The molecules of the gas are in random motion. They collide with each other and also with the walls of the container.

**PRESSURE
EXERTED BY A GAS**

**Expression
for pressure exerted by a gas**

Consider
a monatomic gas of N molecules each having a mass m inside a cubical container
of side *l* as shown in the Figure 9.1
(a).

The
molecules of the gas are in random motion. They collide with each other and
also with the walls of the container. As the collisions are elastic in nature,
there is no loss of energy, but a change in momentum occurs.

The
molecules of the gas exert pressure on the walls of the container due to
collision on it. During each collision, the molecules impart certain momentum
to the wall. Due to transfer of momentum, the walls experience a continuous
force. The force experienced per unit area of the walls of the container
determines the pressure exerted by the gas. It is essential to determine the
total momentum transferred by the molecules in a short interval of time.

A
molecule of mass m moving with a velocity having components (*v*_{x},
*v*_{y}, *v*_{z}) hits the right side wall. Since we have* *assumed that the collision is elastic,
the particle rebounds with same speed and its x-component is reversed. This is
shown in the Figure 9.1 (b). The components of velocity of the molecule after
collision are (—*v _{x}*,

The
x-component of momentum of the molecule before collision = *mv*_{x}

The x-component of momentum of the molecule after collision
= −m*v _{x}*

The
change in momentum of the molecule in x direction

=Final
momentum – initial momentum = −*mv _{x}*
−

According
to law of conservation of linear momentum, the change in momentum of the wall =
2*mv _{x}*

The
number of molecules hitting the right side wall in a small interval of time ∆*t* is calculated as follows.

The
molecules within the distance of *v _{x}*∆

Not
all the n molecules will move to the right, therefore on an average only half
of the n molecules move to the right and the other half moves towards left
side.

The
number of molecules that hit the right side wall in a time interval

In
the same interval of time ∆*t*, the
total momentum transferred by the molecules

From
Newton’s second law, the change in momentum in a small interval of time gives
rise to force.

The
force exerted by the molecules on the wall (in magnitude)

Pressure,
P = force divided by the area of the wall

Since
all the molecules are moving completely in random manner, they do not have same
speed. So we can replace the term *v _{x}*

Since the gas is assumed to move in random direction, it has no preferred direction of motion (the effect of gravity on the molecules is neglected). It implies that the molecule has same average speed in all the three direction. So, . The mean square speed is written as

Using
this in equation (9.5), we get

The
following inference can be made from the above equation. The pressure exerted
by the molecules depends on

**(i) Number density ***n***
**= N/V. It implies that** **if
the number density increases then pressure will increase. For example when we
pump air inside the cycle tyre or car tyre essentially the number density
increases and as a result the pressure increases.

**(ii) Mass of the molecule **Since the pressure** **arises due to momentum transfer to the
wall, larger mass will have larger momentum for a fixed speed. As a result the
pressure will increase.

**(iii) Mean square speed **For a fixed mass** **if we increase the speed, the average speed will also increase. As
a result the pressure will increase.

For
simplicity the cubical container is taken into consideration. The above result
is true for any shape of the container as the area A does not appear in the
final expression (9.6). Hence the pressure exerted by gas molecules on the wall
is independent of area of the wall.

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11th Physics : UNIT 9 : Kinetic Theory of Gases : Expression for pressure exerted by a gas |

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