Chemistry - Ideal Gas Question with Solution | TestHub
A tube of U-shape partly filled by mercury is taken as shown in the figure. The cross-section area of closed arm is A cm2 and that of open arm is 2A cm2. A pin hole is now made at the top of the closed arm by which leakage of gas starts. The leak rate -dP/dt is directly proportional to the pressure of remaining gas inside the tube, over mercury. The proportionality constant is 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 = 1.5(1 - e-kt) atm = 114(1 - e-kt) cm Hg, which equals the change in the level difference between the arms. If the closed arm rises by x and the open arm falls by y, volume conservation gives Ax = 2Ay, i.e. y = x/2, and the level difference changes by x + y = 1.5x. So 1.5x = 114(1 - e-kt), giving x = 76(1 - e-kt) cm. Hence Ans (b).
