Question

A box of 1.00 m³ is filled with nitrogen at 1.50 atm at 300K. 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

The volume of the box `V_1 = 1  m^3`

Initial pressure `P_1 = 1.5  atm`

Final pressure `P_2^' = 1.5 - 0.1 = 1.4  atm`

Air pressure outside box `P_2 = 1  atm`

Initial temperature `T_1 = 300  K`

Final temperature  `T_2 = 300  K`

a = area of hole = 0.01  mm²

= `0.01 xx 10^-6m^2`

= `10^-8m^2`

The initial pressure difference between atmosphere and tyre 

`ΔP = (1.5 - 1)  atm`

Mass of N₂ gas molecule = `(0.028  Kg)/(6.023 xx 10^23)`

= `46.5 xx 10^-27  Kg`

`K_B = 1.38 xx 10^-23`

Assuming `ρ_(i n)` be the initial number of N₂ gas molecules per unit volume at time Δt and also `v_(ix)` be the speed of molecules along the x-axis

At time Δt, the number of molecules colliding to the opposite wall

`1/2 ρ_(i n) [(v_(ix)) Δt]A`

Half is multiple as half molecule will strike the opposite wall

`v_(rms)^2 (N_2  "molecules") = v_(ix)^2 + v_(iy)^2 + v_(iz)^2`

∴ `|v_(ix)| = |v_(iy)| = |v_(iz)|`

Thus, `v_(rms)^2 = 3v_(ix)^2`

K.E. of gas molecule = `3/2 K_BT`

`1/2 mv_(rms)^2 = 3/2 K_BT`

`m3v_(ix)^2 = 3K_BT`

`v_(ix) = sqrt((K_BT)/m`  .....(A)

At time Δt, the number of N₂ gas molecule striking to a wall outward = `1/2 ρ_(i n) sqrt((K_BT)/m) Δt  * a`

The temperature inside the air and box are equal to T

At time Δt, the number of air molecules striking to hole inward = `1/2 ρ_(n2) sqrt((K_BT)/m) Δt  * a` 

Total number of molecules going out from the hole at a time Δt

= `1/2 [ρ_(n1) - ρ_(n2)] sqrt((K_BT)/m) * Δt * a(I)`

Gas equation

`P_1V = μRT`

⇒ `μ = (P_1V)/(RT)`

For box, `μ/V = P_1/(RT)` where μ = number of moles of gas in box

`ρ_(n1) = (N ("Total no. of molecule in box"))/("Volume of box") = (μNA)/V`

 = (P_1N_A)/(RT)` per unit volume

Assuming after time T pressure reduced by 0.1 and becomes `(1.5 - 0.1) = 1.4  atm  P_2^'`

Thus, the new final density of N_A molecule `ρ_(n1)^'`

`ρ_(n1)^' = (P_2N_A)/(RT)` per unit volume (III)

Thus, the total number of molecules going out from volume V

= `(ρ_(n1) - ρ_(n1)^')v`

= `(P_1N_A)/(RT)v - (P^'2N_A)/(RT)v`

= `(N_Av)/(RT) [P_1 - P_2^']`  (IV) (From II, III)

`P_2^'` = Net number of molecules going out in time τ from the hole from (I)

= `1/2 [ρ_(n1) - ρ_(n2)] sqrt((K_BT)/m) τ * a`

`ρ_(n1) - ρ_(n2) = (P_1N_A)/(RT) - (P_2N_A)/(RT)`

∴ `ρ_(n1) - ρ_(n2) = N_A/(RT) [P_1 - P_2]`   .....(P₂ = Press of air out of box)

In τ time the total number of molecules going out from above

= `1/2 N_a/(RT) [P_1 - P_2] sqrt((K_BT)/m) * τ * a`

From (V) and (IV)

`(N_AV)/(RT) (P_1 - P_2^') = 1/2 N_A/(RT) (P_1 - P_2) sqrt((K_BT)/m) * τ * a`

τ = `(N_AV)/(RT) (P_1 - P_2^') (2RT)/N_A 1/((P_1 - P_2)) sqrt(m/(K_BT)) * 1/a`

τ = `(2(P_1 - P_2^'))/((P_1 - P_2)) * V/a sqrt(m/(K_AT))`

= `(2[1.5 - 1.4])/((1.5 - 1)) 1/10^-8 sqrt((46.5 xx 10^-27)/(1.38 xx 10^-23 xx 300))`

= `(2 xx 0.1)/(0.5 xx 10^-8) sqrt((4650 xx 10^(-27+23-2))/(138 xx 3))`

= `0.4 xx 10^+8 sqrt((775 xx 10^-6)/69)`

= `0.4 xx 10^+8 xx 10^-3 x sqrt(11.23)`

= `0.4 xx 10^5 xx 3.35`

τ = `1.34 xx 10^5` sec