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.

Postulates of Kinetic Theory:

  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.
  7. This explains Boyle's law since when volume is reduced, wall collision becomes more frequent and pressure is increased.
  8. 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.
  9. 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.
CRMS2 = (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.
What is the Kinetic theory of gases? How can we formulate kinetic gas equation?
Molecular Velocity and its Components.
Let in a gas molecules, 
N1 have velocity C1
N2 have velocity C2
N3 have 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-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 mCx2/l +2 mCz2/l 
= 2 (m/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,
= (2 mN1C12/l)+(2 mN2C22/l)+(2 mN3C32/l)+ ..
= 2 mN {(N1C12 + N2C22 + N3C13 ..)/N}
= 2 mNCRMS2
Where CRMS2 = 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 mNCRMS2/l
or, P × l3 = 1/3 mNCRMS2
∴ PV = 1/3 mNCRMS2
Here, l3 = Volume of the Cube Contain Gas Molecules.
The other form of the equation is,
PV = 1/3 mN × mN/V × CRMS2
∴ PV = 1/3 d CRMS2
Where mN/V is the density(d) of the gas molecules.
What is the Kinetic theory of gases? How can we derived kinetic gas equation?
Kinetic Gas Equation
This equation are also valid for any shape of the gas container.
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.
We have, N = 1023
m = 10-22 gm = 10-25 Kg, 
V = 1 dm3 = 10-3 m3 
and CRMS2 = 105 cm sec-1 = 103 m sec-1
Therefore, from Kinetic Gas Equation,
PV = 1/3 mNCRMS2
or, P = 1/3 × mN/V × CRMS2
Putting the value we have,
P=(1/3)(10-25 Kg×1023/10-3 m3)×(103 m sec-1)2
∴ P = 0.333 × 107 Pa
Expression of Root Mean Square Velocity(CRMS2) from Kinetic Gas Equation:
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 mNCRMS2
or, RT = 1/3 mNCRMS2
or, CRMS2 = 3RT/M
CRMS2 = √3RT/M
Thus root mean square velocity depends on the molar mass(M) and temperature(T) of the gas.
Calculate the root mean square speed of oxygen gas at 270C.
We know that, CRMS2 = √3RT/M .
Here, M = 32 gm mol-1 ,and T = 270C = (273+27)K = 300 K. 
CRMS2 = √(3 × 8.314 × 107 erg mol-1K-1 × 300 K)/(32 gm mol-1)
= 48356 cm sec-1
Calculate the RMS speed of NH3 at N.T.P.
At N.T.PV = 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, CRMS2 = √3RT/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 m CRMS2.
Again from Kinetic Gas Equation,
PV = 1/3 mNCRMS2
or, PV = (2/3)  N × (1/2) m CRMS2
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 
∴ Ē  = 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) = 3/2RT
Thus Average Kinetic Energy is dependent of T only and it is not dependent on the nature of the gas.
Calculate the kinetic energy of translation of 8.5 gm NH3 at 270C.
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 270C is (0.5 × 900) cal
= 450 cal
Calculate the RMS velocity of oxygen molecules having a kinetic energy of 2 K.cal mol-1. At what temperature the molecules have this value of KE ?
T = 673.9K or 400.90C

Formulation of Kinetic Gas Equation: Postulates of Kinetic Theory, Root Mean Square Speed, Derivation of the Kinetic Gas Equation, Expression of Root Mean Square Velocity and Expression of Average Kinetic Energy.

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