N ( ) (v e )=0 B

dv 2 π k B T dv
 2
−mv −mv 2
d 2 2k T −2 m v 2 2k T 2

(v e )=0
B
[( v e )+(2 v e−mv )]=0
B

dv 2kBT

Solving this and putting v =v mp , we get

1−(
m 2
v )=0
2 k B T mp
v mp =
√ 2 kB T
m

We need to keep one thing in mind, v rms >v average >v mp .

Effect of temperature on the distribution curve : On increasing the temperature, the

sharpness of the curve decreases keeping the area constant.

f(v)= dN/dv
T3>T2 >T1
T1
T2
T3

v
Equation (1) is written in 3 dimensions, similarly this type of equation can be written in 2D
and 1D.
−mv2
dN m
2D → ------ (2)
2k BT
f (v )= =2 π N ( )v e
dv 2π k BT
1 −mv 2
dN m
1D → ------ (3)
2k BT
f (v )= =N( )2 e
dv 2π k BT

☞ Energy distribution of an ideal gas :- In this section we’ll see how much energy an ideal
gas molecule carries. First of all we’ll write the distribution function in terms of energy,
Substitute from E= m v
1
2
2
v=
2E
m
1
√
dE=2 m v dv or dv= dE
√2 m E

Here we’ll assume that the number of molecules having the energy in the range E to E+dE
is equal to the number of molecules having speed in the range v to v+dv. Therefore,

dN E =dN v

3 −E
m 2E k dE
now, dN E =4 π N ( )2 ( )e BT

2π kBT m √2 m E
 1/2 3 −E
dN E E 1 2 k
or we can write, f (E)3 D = =2 N ( π ) ( ) e BT

dE kBT

Now the average energy can be written as,

∞

∫ E f (E) dE
0
⟨ E ⟩3 D = ∞

∫ f (E) dE
0

∞ −E

∫E 3/2
e k BT
dE
0
⟨ E ⟩3 D = ∞ −E

∫E 1/2
e
k BT
dE
0
E
Solve this by substituting x= and solving the gamma function, we get
kB T

3
⟨ E ⟩3 D = k B T
2

Similarly in one dimension we can do this calculation by writing the equation (3) in the
terms of energy,
1 −E
m k T dE
dN E = N ( )2 e B

2π kBT √2 m E

1 −E
dN E N 1
or,
−1/2 k BT
f (E)1 D = = ( )2 E e
d N 2 π kBT

∞ ∞ −E

∫ E f ( E)dE ∫ E1/2 e k T dE
B

0
⟨ E ⟩1 D = ∞ ⟨ E ⟩1 D = ∞0 −E

∫ f ( E)dE ∫ E−1/ 2 e k T dE
B

0 0

like above calculation we can use gamma function to solve this integral, we get

1
⟨ E ⟩1 D = k B T
2

one thing we may conclude from here that energy of a molecule of an ideal gas in one
particular direction is k B T /2 . It means that every dimension has this much of energy. This
is called Equipartition of Energy .
☞ Degrees of freedom :- It is defined as number of modes by which a molecule can keep
energy. Or in other words we can say that it is the number of ways by which a molecule can
move/rotate .
Let us consider a spherical ball. It can move in three dimensions but any kind of rotation will
not change it’s shape or we can say that we can’t recognize any kind of rotation due to
symmetry. So it’s degrees of freedom are three .

vy

vx
vz

Therefore for different kind of molecules we can write the degrees of freedom as,

Mono-atomic → 3 ( translational )
Di-atomic → 5 ( 3 translational + 2 rotational )
Tri-atomic (linear)→ 5 ( 3 translational + 2 rotational )
Tri-atomic (Non-linear)→ 6 ( 3 translational + 3 rotational )

In case of very high temperature ( more than 5000K ), two vibrational degrees of freedom
increase.
If a molecule have f degrees of freedom then it’s energy can be written as,

f
⟨ E ⟩= k B T → average energy of one molecules
2
f
⟨ E ⟩= R T → average energy of one mole
2
for n moles,
f R
⟨ E ⟩= n RT Remember (k B = )
2 N

Heat capacities of gases :- If there’s no external field present, then an ideal gas possesses
only kinetic energy and there’s no potential energy.
Therefore total energy is given by,

f
U= RT
2
Thus we can define the molar heat capacity as the amount of heat required to raise the
temperature of one mole ideal gas by one kelvin. At constant volume it is given as,

∂U f
C V =( ) = R
∂T V 2

Here I’m introducing a relation between molar specific heat capacities at constant volume
and constant pressure, (we’ll prove it later)
 f
C P −C V =R C P =R+C V C P =( +1) R
2

Ratio of Specific heat capacities : Ratio of molar specific heat capacity at constant
pressure to molar specific heat capacity at constant volume is constant.

