Revision: Kinetic Theory of Gases JEE Main Kinetic Theory of Gases

The average distance travelled by the molecule between collisions is called mean free path (λ).

λ = `"kT"/(sqrt(2)π"d"^2"p")`

The minimum number of independent coordinates needed to specify the position and configuration of a thermo-dynamical system in space is called the degree of freedom of the system.

A temperature scale with absolute zero (zero kelvin) as the starting point is called the absolute scale or the kelvin scale.

The volume of a given mass of a dry gas varies inversely as the pressure and directly as the absolute temperature.

V ∝ \[\frac {1}{P}\] × T or \[\frac {PV}{T}\] = k (constant)

If volume changes from V₁​ to V_2​, pressure from P_1​ to P₂​, and temperature from T₁ to T₂​, then:

\[\frac {P_1V_1}{T_1}\] = \[\frac {P_2V_2}{T_2}\] = k (constant)

The energy possessed by a body due to its state of motion is called its kinetic energy.

The motion of a body in a straight line path is called translational motion.

The kinetic energy of the body due to motion in a straight line is called translational kinetic energy.

If a body rotates about an axis, the motion is called rotational motion.

The kinetic energy of the body due to rotational motion is called rotational kinetic energy or simply rotational energy.

If a body moves to and fro about its mean position, the motion is called vibrational motion.

The kinetic energy of the body due to its vibrational motion is called vibrational kinetic energy or simply vibrational energy.

K = \[\frac {1}{2}\] mv²

Kinetic Energy = \[\frac {1}{2}\] mass × (velocity)²

It states that the volume of a given mass of dry gas is inversely proportional to its pressure at a constant temperature.
P₁V₁ = P₂V₂ = k at constant temperature

It states that volume of a given mass of a dry gas is directly proportional to its absolute (kelvin) temperature, if the pressure is kept constant.

OR

The pressure remaining constant, the volume of a given mass of a dry gas increases or decreases by 1/273 of its volume for each 1°C increase or decrease in temperature respectively.

\[\frac {V_1}{T_1}\] = \[\frac {V_2}{T_2}\] = k at constant pressure

Statement:

According to the work-energy theorem, the increase in kinetic energy of a moving body is equal to the work done by a force acting in the direction of the moving body.

Proof:

Let a body of mass m be moving with an initial velocity u. When a constant force F is applied to the body along its direction of motion, it produces an acceleration a, and the body's velocity increases from u to v over a distance S.

Force,

F = ma

Work done by the force,

W = F × S

From the equation of motion,

\[v^2=u^2+2aS\Rightarrow S=\frac{v^2-u^2}{2a}\]

Substituting equations (i) and (iii) into (ii):

W = \[ma\times\frac{v^2-u^2}{2a}=\frac{1}{2}m(v^2-u^2)\]

Now,
Initial kinetic energy, K_i = \[\frac {1}{2}\]mu²
Final kinetic energy, K_f = \[\frac {1}{2}\]mv²

Therefore,

W = K_f − K_i

Conclusion:

Work done on the body = Increase in its kinetic energy.
Hence, the work-energy theorem is proved.