Question

A box of 1.00 m³ is filled with nitrogen at 1.50 atm at 300 K. the box has a hole of an area 0.010 mm² . How much time is required for the pressure to reduce by 0.10 atm, if the pressure outside is 1 atm.

Answer 1

V_ix = speed of molecule inside the box along x direction.

n₁ = number of molecules per unit volume.

In time Δt, particles moving along the wall will collide if they are within (V_ixΔt) distance. Let a = area of the wall.

No, of particles colliding in time Δt = \(\frac{1}{2}\)n_i(V_ixΔt)a (factor of 1/2 due to motion towards wall).

In general, gas Is In equilibrium as the wall is very large as compared to hole.

∴ \(V_1^2x+V_1^2y+V_1^2z=v^2_{rms}\)

∴ \(v_1^2x=\frac{V^2_{rms}}{3}\)

\(\frac{1}{2}mV^2_{rms}=\frac{3}{2}kT\) 

⇒ \(V^2_{rms}=\frac{3kT}{m}\)

∴ \(V^2_1x=\frac{kT}{m}\)

∴ No. of particles colliding in time Δt =\(\frac{1}{2}n_1\sqrt\frac{kT}{m}Δt\,a\).if particles collide along hole, they move out. Similarly outer particles colliding along hole will move in.

∴ Net particle flow in time Δt =\(\frac{1}{2}(n_1-n_2)\sqrt\frac{kT}{m}Δt\,a\).as temperature is same in and out.

pV = uRT 

⇒ μ = \(\frac{PV}{RT}\)

n = \(\frac{μN_A}{V}\)

= \(\frac{PN_A}{RT}\)

After some time τ pressure changes to \(p'_1\)inside

∴ \(n'_1\) = \(\frac{p'_1N_A}{RT}\)

\(n_1\)V − \(n'_1\)V = no. of particle gone out

= \(\frac{1}{2}(n_1-n_2)\sqrt\frac{kT}{m}\tau a\)

∴ \(\frac{P_1N_A}{RT}V-\frac{P'_1N_A}{RT}V\) 

= \(\frac{1}{2}(P_1-P_2)\frac{N_A}{RT}\sqrt\frac{kT}{m}a\)

∴ \(\tau=2(\frac{P_1-P'_1}{P_1-P_2})\frac{V}{a}\sqrt\frac{m}{kT}\)

= \(\tau=2(\frac{1.5-1.4}{1.5-1.0})\frac{5\times1.00}{0.01\times10^{-6}}\sqrt\frac{46.7\times10^{-27}}{1.38\times10^{-23}\times300}\)