The molecules of a gas are in a state of random motion. They continuously collide against the walls of the container. During each collision, momentum is transfered to the walls of the container.

*Pressure exerted by a gas*

The molecules of a gas are in a
state of random motion. They continuously collide against the walls of the
container. During each collision, momentum is transfered to the walls of the
container. The pressure exerted by the gas is due to the continuous collision
of the molecules against the walls of the container.
Due to this continuous collision,
the walls experience a continuous
force which is equal to the total momentum imparted to the walls per second.
The force experienced per unit area of the walls of the container determines
the pressure exerted by the gas.

Consider a cubic container of
side l containing n molecules of perfect gas moving with velocities C1, C2, C3
... Cn (Fig.). A molecule moving with a velocity
C1, will have velocities u1, v1 and w1 as components along the x, y and z axes
respectively. Similarly u2, v2 and w2 are the velocity components of the second
molecule and so on. Let a molecule P (Fig.) having velocity C1 collide against the
wall marked I (BCFG) perpendicular to the x-axis. Only the x-component of the
velocity of the molecule is relevant for the wall I. Hence momentum of the
molecule before collision is mu1 where m is the mass of the molecule. Since the
collision is elastic, the molecule will rebound with the velocity u1 in the
opposite direction. Hence momentum of the molecule after collision is ?mu1.

Change in the momentum of the
molecule

= Final momentum - Initial
momentum

= ?mu1 ? mu1 = ?2mu1

During each successive collision
on face I the molecule must travel a distance 2l from face I to face II and
back to face I.

Time taken between two successive
collisions is = 2*l*/u_{1}

∴ Rate of change of momentum = Change in the momentum / Time
taken

= (-2mu_{1}) / (2*l*/u_{1})

= (-2mu_{1}^{2}) / (2*l*)

= -mu_{1}^{2} / *l*

(i.e) Force exerted on
the molecule = -mu_{1}^{2}
/ *l*

∴ According to Newton?s third law of motion, the force exerted by
the molecule

= -(-mu_{1}^{2}
/ *l) = * mu_{1}^{2} / *l*

Force exerted by all
the n molecules is

F_{x}= mu_{1}^{2}
/ *l + *mu_{2}^{2} / *l* + ?? + mu_{n}^{2} / *l *

Pressure exerted by
the molecules

P_{x}=F_{x}/A

=1/*l*^{2 }( [mu_{1}^{2}]/*l*
+ [mu_{2}^{2}]/*l +
?..+ *[mu_{n}^{2}]/*l *)

=m/*l*^{3}(u_{1}^{2}
+ u_{2}^{2 }+ ?.+ u_{n}^{2 })

Similarly, pressure
exerted by the molecules along Y and Z axes are

P_{y} = m/*l*^{3}(v_{1}^{2}+
v_{2}^{2}+ ???+ v_{n}^{2})

P_{x} = m/*l*^{3}(w_{1}^{2}+
w_{ 2}^{2}+ ???+ w_{ n}^{2})

Since the gas exerts
the same pressure on all the walls of the container

P_{x} = P_{y}
= P_{z} = P

P = [P_{x} + P_{y}
+ P_{z} ]/3

P = 1/3 . m/*l*^{3}. [C_{1}^{2}
+ C_{2}^{2} + .... + C_{n}^{2} ]

where C_{1}^{2}
= (u_{1}^{2} + v_{1}^{2} + w_{1}^{2}
)

P = 1/3 . mn/V . C^{2}

where C is called the
root mean square (RMS) velocity, which is defined as the square root of the
mean value of the squares of velocities of individual molecules.

(i.e.) C = root[ (C_{1}^{2
}+ C_{2}^{2 }+?.+ C_{n}^{2 }) / n ]

**Relation between the pressure exerted by a gas and the mean**

kinetic energy of
translation per unit volume of the gas

Pressure exerted by
unit volume of a gas, P =
mnC^{2} / 3

P = 1/3 . ρC^{2 } (∵ mn = mass per unit
volume of the gas ; mn = ρ, density of the gas)

Mean kinetic energy of
translation per unit volume of the gas

E =1/2 . ρC^{2}

P/E = 2/3

P=2/3 . E

**Average kinetic energy per molecule of the gas**

Let us consider one
mole of gas of mass M and volume V

P = 1/3 . ρC^{2}

P = 1/3 . M/V . C^{2}

PV = 1/3 . M . C^{2}

From gas equation

PV = RT

∴ RT =1/3 . M C^{2}

3/2 . RT = ? . M C^{2}

(i.e) Average kinetic
energy of one mole of the gas is equal to 3/2 . RT

Since one mole of the
gas contains N number of atoms where N is the Avogadro number

we have M = Nm

(1/2 ) . mNC^{2}
= 3/2 . RT

? mC^{2} = 3/2
. R/N . T

=3/2 kT where k =R/N, is
the Boltzmann constant Its value is 1.38 ? 10^{-23} J K^{-1}

∴ Average kinetic energy per molecule of the gas is equal to 3/2
kT

Hence, it is clear
that the temperature of a gas is the measure of the mean translational kinetic
energy per molecule of the gas.

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail

**Related Topics **

Copyright © 2018-2020 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.