A conducting and closed container of capacity 100 liter contains an ideal gas at a high pressure. Now using a pump, the gas is taken out at a constant rate of 5 liter/sec. Find the time taken in which the pressure will decrease to P_initial/100?

Correct option is B. 92sec
The gas is taken out at the rate of 5 lit/sec. The volume of gas ejected in ‘dt’ time is d(vol) = 5 dt
Moles of gas ejected = nV(5dt)
PV = nRT ⇒ (dp)V = (dn)RT ⇒ (dp)V = −(nV5dt)RT ⇒ (dp)V = −5dtnRT
V = (−5dt)P ⇒ dp/p = −5/100dt = −1/20dt
∫dp/p = ∫−1/20dt
From p = pi to p = pf
∫pi^pf dp/p = ∫0^t −1/20dt
⇒ pf = pie^−t/20
⇒ pi/100 = pie^−t/20
ln(1/100) = −t/20
T = 20 ln 100 = 20 × 2 ln 10 = 92sec.