Skip to content
NCERT Exemplar · Q30

Q.A box of 1.00m3^3 is filled with nitrogen at 1.50 atm at 300K. The box has a hole of an area 0.010 mm2^2. How much time is required for the pressure to reduce by 0.10 atm, if the pressure outside is 1 atm.

Punjab PsebLong· 5mImportance★★★★★est
98% · 49/50 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Gas escapes through the tiny hole by effusion, but molecules also enter from outside, so the net leak rate depends on the pressure difference (P−Pout)(P-P_{\text{out}}), not on PP alone. Solving dPdt=−vˉA4V(P−Pout)\dfrac{dP}{dt}=-\dfrac{\bar v A}{4V}(P-P_{\text{out}}) from 1.501.50 atm to 1.401.40 atm gives t≈1.9×105 s≈52 hourst\approx1.9\times10^{5}\ \text{s}\approx\boxed{52\ \text{hours}}.

Setting up the effusion equation

Because the hole is far smaller than the box, gas leaves molecule-by-molecule -- effusion. Kinetic theory says the number of molecules striking (and escaping through) unit area per second is 14nvˉ\tfrac14 n\bar v, where vˉ=8RTπM\bar v=\sqrt{\dfrac{8RT}{\pi M}}. But the outside atmosphere also has molecules entering through the same hole, at rate 14noutvˉ\tfrac14 n_{\text{out}}\bar v. So the net escape rate depends on the density (pressure) difference -- the process stops when pressures equalise, not when the box empties.

Deriving the pressure decay law

dNdt=−14vˉA (n−nout)\frac{dN}{dt} = -\frac14\bar v A\,(n - n_{\text{out}})

Using N=PVkBTN = \dfrac{PV}{k_BT} and n=PkBTn = \dfrac{P}{k_BT}, the kBTk_BT cancels:

dPdt=−vˉA4V (P−Pout)\frac{dP}{dt} = -\frac{\bar v A}{4V}\,(P - P_{\text{out}})

This is exactly the form of Newton's law of cooling.

Step-by-step calculation

1. Mean speed of N2_2 at T=300 KT=300\ \text{K} (M=28.0×10−3 kg/molM = 28.0\times10^{-3}\ \text{kg/mol}):

vˉ=8RTπM=8(8.314)(300)π(0.028)≈476 m/s\bar v = \sqrt{\frac{8RT}{\pi M}} = \sqrt{\frac{8(8.314)(300)}{\pi(0.028)}} \approx 476\ \text{m/s}

2. Rate constant. With A=0.010 mm2=1.0×10−8 m2A = 0.010\ \text{mm}^2 = 1.0\times10^{-8}\ \text{m}^2, V=1.00 m3V=1.00\ \text{m}^3:

k=vˉA4V=(476)(1.0×10−8)4(1.00)≈1.19×10−6 s−1k = \frac{\bar v A}{4V} = \frac{(476)(1.0\times10^{-8})}{4(1.00)} \approx 1.19\times10^{-6}\ \text{s}^{-1} …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.