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**__Kinetic Theory of Gases____Ideal Gas__.###
__Postulates of Kinetic Theory:__

__Postulates of Kinetic Theory:__

- 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.
- The molecules are moving in all directions with verity of speeds. Some are very fast while other are slow.
- 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__ - The gas molecules are assumed to be point masses, that is their size is very small in comparison to the distance they travel.
- There exist no inter molecular attraction specially at law pressure, that is one molecule can exert pressure independent of the influence of other molecules.
- 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.
- 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

This explains

**since when volume is reduced, wall collision becomes more frequent and pressure is increased.**__Boyle's law__**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