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Nov 8, 2018

Formulation of Kinetic Gas Equation

Formulation of Kinetic Gas Equation:

After knowing the 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 Kinetic Theory of Gases based upon the certain postulates which are supposed to be applicable to an ideal gas.
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(є) 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, RMS = (N₁C₁² + N₂C₂² + N₃C₃²......)/N
Derivation of the Kinetic Gas Equation:
Let us take a cube of edge length l containing 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
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, 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/l and 2 mCz²/l respectively.
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 mNC²/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 × 6l² = 2 mNC²/l
or, P × l³ = (1/3)mNC²
or, PV = (1/3)mNC²
The other from of the equation is P = (1/3)(mN/V)C²
P = (1/3)dC²
Where (mN/V) is the density(d) of the gas molecules.
The Kinetic Gas Equation
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)mNC²
RT = (1/3)MC²
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(Ē):
The average kinetic energy(Ē) is defined as Ē = (1/2)mC².
Again from Kinetic Gas Equation PV = (1/3)mNC²
or, PV = (2/3)N{(1/2)mC²}
or, PV = (2/3) N Ē
For 1 mole ideal gas PV = RT and N = N₀
Thus RT = (2/3)N₀ Ē
or, Ē = (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, Etotal = N₀(Ē) = (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

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