- From equation (5) we have

PV = (1/3)Nmv_{mq}^{2}

where N is the number of molecules in the sample. Above equation can also be written as

PV = (2/3)N(1/2)Nmv_{mq}^{2}(7)

- The quantity (1/2)Nmv
_{mq}^{2}in equation (7) is the Kinetic energy of molecules in the gas. Since the internal energy of an ideal gas is purely kinetic we have,

E=(1/2)Nmv_{mq}^{2}(8)

- Combining equation 7 and 8 we get

PV=(2/3)E

- Comparing this result with the ideal gas equation (equation (4) ) we get

E=(3/2)K_{B}NT

or, E/N=(1/2)mv_{mq}^{2}=(3/2)K_{B}T (9)

Where, K_{B}is known as Boltzmann constant and its value is K_{B}=1.38 X 10^{-23}J/K

- From equation (11) we conclude that the average kinetic energy of a gas molecule is directly proportional to the absolute temperature of the gas and is independent of the pressure , volume and nature of the gas.

- Hence average KE per molecule is

(1/2)mv^{2}¯=(3/2)K_{B}T

from this since v^{2}¯=(v_{rms})^{2}, rms velocity of a molecule is

v_{rms}=√(3K_{B}T/m) (10)

This can also be written as

v_{rms}=√(3K_{B}NT/Nm)

=√(3RT/M) (11)

where, M=mN is the molecular mass of the gas.

Assignment

(1)Gas laws (Boyle's and Charle's law) and perfect gas equation can be derived using kinetic theory of gases. try to derive them.

- Moleular nature of matter
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- Behaviour Of Gases
- Kinetic Theory of an ideal gas
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- Pressure of gas
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- Kinetic interpretation of temperature
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- Law of Equipartition of energy
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- Specific Heat Capacity Of Gases
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- Specific heat Capacity of Solids
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- Mean free Path
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- Solved examples

Class 11 Maths Class 11 Physics Class 11 Chemistry

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