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Q.Derive the expression for pressure of an ideal gas enclosed in a container. From this, discuss the kinetic interpretation of temperature.

Assam AhsecAHSEC Higher Secondary (HS) 1st Year Examination 2025Subjective· 5mImportance★★★★★
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Kinetic theory gives gas pressure P = (1/3)ρ⟨v²⟩; comparing with the ideal gas law shows average molecular kinetic energy ⟨KE⟩ = (3/2)kT, giving temperature its kinetic meaning.

Derivation of pressure:

Consider N molecules, each of mass m, inside a cubical container of side L (volume V=L3V=L^3), moving randomly with velocities having components (vx,vy,vz)(v_x,v_y,v_z).

Consider one molecule moving with x-component of velocity vxv_x, colliding elastically with the wall perpendicular to the x-axis. Its momentum reverses: change in momentum per collision =2mvx= 2mv_x.

Time between two successive collisions with the same wall (travelling distance 2L2L at speed vxv_x): Δt=2Lvx\Delta t = \dfrac{2L}{v_x}

Force exerted by this molecule on the wall (average, over many collisions):

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

Summing over all N molecules and using the mean-square value ⟨vx2⟩\langle v_x^2\rangle:

F=Nm⟨vx2⟩LF = \dfrac{Nm\langle v_x^2\rangle}{L}

Pressure:

P=FA=FL2=Nm⟨vx2⟩L3=Nm⟨vx2⟩VP = \dfrac{F}{A} = \dfrac{F}{L^2} = \dfrac{Nm\langle v_x^2\rangle}{L^3} = \dfrac{Nm\langle v_x^2\rangle}{V}

By symmetry of random motion, ⟨vx2⟩=⟨vy2⟩=⟨vz2⟩=13⟨v2⟩\langle v_x^2\rangle=\langle v_y^2\rangle=\langle v_z^2\rangle=\dfrac{1}{3}\langle v^2\rangle (since v2=vx2+vy2+vz2v^2=v_x^2+v_y^2+v_z^2):

P=13Nm⟨v2⟩V=13ρ⟨v2⟩P = \dfrac{1}{3}\dfrac{Nm\langle v^2\rangle}{V} = \dfrac{1}{3}\rho\langle v^2\rangle

where ρ=NmV\rho = \dfrac{Nm}{V} is the density of the gas. Equivalently, PV=13Nm⟨v2⟩PV = \dfrac{1}{3}Nm\langle v^2\rangle.

Kinetic interpretation of temperature:

Rewrite: PV=23N(12m⟨v2⟩)=23N⟨KE⟩PV = \dfrac{2}{3}N\left(\dfrac{1}{2}m\langle v^2\rangle\right) = \dfrac{2}{3}N\langle KE\rangle

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