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

Given volume of the box `V =1 .00m^(3)`
Area `= a = 0.0010mm^(2)`
`=8.01 xx 10^(-6) m^(2)`
`=10^(-8) m^(2)`

Temperature outside = Temperature inside
initial pressure inside the box =1.50 atm.
Final pressure inside the box =0.10 atm. Assuming
`V_(ix) =` Speed of nitrogen molecule inside the box along x- direction `n_(i)` = Number of molecules per unit volume in a time interval of of `DeltaT.` all the particles at a distance `(V_(ix) DeltaT)` will collide the hole and the wall the particle colliding along the hole will escape out reducing the pressure in the box.
Let area of the wall number of particles colliding in time
`Deltat=1/2 n_(i)(V_(ix)Deltat) A`
`1/2` is the factor because all the particles along x- direction are behaving randomly . Hence half of these are colliding against the walls on either side.
`"Inside the box "" " V_(ix)^(2) +V_(ix)^(2) +V_(iz)^(2) =V_("rms")^(2)`
`:. " "V_(ix)^(2) =(V_("rms")^(2))/(3)" "(V_(ix) =V_(iy)=V_(iz))`
`"or "" "1/2 mv_("rms")^(2) =3/2 k_(B) T " "[underset(T= "Temperature")underset(K_(B) ="Boltzmann constant")(underset(V_("rms") ="Root mean squre velocity"]`
`rArr " " V_("rms")^(2) (3k_(B)T)/(m)`
`rArr " " V_("rms") =sqrt((3K_(B)T)/(m))`
[According to kinetic theory of gases]
`"Nos"" " v_(ix)^(2) =(V_("rms")^(2))/(3) =1/3 xx (3K_(B) T)/(m)`
`"or "" " v_(ix)^(2) =(K_(B)T)/(m)`
`:.` Number of particles colliding in time
`Deltat=1/2 n_(i) sqrt((K_(B) T)/(m)) DeltatA`
If particles collide along hole, they move out. Simailrly, outer particles colliding along hole will move in.
If a = area of hole
Then , net particle flow in time `Deltat =1/2 (n_(1)- n_(2)) sqrt((k_(B) T)/(m)) DeltatA`
[Temperature inside and outside the box are equal]
`pV =muRT rArr (pV)/(RT)`
Let n=number density of nitrogen `=(nuN_(A))/(V) = (pN_(A))/(RT)" "[:. (mu)/(V) =(p)/(RT)]`
Let `N_(A)=` Avogardro's number
If after time `tau` pressure inside changes from p to `p_(1)^(1)`
`:. " "n_(1) =(vN_(A))/(RT)`
Now number of molecules gone out `=n_(1)V-n_(1)V`
`=1/2 (n_(1) -n_(2)) sqrt((k_(B)T)/(m)) taua`
`:. " "(p_(1)N_(A))/(RT) V- (vN_(A))/(RT) V =1/2 (P_(1) -P_(2)) (N_(A))/(RT) sqrt((K_(B)T)/(m)) taua`
`"or "" "(p_(1)N_(A))/(RT) V-(vN_(A))/(RT) V=1/2 (p_(1) -p_(2)) (N_(A))/(RT) sqrt((k_(B)T)/(m)) taua`
`:. " "tau =2 ((p_(1)-pv_(1))/(p_(1)-p_(2))) v/a sqrt((m)/(k_(B)T))`
Putting hte values from the data given,
`tau =2 ((1.5-1.4)/(1.5 -1.0)) (1xx1.00)/(0.01xx10^(-6))sqrt((46.7xx10^(-27))/(1.38xx10^(-23)xx300))`
`=((0.1)/(0.5)) (1)/(10^(-8)) sqrt((4.7)/(1.38xx 3)) xx 10^(-6)`
`=2((1)/(5)) 1xx 10^(8) xx 10^(-3) xx sqrt((46.6)/(4.14)) =2/5 xx 10^(5) sqrt((45.7)/(4.14))`
`=2/5 xx 10^(5) sqrt(11.28)`
`=2/5 xx 3.358 xx 10^(5) =(6.717)/(5) xx 10^(5) =1.343 xx 10^(5) s`