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III. Long Answer Questions · Q2

Q.Derive the expression of pressure exerted by the gas on the walls of the container.

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Step 1. Consider NN molecules of mass mm in a cubical container of side ll. A molecule with velocity components (vx,vy,vz)(v_x,v_y,v_z) strikes the right-hand wall; since the collision is elastic, it rebounds with −vx-v_x while vy,vzv_y,v_z stay unchanged.

Step 2. The molecule's momentum change is −2mvx-2mv_x, so by conservation of momentum the wall gains momentum 2mvx2mv_x per collision.

Step 3. In time Δt\Delta t, only molecules within a slab of volume AvxΔtAv_x\Delta t next to the wall, moving toward it, can strike it; with random motion, half of these are moving the right way, giving n2AvxΔt\tfrac{n}{2}Av_x\Delta t collisions in time Δt\Delta t (n = number density).

Step 4. Total momentum transferred: Δp=n2AvxΔt×2mvx=Anmvx2Δt\Delta p=\tfrac{n}{2}Av_x\Delta t\times2mv_x=Anmv_x^2\Delta t.

Step 5. Force: F=Δp/Δt=nmAvx2F=\Delta p/\Delta t=nmAv_x^2; pressure: P=F/A=nmvx2P=F/A=nmv_x^2.

Step 6. Averaging over all molecules and using isotropy (vx2‾=13v2‾\overline{v_x^2}=\tfrac13\overline{v^2}, since v2‾=vx2‾+vy2‾+vz2‾=3vx2‾\overline{v^2}=\overline{v_x^2}+\overline{v_y^2}+\overline{v_z^2}=3\overline{v_x^2}) gives the final result P=13nmv2‾=13NVmv2‾P=\tfrac13nm\overline{v^2}=\tfrac13\tfrac{N}{V}m\overline{v^2}.

Step 7. Although a cubical container was used purely for calculational convenience, the wall area cancels out of the final formula, so the result holds for a container of any shape.

✓Final answer

P=13NVmv2‾P=\dfrac13\dfrac{N}{V}m\overline{v^2}.

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