Kinetic Molecular Theory of Gases

KINETIC MOLECULAR THEORY OF GASES

The gas laws give us the general behavior of gases. These laws are based on experimental observations and they are almost independent of the nature of gas.

When we want to discuss the behavior of gases quantitatively, then we should appreciate the kinetic theory of gases given by Bernoulli in 1738. He was the first person who proposed that the pressure of the gas is due to the collisions of the gas molecules at the walls of the vessel. These proposals of Bernoulli led Clausius (1857) to derive the kinetic equation. From his kinetic equation of gases, he was able to derive the gas laws. This theory was elaborated and extended by Maxwell, Boltzmann and Van der Waals

Postulates of Kinetic Molecular Theory

  1. The gases are consisted of tiny particles called molecules. It means that gases like He, NC, Ar etc., are said to be monoatomic gases.
  2. The molecules of a gas move randomly, collide among themselves, collide with the walls of the vessel and change their directions.
  3. The pressure on the walls of the vessel is due to collisions on the walls.
  4. The collisions of the molecules among themselves and on the walls of the vessel are perfectly elastic.
  5. The molecules of the gases are widely separated from each other at ordinary temperature and pressure.
  6. There are no forces of attractions among the molecules of the gases.
  7. The actual volume of the gas molecules is negligible as compared to the volume of the vessel.
  8. The force of gravity has almost no influence on the molecules of the gas.
  9. The kinetic energy of the gas molecules is proportional to the absolute temperature of gas.

Kinetic Equation of Gases

The kinetic equation of gases is derived by keeping in view the above postulates.It is

PV = mn2

Wherec-2 is the mean square velocity of gas molecules ‘m’ is mass of one molecule and number of molecules in the vessel.

Molecular Velocities

It a known fact that the molecules of a gas move haphazardly, collide among themselves and with the walls of the vessel. So the molecules have certain velocities. All themolecules of a gas do not have the same velocities at a given temperature. For this reasons various types of velocities have been designed which are as follows

(a) Root mean square velocity

(b) Average velocity

(c) Most probable velocity.

(a) Root Mean Square Velocity

The mathematical expression for root mean square velocity can be derived from kinetic equation of gases, i.e.,

PV =mn2

Suppose, that gas is one mole, so

n = NA

PV =m NA2 (1)

Equation (1) is the kinetic equation of gases for one mole of the gas General gas equation for one mole is

PV = RT  (2)

Comparing equation (1) and (2)

m NA2= RT  mNA = M

M2= RT

or 2 = (3)

According to equation (3) the mean square velocity of gas molecules depends upon the temperature and the molar mass. If we take the square root of the equation (3), then we get theroot mean square velocity.

=

cr.m.s=== 1.732

cr.m.s= 1.732 (4)

According to equation (4), the root mean square velocity cr.m.sof the gas molecules is directly proportional to the square root of absolute temperature and inversely proportional to thesquare root of molar mass.

It means heavier molecules with greater molar mass move slowly. Moreover, we come to know that the elevated temperature makes the molecules to move faster

(b) Average Velocity

This is also called mean velocity and It is “the mean of the all the possible velocities” denoted by

=

The expression for c-2 can be obtained from Maxwell’s law of distribution of velocities

=== 1.59

= 1.59  (5)

Average velocity also depends upon the temperature and the molar mass of the gas just like root mean square velocity given by equation (4).

(c) Most Probable Velocity

It is defined as, “the velocity possessed by greatest fraction of the molecules in a gas at temperature.”

Actually, all the molecules of the gas have not equal velocities. Molecules are distributed in groups of velocities. These groups may be called as fractions. All these fractions are not equal. That fraction which is biggest possesses a certain velocity and is called most probable velocity. It is obtained by plotting a graph between velocities of molecules on x-axis and fractions of molecules on y-axis. Its expression is as follows.

cm.p=== 1.414

cm.p= 1.414 (6)

Equation (6) also tells us that most probable velocity is directly proportional to the square root of absolute temperature and inversely proportional to the square root of molar mass of the gas.

Comparison of Velocities

By using above formula for three types of velocities, we can calculate their values by putting the values of temperature and molar mass. Anyhow, the ratio of the velocities will remain the same, if the molar mass is same and they are maintained at same temperature.

cr.m.s::cm.p

1.732 : 1.59: 1.414

1.732         :         1.59         :         1.414

It means that root mean square velocity of a gas is of maximum value and most probablyis the least value among the three velocities.

Example (1)

Calculate the Cr.m.s, and Cm.p of O2 molecule at 25C. What do you expect these velocities for H2 will be at 25C.

Solution

The molar mass of ‘O2’ molecule should be taken in kg, because the value of “R” is in J K-1mol-1

M = 32 x 10-3= 0.032 kg mol-1

T = 25C + 273 = 298K

R = 8.314 J K-1

Putting the values in equation for various velocities

crms=

=

=

= 481.9 ms-1Ans.

=

=

= = 441.13ms-1Ans.

cm.p==

 

cm.p = = 393.5ms-1Ans.

Because the molar mass of 1-12 is 16 times less than that of 02, so the above calculated velocities for 1-12 will be four times higher than that of 02 at 250C.

Example (2)

At what temperature will the Crms of SO2 be the same as that of O2 at 27C.

Solution

Let us suppose that the temperature is ‘T1‘ at which SO2 would have the same root mean square velocity as O2 at 27C.

Since we donot have to put the value of R in any given units in the numerical, so the unit of 11101ar mass may be in g mol-1

crms (SO2) =

For O2 the crms velocity at 27C (300 K) is

crms (O2) =

Since velocities of both are same, so

So,  =

Squaring both sides, we get

=

T1= = 600 K

or T1 = 600 – 273 = 327C Ans.

Kinetic Energy of One Mole of a Monoatomic Ideal Gas

The kinetic equation of gases can help us to derive the formula for kinetic energy possessed by one mole of an ideal gas irrespective of the nature of the gas. The basic condition is that the gas should be monoatomic. We know that

PV =mn2

If the gas is one mole, then

n = NA

PV =mNA2            mNA=M (molar mass)

PV =M2 (1)

The general gas equation for one mole of gas is

PV = RT  (2)

Comparing equation (1) and equation (2)

M2= RT (3)

Multiply and divide L.H.S of equation (3) with ‘2’

= RT

Ek= RT (3)

According to equation (4), the kinetic energy of one mole of an ideal gas depends upon the temperature of the gas and not upon its nature.

Remember that, if the gas is diatomic or polyatomic, then some extra energy is required to maintain the rotational and vibrational motions of the monoatomic gas.

Example (3)

Calculate translational kinetic energy for 2 moles of a gas at 27C.

Solution

For 1 mole of a gas, K.E = RT

So, for 2 moles of the gas,

K.E = 2

Since         R = 8.314 JK-1mol-1

And            T = 27C = (27 + 273) K = 300 K

Kinetic Energy =  8.3143 JK-1 mol-1 300 K  2 moles

= 7482.6 Joules Ans.

Kinetic Energy of One Molecule of an Ideal Gas

If we divide the equation (4) with Avogadro’s number NA, then we get the equation for one molecule of gas i.e. k.

EK= RT

k= == k (Bltsmann’s constant)

k=kT (5)

The value of k =  is, 1.38 x 10-23 JK-1 molecule-1

This equation (5) gives us the translational energy of a monoatomic gas.