Question

How does the kinetic-molecular theory explain the pressure exerted by gases?

Answer 1

Hint: In the kinetic theory of gases, gas molecules move around in the most arbitrary manner impacting and chancing upon one another and afterward skipping off the other way. The molecules apply no intermolecular powers on one another and they have no potential energy related with them in their movement. Consequently, the energy is entirely kinetic.
The idea of pressure is clarified in kinetic theory as an outcome of kinetic energy of gases.

Complete step by step answer:
Think about an ideal gas (the ones with which kinetic theory is worried) in a shut holder.
Because of the inconvenient movement of the gas atoms, they will crash into one another, some of them will crash into the compartment dividers and afterward bob back and this cycle proceeds.
Presently if $m\vec v$ be the momentum of a solitary molecules before it finds the divider and it makes a point \[\theta \] with the divider and since the divider is thought to be entirely smooth it bounces back at a point \[\theta \] once more (coordinated inverse to the course from which the gas atom at first came.
Along these lines, it moves a momentum, $mv\cos \theta $ and goes through an energy change and bounces back with a last momentum $ - mv\cos \theta $.
In this way the momentum change ends up being,
$ - mv\cos \theta + \left( { - mv\cos \theta } \right) = - 2mv\cos \theta $
This suggests that there is an energy move of \[2mvcos\theta \] to the divider from the molecule.
Accordingly considering various molecules striking the dividers, they move momentum.
Presently power is characterized to be the time pace of progress of energy.
In this way the molecules apply a power on the dividers.
Further since, pressure is characterized to be power per unit area, at that point the molecules apply a tension on the compartment dividers.
This pressure isn't the hydrostatic pressure of a liquid in an outside field since we expected that there is no potential energy related to them. Or maybe this pressure is related with dynamic energy of movement (the more is the motor energy, the more prominent is the movement and thus, a greater number of impacts happen and from contentions as before a more noteworthy momentum is moved in unit area per unit time.
Subsequently, a greater pressure is obtained.
That is the kinetic understanding of pressure.

Note: Using some numerical improvements and averaging over all molecules, one can show that the pressure is given by,
$p = \dfrac{{mn{v^2}}}{3}$
Where \[m\] is the mass of every molecule, \[n\] is the quantity of atoms in unit volume (subatomic thickness) and \[v\] is the rms speed.