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__Formulation of Kinetic Gas Equation:__

After knowing the__Formulation of Kinetic Gas Equation:__

__experimental gas lows__, it is of interest to develop a

__theoretical model__based on the

__structure of gases__, which can correlated to the experimental facts. Fortunately such theory has been developed and known as the

**based upon the certain postulates which are supposed to be applicable to an ideal gas.**

__Kinetic Theory of Gases__

__The essential Postulates are:__1.The gas is composed

__very small discrete particles__, now called

__molecules__. For a gas the

__mass and size of the molecules are same and different for different gases.__

2.The molecules are

__moving in all directions with verity of speeds__. Some are very fast while other are slow.

3.Due to random motion, the molecules are executing collision with the walls of the container (

__wall collision__) and with themselves (

__inter molecular collision__). These collision are

__perfectly elastic__and so there occurs no loss of kinetic energy or momentum of the molecules by this collision.

4.The gas molecules are assumed to be

__point masses__, that is their size is very small in comparison to the distance they travel.

5.There exist

__no inter molecular attraction specially at law pressure__, that is one molecule can exert pressure independent of the influence of other molecules.

6.The pressure exerted by a gas is due to the uniform wall collisions. Higher the frequency of the wall collision greater will be the pressure of the gas.

__This explains Boyle's Law since when volume is reduced, wall collision becomes more frequent and pressure is increased.__

7.Through the molecular velocity are constantly changing due to inter molecular collision,

__average kinetic energy__(

**Ñ”**) of the molecules remains

__fixed at a given temperature__. This explain the

__Charl's law__that when

**T**is

__increased__, velocity are increased, wall collision become more frequent and pressure(

**P**) is increased when

**T**kept constant or Volume(

**V**) is increases when

**P**kept constant.

__Root Mean Square Speed:__
Root
mean square(RMS) speed is defined as the

__square root of the average of the squares of speeds.__
That
is,

**C²RMS**= (N₁C₁² + N₂C₂² + N₃C₃²......)/N

__Derivation of the Kinetic Gas Equation:__Let us take a cube of

__edge length__

*containing*

**l****N**

__molecules__of a gas of

__molecular mass__

**m**and

**RMS**speed is

**CRMS**at

__temperature__

**T**and

__Pressure__

**P**.

Molecular Velocity and its Components |

Let, in a gas molecules,

**N₁**have the velocity**C₁**,**N₂**have the velocity**C₂**,**N₃**have the velocity**C₃**, and so on.
Let us concentrate our discussion to a

__single molecule__among**N₁**that have__resultant velocity__**C₁**and the__component velocities__are**Cx**,**Cy**and**Cz**.
So that,

**C₁²**=**Cx²**+**Cy²**+**Cz²**
The Molecule will collide walls

**A**and**B**with the__Component Velocity__**Cx**and other opposite faces by**Cy**and**Cz**.Change of Momentum along X, Y and Z Direction |

Change of momentum along

**X**-direction for a single collision =**mCx - (-mCx) = 2 mCx**.
Rate of change of momentum of the above type of collision =

**2 mCx × (Cx/**.*l*) = 2 mCx²/*l*
Similarly along

**Y**and**Z**directions, the rate of change of momentum of the molecule are**2 mCy²/**and

*l***2 mCz²/**respectively.

*l*
Total rate change of momentum for the molecule,

= (2 mCx²/

*l*) + (2 mCy²/*l*) + (2 mCz²/*l*)
= 2m/

*l*(Cx² + Cy² + Cz²)
= 2 mC₁²/

*l*
For similar N₁ molecules, it is 2 mN₁C₁²/l.