CP 2
γ= =1+
CV f

2 2
For monoatomic gas , γ =1+ =1+ =1.67
f 3
2 2
For diatomic gas, γ =1+ =1+ =1.40
f 5

2 2
For triatomic gas, γ =1+ =1+ =1.34
f 6

Calculation for mixture of gases :- Let us suppose that n1 moles of a gas ( degrees of
freedom are f1 ) is mixed with n2 moles ( degrees of freedom are f2 ) of other gas. We need
to find the degrees of freedom and specifics heats of the new gas.

Change in total energy can be written as,

f1 f f n f n
ΔU= n1 R Δ T + 2 n 2 R Δ T =( 1 1 + 2 2 ) R Δ T ---- (a)
2 2 2 2

f mix
for the mixture, ΔU= (n1 +n2 ) R ΔT ---- (b)
2

Comparing both equations (a) and (b), we get

f 1 n1 +f 2 n2
f mix =
n1 +n2

2
from here we can write, f mix (n 1 +n2 )=f 1 n1 + f 2 n2 , put f=
γ−1

2 2 2
(n +n )= (n )+ (n ) (C V )mix R(n1 +n2 )=(C V )1 R n 1 +(C V )2 R n2
γ mix −1 1 2 γ 1 −1 1 γ 2−1 2

(C V )1 n1 +(C V )2 n2
Therefore, (C V )mix =
(n1 +n2 )
 (C P )1 n1 +(C P )2 n2
Similarly, (C P )mix =
(n1 +n 2)

Thus ratio of specific heats for the mixture can be written as,

(C P )mix (C P )1 n1 +(C P )2 n2
γ mix = =
(C V )mix (C V )1 n1 +(C V )2 n 2

Real gas equation :- We make some corrections in ideal gas equation depending upon the
practical cases. Now we’ll consider that the molecules of the gas are of the finite size and
there’s a intermolecular attraction between them. Therefore the real gas equation becomes
like this,
a
(P+ 2 )(V −b)= R T
V

here a is pressure correction term and b is volume correction term. And this equation is
called as Van der Waals equation of state .
For n moles this equation can be written as,

2
n a
(P+ 2
)(V −n b)=n RT
V

Mean free path :- The mean free path is the average distance travelled by a molecule
between two successive collisions. If λ1 , λ2 , ..... , λN denote the successive free paths
traversed in time t and N is total number of collisions suffered in this period, therefore

Average velocity λ 1 + λ 2 + λ 3 +....+ λ N

λ= =
Average time N

To calculate the mean free path we can assume two situations here ;

☞ Zeroth order approximation : In this approximation we assume that only one molecule
is in motion and all the other molecules are at rest.
Let’s suppose that the diameter of the molecule is d. Since only one molecule is moving
, therefore it’ll collide with the molecules whose centres lie in the cylinder of radius d . The
centre of this imaginary cylinder and the moving molecule coincide.

d
d v
Length covered by the molecule in time Δt, L=v Δt

Therefore volume covered in the same time 2

V =πd v Δt

N
Let us suppose that the n is the number of molecules per unit volume (n= ) . Therefore
V
number of collisions N in time Δt

N=n πd 2 v Δt

N
Number of collisions per unit time,
2
=n πd v
Δt
1
Hence average time between two collisions T= 2
n πd v

And average distance travelled in this time or mean free path,

v 1
λ= =
T n πd 2

This is approximated result in which only one molecule is moving. Now assuming that all
the molecules are moving ( First order approximation ) we get,

λ=
1
=
V ( I’m not deriving this result, that’s
√ 2 n πd √ 2 N πd 2
2
not necessary )
 Exercise - 1

Q.1 : Calculate the most probable speed, average speed and the root mean square speed
for oxygen molecules at 27oC . m(O 2)=5.31×10−26 kg and k B =1.38×10−23 JK −1 .

Q.2 : Find the ratio of the Root mean square speed, average speed and most probable
speed.

Q.3 : Starting from the maxwell velocity distribution function in two dimensions,
2
−mv
dN m 2k
BT
f (v )= =2 π N ( )v e
dv 2π k BT

Prove that average energy of a molecule in two dimensions is k B T .

Q.4 : Plot the P-V , P-T and V-T diagrams for each constant temperature, constant volume
and constant pressure.

Q.5 : Consider a cyclic process in the P-T diagram as shown in figure below. Draw the P-V
and V-T diagrams.
P
B
C

Q.6 : Two closed vessels of equal volume contain air at 105 kPa, 300 K are connected
through a narrow tube. If one of the vessels is now maintained at 300 K and other is at 400
K , what will be the pressure in the vessels.

Q.7 : Consider a cyclic process in V-T diagram. Draw the P-V and P-t diagrams for it.

