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

**based upon the certain postulates which are supposed to be applicable to an ideal gas.**__Kinetic Theory of Gases__###
__Postulates of Kinetic Theory:__

__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 (

**) and with themselves (**__wall collision__**). These collision are perfectly elastic and so there occurs no loss of kinetic energy or momentum of the molecules by this collision.**__inter molecular 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

**since when volume is reduced, wall collision becomes more frequent and pressure is increased.**__Boyle's Law__
7.Through the molecular velocity are constantly changing due to inter molecular collision,

**of the molecules remains fixed at a given temperature.**__average kinetic energy__(Ñ”)
This explain the

**that when**__Charl's law__**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.C_{RMS}^{2} = (N_{1}C_{1}^{2} + N_{2}C_{2}^{2} + N_{3}C_{3}^{2} + ..)/N |

###
__Derivation of the Kinetic Gas Equation:__

__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**C**at temperature_{RMS}**T**and Pressure**P**.Molecular Velocity and its Components. |

Let in a gas molecules,

**N**have velocity

_{1}**C**,

_{1}**N**have velocity

_{2}**C**,

_{2}**N**have velocity

_{3}**C**, and so on.

_{3}
Let us concentrate our discussion to a single molecule among

**N**that have resultant velocity_{1}**C**and the component velocities are_{1}**C**,_{x}**C**and_{y}**C**._{z}So that, C_{1}^{2} = C_{x}^{2} + C_{y}^{2} + C_{z}^{2} |

The Molecule will collide walls

**A**and**B**with the Component Velocity**C**and other opposite faces by_{x}**C**and_{y}**C**._{z}
Change of momentum along

**X**for a single collision,__-direction__
= mC

_{x}- (- mC_{x})**= 2 mC**_{x}
Rate of change of momentum of the above type of collision,

= 2 mC

_{x}× (C_{x}/*l*)**= 2 mC**

_{x}^{2}/*l*
Similarly along

**Y**and**Z**directions, the rate of change of momentum of the molecule are**2 mC**and_{y}2/*l***2 mC**respectively._{z}2/*l*
Total rate change of momentum for the molecule,

= 2 mC_{x}^{2}/l + 2 mC_{x}^{2}/l +2 mC_{z}^{2}/l = 2 (m/ l) (C_{x}^{2} + C_{y}^{2} + C_{z}^{2})= 2 mC
_{1}^{2}/l |

For similar

**N**molecules, it is 2 m_{1}**N**/_{1}C_{1}^{2}**l**
Taking all the molecules of the gas, the total rate of change of momentum,

= (2 mN_{1}C_{1}^{2}/l)+(2 mN_{2}C_{2}^{2}/l)+(2 mN_{3}C_{3}^{2}/l)+ ..= 2 m N {(N_{1}C_{1}^{2} + N_{2}C_{2}^{2} + N_{3}C_{1}^{3} ..)/N}= 2 mNC_{RMS}^{2} |

Where

**C**= Root Mean Square Velocity of the Gases._{RMS}^{2}
According to the

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

*l*^{2}= 2 mN**C**/_{RMS}^{2}*l*
or, P ×

*l*3 = 1/3 mN**C**_{RMS}^{2}∴ PV = 1/3 mNC_{RMS}^{2} |

Here,

**l**= Volume of the Cube Contain Gas Molecules.^{3}
The other form of the equation is,

PV = 1/3 mN × mN/V × C

_{RMS}^{2}∴ PV = 1/3 d C_{RMS}^{2} |

Where

**mN/V**is the density**(d)**of the gas molecules.Kinetic Gas Equation |

__This equation are also valid for any shape of the gas container.__
Calculate the pressure exerted by

**10**gas particles each of mass^{23}**10**gm in a container of volume^{-22}**1 dm**. The root mean square speed is^{3}**10**.^{5}cm sec^{-1}
We have, N = 10

^{23},
m = 10

^{-22}gm = 10^{-25}Kg,
V = 1 dm

^{3}= 10^{-3}m^{3}
and

**C**= 10_{RMS}^{2}^{5}cm sec^{-1}= 10^{3}m sec^{-1}.
Therefore, from

**Kinetic Gas Equation**,
PV = 1/3 mNC

_{RMS}^{2}
or, P = 1/3 × mN/V × C

_{RMS}^{2}
Putting the value we have,

P=(1/3)(10

^{-25}Kg×10^{23}/10^{-3}m^{3})×(10^{3}m sec^{-1})^{2}**∴ P = 0.333 × 10**

^{7}Pa

__Expression of Root Mean Square Velocity__(C_{RMS}^{2})

__from Kinetic Gas Equation:__
Let us apply the kinetic equation for

**1 mole Ideal Gas**.
In that case

**mN**=**mN**=_{0}**M**and**PV**=**RT**.
Hence from Kinetic Gas Equation,

PV = 1/3 mNC

_{RMS}^{2}
or, RT = 1/3 mNC

_{RMS}^{2}or, C_{RMS}^{2} = 3RT/M∴ C_{RMS}^{2} = √3RT/M |

Thus root mean square velocity depends on the

**and**__molar mass(M)__**of the gas.**__temperature(T)__
Calculate the

**root mean square speed**of oxygen gas at**27**.^{0}C
We know that,

**C**._{RMS}^{2}= √3RT/M
Here,

**M**= 32 gm mol^{-1},and**T**= 27^{0}C = (273+27)K = 300 K.
∴

**C**= √(3 × 8.314 × 10_{RMS}^{2}^{7}erg mol^{-1}K^{-1}× 300 K)/(32 gm mol^{-1})
= 48356 cm sec

^{-1}
Calculate the

**RMS**speed of**NH**at_{3}**N.T.P**.
At

**N.T.P**,**V**= 22.4 dm^{3}mol^{-1}= 22.4 × 10^{-3}m^{3}mol^{-1},**P**= 1 atm = 101325 Pa

and

**M**= 17 × 10^{-3}Kg mol^{-1}.
Thus,

**C**_{RMS}^{2}= √3RT/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**.

_{RMS}^{2}
Again from Kinetic Gas Equation,

PV = 1/3 mNC

_{RMS}^{2}
or, PV = (2/3) N × (1/2) m C

_{RMS}^{2}**or, PV = 2/3 N Ä’**

For

**1 mole ideal gas**,**PV = RT**and

**N = N**

_{0}
Thus RT = 2/3 N

_{0}Ä’or, Ä’ = (3/2)(R/N _{0})T∴ Ä’ = 3/2 kT |

Where

**k = R/N**and is known as the Boltzmann Constant. Its value is 1.38 × 10_{0}^{-23}JK^{-1}.
The total kinetic energy for 1 mole of the gas is,

E_{Total} = N_{0} (Ä’) = 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****NH**at_{3}**27**.^{0}C
We know that total kinetic energy for 1 mole of the gas is,

**E**

_{Total}= (3/2)RT
= (3/2)(2 cal mol

^{-1}K^{-1}× 300 K)**= 900 cal mol**

^{-1}
Again 8.5 gm NH

_{3}= (8.5/17) mol = 0.5 mol
Thus Kinetic Energy of 8.5 gm NH

_{3}at 27^{0}C is (0.5 × 900) cal**= 450 cal**

Calculate the

**RMS**velocity of oxygen molecules having a kinetic energy of**2 K.cal mol**. At what temperature the molecules have this value of^{-1}**KE**?**T = 673.9K or 400.9**

^{0}C