A box of `1.00 m^(3)` is filled with nitrogen at `1.50` atm at 300 k. The box has a hole of an area 0.010 `mm^(2).` How much time is required for the pressure to reduce by 0.10 atm, if the pressure outside is 1 atm.

Explanation

Here ` V = 1.00 m^(3) , P_(i) = 1.50 -0.10 = 1.40 atm, P_(0)= 1atm`
let `n_(i)` = number of molecules//volume inside the box, initially
`n'_(i)` = number of molecules//volume inside the box, finally
`n_(0)` = number of molecules//volume outside the box.
If `upsilon_(1x)` is speed of a gas molecules along x direction, then as wall of box is very large compared to the hole in it.
`:. upsilon_(1x)^(2)+upsilon_(1y)^(2) + upsilon_(1z) = upsilon_(rms)^(2)`, which implies
`upsilon_(1x)^(2) = (upsilon_(rms)^(2))/(3)`
As, `1/2 m upsilon_(rms)^(2) = 3/2 k_(B)T :. upsilon_(1x) = sqrt((k_(B)T)/(m))`
Now, number of inside molecules per unit volume colliding with the wall in time `tau`
=collision frequency `xx tau`
`=1/2 An_(i) upsilon_(1x)*tau = 1/2 An_(i) sqrt((k_(B)T)/(m)) xx tau` ...(i)
The molecules colliding along the hole in the box move out. similarly, outside molecules colliding along the hole move in. As temperature inside and outside the box is the same, therefore, net flow of molecules out of the hole in time `tau`
`1/2 A(n_(i)-n_(0)) sqrt((k_(B)T)/(m)) xx tau`
After this time, number of molecules//volume inside the box change from `n_(i) to n'_(i)` and pressure inside change from `P_(i) to P'_(i)`
`:. ` Number of molecules escaping out of the hole in time `tau = (n_(i)-n_(i)^('))V` ...(ii)
From (i) and (ii) , `(n_(i)n_(i)) V = 1/2 A (n_(i)-n_(0))sqrt(k_BT)/(m) xx tau` ..(iii)
from `P= n k_(B)T, n = (P)/(k_(B)T)` , Putting in (iii), we get
`:. ((P_i)/(k_(B)T) - (P_(i)^('))/(k_(B)T)) V = 1/2 A [(P_i)/(k_(B)T) - (P_(0))/(k_(B)T)] sqrt((k_(B)T)/(m)) xx tau or tau = (2(P_(i)-P_(i)^(')))/((P_(i)-P_(0))) (V/A) sqrt((m)/(k_(B)T))`
Putting the given value, we get
`tau = 2 [(1.50 atm - 1.40 atm)/(1.50 atm - 1 atm)] ((1.00m^(3))/(0.01 xx 10^(-6)m^2)) ((28 xx 1.67 xx 10^(-27) kg)/(1.38 xx 10^(-23) JK xx 300 K))^(1//2)`
`tau = 1.34 xx 10^(5)s`.