Chemistry - Ideal Gas Question with Solution | TestHub
A U-shaped tube partly filled with mercury is taken as shown in the figure. The cross-section area of the closed arm is A cm² and that of the open arm is 3A cm²; initially the mercury level in the open arm stands 76 cm higher than in the closed arm. A pin hole is now made at the top of the closed arm and the trapped gas starts leaking out. The leak rate -dP/dt is directly proportional to the pressure of the remaining gas above mercury, the proportionality constant being K s⁻¹.
What will be the increase in level of mercury in the closed arm at any time, t?
Options:
Answer:
Solution:
The pressure drop at time t is Δ P = 2(1 - e-kt) atm = 152(1 - e-kt) cm Hg, equal to the change in level difference. If the closed arm rises by x and the open arm falls by y, volume conservation gives Ax = 3Ay, i.e. y = x/3, and the difference changes by x + y = 4x/3. So 4x/3 = 152(1 - e-kt), giving x = 114(1 - e-kt) cm. Hence Ans (c).
