Chemistry - Ideal Gas Question with Solution | TestHub
ChemistryIdeal GasProblems related with container/piston/SPMMedium2 minQB
ChemistryMediumsingle choice
Passage / Comprehension
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 expression for pressures of gas above mercury in the closed arm at any time, t?
Options:
Answer:
B
Solution:
Initially the open-arm mercury level is 38 cm above the closed-arm level, so the trapped gas pressure is P0 = 76 + 38 = 114 cm Hg = 1.5 atm. The leak law -dP/dt = kP integrates to P = P0 e-kt = 1.5 e-kt atm. Hence Ans (b).
Stream:JEESubject:ChemistryTopic:Ideal GasSubtopic:Problems related with container/piston/SPM
⏱ 2mℹ️ Source: QB
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