Chemistry - Ideal Gas Question with Solution | TestHub
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Passage / Comprehension
A glass tube AD of uniform cross-section of length 56 cm contains two columns of air (assume ideal behaviour) AB and CD, separated by a column of mercury (BC) of length 8 cm. When the tube is horizontal, AB = 20 cm and CD = 28 cm. When the tube is held vertically with the end 'A' up, the mercury column moves down 10 cm.
What is the pressure of air in the air columns, when the tube was horizontal?
Options:
Answer:
C
Solution:
Horizontal: PAB = PCD = P₀. Vertical (A up): PAB' = 20P₀/30 = 2P₀/3; PCD' = 28P₀/18 = 1.5556P₀. Balance: PCD' = PAB' + 8 ⇒ 0.8889P₀ = 8 ⇒ P₀ = 9 cm Hg. Hence (c).
Stream:JEESubject:ChemistryTopic:Ideal GasSubtopic:Problems related with container/piston/SPM
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