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Q.(a) State the postulates of kinetic theory of gases. Derive an expression for the pressure exerted by the gas on the basis of kinetic theory of gases.

(OR)
(a) If 'X' is the kinetic energy of an ideal gas at 127 C. What will be its kinetic energy if its temperature is raised to 527 C? [2 1/2 marks]
(b) The root mean square velocity of oxygen gas at 327 C is 'V'. Find the root mean square velocity of hydrogen at same temperature. [2 1/2 marks]
Himachal HpboseHPBOSE Himachal Pradesh Class 11 Board Exam 2025Subjective· 5mImportance★★★★★
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Kinetic theory postulates model gas molecules as point particles in random elastic motion; applying Newton's laws to wall collisions gives P = (1/3)ρ⟨v²⟩.

Postulates of the kinetic theory of gases:

  1. A gas consists of a very large number of identical molecules, which are treated as perfectly elastic, rigid spheres so small that their own size (volume) is negligible compared to the volume of the container.
  2. The molecules are in a state of continuous, random motion, moving in all directions with different speeds, colliding with each other and with the walls of the container.
  3. All collisions — molecule-molecule and molecule-wall — are perfectly elastic, so no kinetic energy is lost in a collision (though individual molecular speeds may change).
  4. Between collisions, molecules travel in straight lines with constant velocity (no forces act on them except during the brief instant of collision — intermolecular forces are neglected).
  5. The time spent in a collision is negligible compared to the time between successive collisions.
  6. The molecules obey Newton's laws of motion.
  7. The density and distribution of molecules and their velocities are uniform and isotropic (no preferred position or direction) throughout the container.

Derivation of pressure exerted by a gas:

Consider N molecules, each of mass m, enclosed in a cubical container of side L (volume V=L3V = L^3).

Take one molecule moving with velocity components (vx,vy,vz)(v_x, v_y, v_z). Consider its collisions with the wall perpendicular to the x-axis (area L2L^2).

Before collision, its momentum along x is mvxmv_x. Since the collision is perfectly elastic and the wall is rigid, the molecule rebounds with the same speed but reversed x-component: −mvx-mv_x.

Change in momentum of the molecule (per collision with this wall) =−mvx−(mvx)=−2mvx= -mv_x - (mv_x) = -2mv_x.

By Newton's third law, the momentum imparted to the wall per collision =2mvx= 2mv_x.

After rebounding, the molecule travels a distance 2L2L (to the opposite wall and back) before it strikes this same wall again, taking time Δt=2Lvx\Delta t = \dfrac{2L}{v_x}.

So the force exerted by this one molecule on the wall (rate of momentum transfer):

f=2mvxΔt=2mvx2L/vx=mvx2Lf = \dfrac{2mv_x}{\Delta t} = \dfrac{2mv_x}{2L/v_x} = \dfrac{mv_x^2}{L}

Summing over all N molecules, the total force on this wall:

F=mL∑vx2=mN⟨vx2⟩LF = \dfrac{m}{L}\sum v_x^2 = \dfrac{mN\langle v_x^2\rangle}{L}

where ⟨vx2⟩\langle v_x^2\rangle is the average of vx2v_x^2 over all molecules.

Since molecular motion is random and isotropic (no preferred direction), ⟨vx2⟩=⟨vy2⟩=⟨vz2⟩\langle v_x^2\rangle = \langle v_y^2\rangle = \langle v_z^2\rangle. Also v2=vx2+vy2+vz2v^2 = v_x^2+v_y^2+v_z^2, so on averaging: ⟨v2⟩=3⟨vx2⟩\langle v^2\rangle = 3\langle v_x^2\rangle, giving ⟨vx2⟩=13⟨v2⟩\langle v_x^2\rangle = \dfrac{1}{3}\langle v^2\rangle.

So:

F=mNL⋅⟨v2⟩3=mN⟨v2⟩3LF = \dfrac{mN}{L}\cdot\dfrac{\langle v^2\rangle}{3} = \dfrac{mN\langle v^2\rangle}{3L}

Pressure is force per unit area, and the area of one face is L2L^2:

P=FL2=mN⟨v2⟩3L3=mN⟨v2⟩3VP = \dfrac{F}{L^2} = \dfrac{mN\langle v^2\rangle}{3L^3} = \dfrac{mN\langle v^2\rangle}{3V} …

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