Taking all the molecules of the gas, the total rate of change of momentum equal to

(2 mN₁C₁²/

*l*) + (2 mN₂C₂²/*l*) + (2 mN₃C₃²/*l*) + .......
Thus Rate of Change of Momentum = 2mN/

*l*{(N₁C₁² + N₂C₂² + N₃C₃²)/N} = 2 mN**C**²/*l*
Where

**C =**Root Mean Square Velocity of the Gases.__According to the__

**Newton's**2nd Low of Motion, rate of change of momentum due to wall collision is equal to force developed within the gas molecules.
That is, P × 6

*l*² = 2 mN**C**²/*l*
or, P × l³ = (1/3)mN

**C**²
or,

**PV =****(1/3)mNC²**
The other from of the equation is P = (1/3)(mN/V)

**C**²
P = (1/3)d

**C**²
Where (mN/V) is the density(d) of the gas molecules.

The Kinetic Gas Equation |

__These equation are also valid for any shape of the gas container.__

__Problem:__
Calculate the pressure exerted by 10²³ gas particles each of mass 10⁻²² gm in a container of volume 1 dm³. The root mean square speed is 10⁵ cm sec⁻¹.

__Solution:__
From the given data, we have N = 10²³, m = 10⁻²² gm = 10⁻²⁵ Kg, V = 1 dm³ = 10⁻³ m³ and Root Mean Square(C) = 10⁵ cm sec⁻¹ = 10³ m sec⁻¹.

Therefore, from

__Kinetic Gas Equation__, PV = (1/3)mNC²
or, P = (1/3)(mN/V)C²

Putting the value We have P = (1/3)(10⁻²⁵ Kg × 10²³/10⁻³ m³) × (10³ m sec⁻¹)²

**or, P = 0.333 × 10⁷ Pa**

__Expression of Root Mean Square Velocity:__
Let us apply the kinetic equation for 1 mole Ideal Gas. In that case mN = mN₀ = M

and PV = RT.

Hence from Kinetic Gas Equation PV = (1/3)mN

**C**²
RT = (1/3)M

**C**²
or,

**C² = 3RT/M**
or

**,**C = √(3RT/M)__Thus root mean square velocity depends on the molar mass(M) and temperature(T) of the gas.__

__Problem:__
Calculate the root mean square speed of oxygen gas at 27°C.

__Solution:__
We know that Crms = √(3RT/M).

Here, M = 32 gm mol⁻¹ and T = 27°C = (273+27)K = 300 K.

Thus Crms = √{(3 × 8.314 × 10⁷ erg mol⁻¹K⁻¹ × 300 K)/(32 gm mol⁻¹)}

=

**48356 cm sec⁻¹**

__Problem:__
Calculate the rms speed of ammonia in N.T.P.

__Answer:__
Here V = 22.4 dm³ mol⁻¹ = 22.4 × 10⁻³ m³ mol⁻¹, P = 1 atm = 101325 Pa and M = 17 × 10⁻³ Kg mol⁻¹.

Thus Crms = √(3PV/M) = √{(3 × 101325 × 22.4 × 10⁻³)/(17 × 10⁻³)}

=

**632 m sec⁻¹**

__Expression of Average Kinetic Energy(Ä’):__
The average kinetic energy(Ä’) is defined as Ä’ = (1/2)m

**C**².
Again from

__Kinetic Gas Equation__PV = (1/3)mN**C**²
or, PV = (2/3)N{(1/2)m

**C**²}
or, PV = (2/3) N Ä’

For 1 mole ideal gas PV = RT and N = N₀

Thus RT = (2/3)N₀ Ä’

or, Ä’ = (3/2)(R/N₀)T = (3/2)kT

Where k = R/N₀ and is known as the Boltzmann Constant. Its value is 1.38 × 10⁻²³ JK⁻¹.

The total kinetic energy for 1 mole of the gas is,

**E**total = N₀(Ä’) =**(3/2)RT.**__Thus Average Kinetic Energy is dependent of T only and it is not dependent on the nature of the gas.__

__Problem:__
Calculate the kinetic energy of translation of 8.5 gm NH₃ at 27°C.

__Solution:__
We know that total kinetic energy for 1 mole of the gas is, Etotal = (3/2)RT

= (3/2)(2 cal mol⁻¹ K⁻¹ × 300 K) = 900 cal mol⁻¹

Again 8.5 gm NH₃ = (8.5/17) mol = 0.5 mol

Thus Kinetic Energy of 8.5 gm NH₃ at 27°C is (0.5 × 900) cal = 450 cal

__Related Post:____Ven der Waals Equation__