V
D C

B
A

T
 Exercise - 2

Q.1 : The molar specific heat of a gas as given from the kinetic theory is 5R/2 . If it is not
specified whether it is CP or CV , one could conclude that the molecules of the gas
( IIT JAM 2005 )
(a) are definitely monatomic (b) are definitely rigid diatomic
(c) are definitely non-rigid diatomic (d) can be monatomic or rigid diatomic

Q.2 : A box containing 2 moles of a diatomic ideal gas at temperature T o is connected to

another identical box containing 2 moles of a monatomic ideal gas at temperature 5T o .
There are no thermal losses and the heat capacity of the boxes is negligible. Find the final
temperature of the mixture of gases (ignore the vibrational degrees of freedom for the
diatomic molecules). ( IIT JAM 2009 )

(a). To (b). 1.5To (c). 2.5To (d). 3To

Q.3 : A gas of molecules each having mass m is in thermal equilibrium at a temperature T .

Let vx , vy and vz be the Cartesian components of velocity, v of a molecules. The mean value
of (v x −α v y + β v z )2 is
( IIT JAM 2010 )
kB T kB T
(a). (b).
2 2 2 2
(1+α + β ) (1−α + β )
m m

kBT kBT
(c). (d).
2 2 2 2
( β −α ) (β +α )
m m

Q.4 : A gas of molecular mass m is at temperature T . If the gas obeys Maxwell-Boltzmann

velocity distribution, the average speed of molecules is given by
( IIT JAM 2011 )

(a).
√ kB T
m
(b).
√ 2kB T
m
(c).
√ 2kB T
πm
(d).
√ 8 kB T
πm

Q.5 : A rigid triangular molecule consists of three non-collinear atoms joined by rigid rods.
The constant pressure molar specific heat ( CP ) of an ideal gas consisting of such molecules
is
( IIT JAM 2015 )
(a). 6R (b). 5R (c). 4R (d). 3R

Q.6 : A spherical closed container with smooth inner wall contains a mono-atomic ideal gas.
If the collisions between the wall and the atoms are elastic, then the Maxwell speed
dnv
distribution function ( ) for the atoms is best represented by : ( IIT JAM 2016 )
dv
 dnv dnv
(a). ( ) (b). ( )
dv dv

0
0 v
v

dnv dnv
(c). ( ) (d). ( )
dv dv

0
v 0 v

Q.7 : One mole of an ideal gas with average molecular speed vo is kept in a container of
fixed volume. If the temperature of the gas is increased such that the average speed gets
doubled, then ( IIT JAM 2016 )
(a) the mean free path of the gas molecule will increase
(b) the mean free path of the gas molecule will not change
(c) the mean free path of the gas molecule will decrease
(d) the collision frequency of the gas molecule with wall of the container remains
unchanged

Q.8 : Two boxes A and B contain an equal number of molecules of the same gas. If the
volumes are VA and VB and λA and λB denote respective mean free paths, then
( IIT JAM 2018 )
λ A λB λA λ
(a). λ A =λ B (b). = (c). = B1/2 (d). λ A V A= λ B V B
V A VB VA
1/2
VB

Q.9 : An ideal gas consists of three dimensional polyatomic molecules. The temperature is
such that only one vibrational mode is excited. If R denotes the gas constant, then the
specific heat at constant volume of one mole of the gas at this temperature is
( IIT JAM 2018 )
(a). 3R (b). 7R/2 (c). 4R (d). 9R/2

Q.10 : Two gases having molecular diameters D1 and D2 and mean free paths λ1 and λ2 ,
λ
respectively, are trapped separately in identical containers. If D2 = 2D1 , then 1 =?
λ2
( IIT JAM 2019 )

Q.11 : The root mean square ( rms ) speeds of Hydrogen atoms at 500K ,VH and Helium
atoms at 2000K , VHe are related as
( IIT JAM 2020 )
(a). v H >v He (b). v H <v He (c). v H =v He (d). can ' t predict

Q.12 : One mole of an ideal gas having specific heat ratio (γ) of 1.6 is mixed with one mole
of another ideal gas having specific heat ratio of 1.4 . If CV and CP are the molar specific
heat capacities of the gas mixture at constant volume and pressure, respectively, which of
the following is/are correct? ( R denotes thee universal gas constant).
( IIT JAM 2020 )

(a). C V =2.08 R (b). C P =2.9 R (c). C P =1.48 C V (d). C P =1.52 C V

Answers ( Exercise-2 ) : 1- (d) , 2- (c) , 3- (a) , 4- (d) , 5- (c) , 6- (c) , 7- (b) , 8- (b) ,
9- (c) , 10- 4 , 11- (c) , 12- (a), (b), (c)