Chemistry KINETIC THEORY AND CONCEPT OF AVERAGE, ROOT MEAN SQUARE AND MOST PROBABLE VELOCITIES

Velocity Distribution of Gas molecules :

Speed of a gas molecule changes continuously due to the intermolecular collisions and their collisions with the wall of container. Thus, speed of an individual molecule is not constant. Also, the observable properties of gas such as volume, pressure and temperature are constant with time. It is expected to be applicable to the distribution of molecular speed and we consider a statistical average of the speed of whole of the collection of gas molecules to remain constant with time. That is the fraction or the total molecules having speed between any definite ranges must be constant, even though the speeds of the individual molecules may be changing as a result of molecular collisions. Distribution of molecular speed over a possible range was first investigated by Maxwell using the theory of probability. Results were expressed as the Maxwell law for distribution of molecular speed as shown in fig.1.

`dN_u = 4 pi N (M/(2 pi RT))^(3/2) e^((-m u^2)/(2RT)) u^2 du`

`= 4 pi N (M/(2 pi kT))^(3/2) e^((-m u^2)/(2kT)) u^2 du`

The above expression gives the number of molecules `dN_u` having speeds between `u` and `u + du` in term of total number of molecules present (`N`), molar mass (`M`) of gas and temperature (`T`). According to expression the fraction of molecules `((dN_u) /N)` having the speeds in the range of `u` and `u + du`, for a gas of molar mass (`M`) depends only on temperature. Hence, for a given temperature, this fraction is constant. A plot of fraction of molecules in the speed range `u` and `u + du`, `1/N ((dNu)/(du))` vs `u` is described in the graph. The peaks in the curve correspond to a speed, which is possessed by maximum fraction of molecules, called "most probable speed".

`ast` `text(Some conclusive points for distribution of molecular Speeds are)` :

(i) The fraction of molecules having either very high or very low speeds is very small. Majority of molecules have speed near to `U_(rnps)` in the middle of the range of molecular speeds. See fig.2.

(ii) The total area under the curve in fig.3 is a measure of total number of molecules in collection.

(iii) Since, rise in temperature raises the kinetic energy of gas molecules, it follows that fractions of molecules having lower speed range decreases and fraction of molecules having higher speed range increases. Also the curve at higher temperature has its `U_(mps)` shifted to higher value but the corresponding fractions of molecules have decreased. Thus in general, the distribution of speeds is wider at higher temperature than at lower temperature as shown in figure `(T_2 >T_1)`.

(iv) Distribution of molecular speed also depends upon the mass of gas molecule. At similar temperature, a heavier gas molecule has a narrow distribution of speeds than a lighter gas molecule as shown in the following diagram `(M_x > M_y)` :

As per kinetic theory of gases, each molecule is moving with altogether different velocity. Let `n` molecules be present in a given mass of gas, in which `n_1` molecules have a speed of `v_1`, `n_2` molecules have a speed of `v_2`, `n_n` molecules have a speed of `v_n`.

The average velocity or `U_(av) =` average of all such velocity terms.

Average velocity `= (n_1u_1 + n_2u_2 + n_3u_3 + ............+n_n u_n)/n`

Using Maxwell distribution average velocity is determined as `U_(av) = srqt((8RT)/(pi M))`

Another mean velocity proposed by Maxwell was `U_(rms)` as the square root of means of square of all such velocities.

`U_(rms) = sqrt((n_1 u_1^2 + n_2 u_2^2 +.....................+ n_n u_n^2)/n)`

Using Maxwell distribution `rms` velocity is determined as `U_(rms) = sqrt((3RT)/M)`

Most probable speed is the velocity which is possessed by maximum no. of molecules. `U_(mp) = sqrt((2RT)/M)`

Furthermore `U_(mp) : U_(av) : U_(rms) : : sqrt((2RT)/M) : sqrt((8RT)/(piM)) : sqrt((3RT)/M) = sqrt2 : sqrt(8/pi) : sqrt3 = 1 : 1.128 : 1.224`

Also `U_(av) = U_(rms) xx0.9213`

`ast` `text(For calculating) V_(rms), V_(mp), text(or) V_(av)` by above relations the value of `M` should be used in `kg`/`mol` and `R` should be taken as `8.314` `J`/`K` `mol` to get velocity in `m`/`s`.

Kinetic Interperetation of Pressure :

Imagine a cube of edge-length `l`, containing `N` molecules, each having a mass of `m`. Molecules are moving at random in all directions, with speed covering a considerable range of values. See fig.

The velocity `u_1` or any molecule may be resolved into three-component velocities designated as `u_x`, `u_y` and `u_z`. These are in the three directions at right angles to each other and parallel to the sides of the cube as shown in figure. The component velocities are related by the expression

`u_1^2 = u_x^2 +u_y^2 +u_z^2.................(1)`

Considering the `x`-component motion of a molecule, we will have

Momentum of molecule before collision with the side `ABCD` `= m u_x`.

Momentum of the molecule after collision with the side `ABCD` `= - m u_x`.

Change of momentum of the molecule in a single collision with the side `ABCD` `= |2m u_x |`

Since `l` is the edge length of the cube, the molecule has to travel a distance `2l` to arrive back at the wall `ABCD`. The number of collisions per unit time with the wall `ABCD` will be equal to `u_x /(2l)`.

The total change of momentum per unit time due to such impacts is `2m u_x(u_x/(2l)) = (m u_x^2)/l`

According to Newton's second law of motion

Force = mass `xx` acceleration = mass `xx (d(text(velocity)))/(dt) = d/(dt) (text(mass) xx text(velocity))`

`= d/(dt)(text(momentum)) = ` rate of change of momentum

Hence, total force due to impacts of a single molecule with the wall `ABCD` of the vessel is `(m u_x^2)/l`. The area of the wall is `l^2`. Hence, the pressure exerted due to the collision of `x`-component velocity of a single molecule with the side `ABCD` is

`p_x =((m u_x^2)/l)/l^2 = (m u_x^2)/V` where `V` is the volume of the vessel.

Since each molecule will exert similar pressure, the total pressure exerted on the wall `ABCD` will be

`p = sum_(i=1)^(N) p_(ix) = M/V sum_(i=1)^(N) u_x^2`

Defining the mean square speed as `bar(u_x^2) = 1/N sum_(i=1)^(N) u_(ix)^2`

We can write `p = (MN)/V bar(u_x^2) ...(2)`

Since the directions `x`, `y` and `z` are equivalent, we will also have `bar(u_x^2) = bar(u_y^2) = bar(u_z^2)........(3)`

But from Eq. (1), we will have `bar(u^2) = bar(u_x^2) + bar(u_y^2) + bar(u_z^2)`

from Eqs (1) and (3), we can write `bar(u_x^2) = bar(u_y^2) = bar(u_z^2) = 1/3 bar(u^2)`

Substituting this in Eq. (2). we get `p = (mN)/V(1/3 bar(u^2))`

or `pV = 1/3 mN bar(u^2)`

From Maxwell distribution we already know

`bar(u^2) = (3RT)/M`

So `pV = 1/3 mN xx (3 RT)/M = nM xx (RT)/M = nRT => pV - nRT`

� Average Translational kinetic energy per molecule = `1/2 m u_(rms)^2 = 3/2kT`

� Average Translational Kinctic energy per mole `= N_A xx 1/2 m u_(rms)^2 = 3/2 RT`

� k (Boltzmrum constant) `= R/N_A = 1.38 xx 10^(-23) J K^(-1)`

 
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