Chemistry - Ideal Gas Question with Solution | TestHub

ChemistryIdeal GasProblems related with container/piston/SPMHard2 minQB
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Match Column I with Column II (Pressure exerted by gas in figures) (Consider atmospheric pressure equal to 1 atm and density of mercury = 13.6 g/ml, if needed)

Column I

Column II

A) 1.25 atm

P)

B) 2.0 atm

Q)

C) 0.25 atm

R)

 

S)

Options:

Answer:
D
Solution:

(P) Open-end manometer, open arm 19 cm higher: Pgas = 76 + 19 = 95 cm Hg = 1.25 atm (A). (Q) Massless piston with 76 cm Hg column above: Pgas = 76 + 76 = 152 cm Hg = 2.0 atm (B). (R) Closed-end manometer with evacuated arm 19 cm higher: Pgas = 19 cm Hg = 0.25 atm (C). (S) Glycerine (ρ = 2.72 g/ml), open arm 0.95 m higher: equivalent Hg height = 95 × 2.72/13.6 = 19 cm, so Pgas = 76 + 19 = 95 cm Hg = 1.25 atm (A). Hence AP,S; BQ; CR.

Stream:JEESubject:ChemistryTopic:Ideal GasSubtopic:Problems related with container/piston/SPM
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