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ChemistryIdeal GasIdeal Gas Equation & problemsHard2 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) 1.50 atm

P)

B) 1.75 atm

Q)

C) 0.50 atm

R)

 

S)

Options:

Answer:
C
Solution:

(P) Open-end manometer, open arm 38 cm higher: Pgas = 76 + 38 = 114 cm Hg = 1.50 atm (A). (Q) Massless piston with 57 cm Hg column above: Pgas = 76 + 57 = 133 cm Hg = 1.75 atm (B). (R) Closed-end manometer with evacuated arm 38 cm higher: Pgas = 38 cm Hg = 0.50 atm (C). (S) Glycerine (ρ = 2.72 g/ml), open arm 1.9 m higher: equivalent Hg height = 190 × 2.72/13.6 = 38 cm, so Pgas = 76 + 38 = 114 cm Hg = 1.50 atm (A). Hence A→P,S; B→Q; C→R.

Stream:JEESubject:ChemistryTopic:Ideal GasSubtopic:Ideal Gas Equation & problems
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