Solution:

pressure `P = 1/3 (nmc^2)/V`
Multiplying and dividing the right side by 2,

`PV = (2/3) 1/2nmc^2 = 2/3 E`

where E is the average kinetic energy.
Also, `PV = RT` for 1 mole of a gas

`:. RT= 2/3E`

. . Kinetic energy (average) is directly proportional to the absolute temperature

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Solution:

1. The kinetic theory of gases is based on the following assumptions :
(i) A gas consists of a very large nu!l}ber of molecules which are perfect elastic spheres are identical in all respects for a given gas and are different for different gases.
(ii) The molecules of a gas are in a state of continuous, rapid and random motion in all directions with different speeds, :ranging from zero to infinity and obey Newton's laws of motion.
(iii) The size of the gas molecules is very small as compared to the distance between them. Hence volume occupied by the molecules is negligible in comparison to the volume of the gas.
(iv) The molecules do not exert any force of attraction or repulsion on each other, except during collision.
(v) The collisions of the molecules with themselves and with the walls of the vessels are perfectly e!astic. i.e., the momentum and the kinetic energy of the molecules are conserved during collisions.

Expression for pressure due to an ideal gas: Consider an ideal gas contained in a cubical container OPQRSTKL of each side a and having a volume V. Clearly, volume of the gas, V = volume of the container `= a 3` i.e., `V = a^3` Let there be n molecules of the gas in the container each of mass m. Then total mass of the gas in the container is

`M=mxxn`

Let the random velocities of the gas molecules
`A_1 A_2, .... A_n be c_1 c_2, ..... C_n` respectively. Let
`(x_ Y_1 z_l), (x_2, Y_2, z_2), ...... (X_m Y_m Z_n)` be the rectangular components of the velocities `c_1, c_2,` .... en respectively, along three mutually perpendicular directions `OX, OY` and `OZ.`

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Solution:

`P = 1/3 (mnc^2)/V`

`= 2/3V . 1/2 nmc^2 = 2/3V E,`

where V is volume and E is the total K.E. of the molecules.

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Solution:

According to kinetic theory, molecules of a liquid are in a state of continuous random motion. The molecules near the surface of liquid may have enough K.E. so as to overcome the intermolecular attraction of other molecules on the surface and hence manage to escape. Such molecules would move around freely in the space above the liquid. This is the phenomenon of evaporation which may occur at all temperatures.

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Solution:

The average distance through which a molecule moves freely between two collisions is called the mean free path.

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Solution:

It may be defined as the temperature at which the velocities of the gas molecules are reduced to zero.

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Solution:

`"Law of equipartition of energy."` For any dynamical system in thermal equilibrium, the total energy is distributed equally amongst all degrees of freedom and the energy associated with each molecule per degree of freedom is
`1/2K_BT` where `K_B` is Boltzmann constant and T is temperature of the system.

li'or monoatomic gas there are only three degrees of freedom. For a gas in thermal equilibrium at temperature T, the average value of translational energy of molecule is

`< E_t > = 1/2m u_(x)^(2)> + <1/2 mv_(y)^(2)>+ <1/2mv_(z)^(2)>`

Therefore energy associated with monoatomic molecule is `3/2 K_B T.`

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Solution:

For a given mass of a gas, the pressure of a gas is inversely proportional to its volume at a fixed temperature.

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Solution:

It states that equal volumes of all gases under identical conditions of temperature and pressure, contain the same number of molecules.

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Solution:

`PV = nRT`
`R = (PV)/(nT)`

(n is a number of molecules)
`R = ([ML^(-1)T^(-2)]L^3)/K`

`=[ML^2T^(-2)K^(-1)]`

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Solution:

According to law of equipartition of energy, for any dynamical system in thermal equilibrium the total energy is distributed equally amongst all the degrees of freedom and the energy associated with each molecule per degree of freedom is `1/2 K_BT` where `K_B` is Boltzmann constant and T is temperature of the system.

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