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A box of 1.00m³ 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.

Answers (1)

Let v1 = volume of box = 1 m³

Let p1 be the initial pressure = 1.5 atm

Let p2’ be the final pressure =1.5 - 0.1 = 1.4 atm

t1 = initial temperature = 300K, t2 = final temperature = 300 K

let a be the area of hole = $10 ^{-8} m^{2}$

The pressure difference initially between the tyre and atmosphere, ΔP = (1.5 -1) = 0.5 atm

Mass of N2 molecule

$=\frac{0.028}{6.023 \times 10^{23}} = 46.5 \times 10^{-27} kg$

$Kb = 1.38 \times 10^{-23}$

Now since

$|v_{ix}|= | v_{iy} |= \left |v_{iz} \right |$

Hence, $v^2_{rms} = 3v_{ix}^2$

For a gas molecule, Kinetic energy = $\frac{3}{2}KbT$

Now,

$\frac{1}{2}mv^{2} = 3v_{ix}^{2}$

$v_{ix }= \sqrt{\frac{Kb T}{m}}$ (A)

The N2 molecules striking the wall in time $\Delta t$ outwards

$\frac{1}{2} P_{n1} \sqrt{\frac{Kb T}{m}}$ (A) outwards.

The temperature of the air and inside the box is equal to T.

The N2 molecules striking the wall in time $\Delta t$ inwards = $\frac{1}{2} P_{n2} \sqrt{\frac{Kb T}{m}}$ (A) inwards.

Here, Pn2 = density of air molecules

So, we can calculate the net number of molecules going outward as:

$\frac{1}{2} \left [ P_{n1}-P_{n2} \right ] \sqrt{\frac{K_b T}{m}}(A)\Delta t$

We know that, $PV = nRT, n= \frac{PV}{RT}$

Pn1 = total number of molecules in the box/ volume of the box

$=\frac{nNa}{V}= \frac{PNa}{RT}$ per unit volume.

After a time, T, the pressure reduces to (1.5-0.1) = 1.4 atm = P’2

The new density of the Na molecules $=P^{'}n1 = \frac{P2^{'}Na}{RT}$ per unit volume.

The number of molecules going out from volume $V = (P1 - P^{'}1) v = \frac{P1Na}{RT} (V) -\frac{ P2Na}{RT}(V)$

= $\frac{Na V}{RT} [P1 - P^{'}2]$ where P’2 = final pressure of the box

The total number of molecules which exit the hole in time t’ is:

$\frac{1}{2}[Pn1 - Pn2] \sqrt{\frac{Kb T}{m t^{'}}} (A)$

$Pn1 - Pn2 = \frac{P1Na}{RT} -\frac{ P2Na}{RT}= \frac{Na}{RT} [P1-P2]$

The net number of molecules exiting in time t' is

$\frac{Na}{2RT} [P1 - P2] \sqrt{\frac{Kb T}{m t^{'}}} (A)$

Now we can write from the above results,

$t^{'} = 2\frac{(P_1 - P^{'}_2)}{(P_1-P_2)}. \left ( \frac{V}{a} \right ).\sqrt{ \frac{m}{Kb T}}$

$t^{'} = \frac{2[1.5-1.4]}{ [1.5-1]}.\frac{1}{10^{-8}}. \sqrt{\frac{46.5 \times 10^{-27}}{1.38 \times 10^{-23}\times 300} }$

$t^{'}= 0.4 \times 10^{8}\sqrt{ \frac{775 \times 10^{-6}}{69}}$

$t^{'} = 0.4 \times 10^{5} \times 3.35 = 1.34 \times 10^{5}$ seconds.

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