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Question

A narrow glass tube of length 100 cm is closed at both ends. It lies horizontally with 20 cm of mercury column in the middle dividing the tube into two compartments I and II of equal length. The air in each compartment is at standard temperature and pressure. The tube is then turned to a vertical position. By what distance will the mercury column be displaced?


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Solution


P2P1=20(cm of Hg)----(1)

For chamber 1

P0V0=P1V1

P1=P0V0V1---(2)

For chamber 1I

P0V0=P2V2

P2=P0V0V2 ---(3)

We have P2P1=20 (cm of Hg)

P0V0V2P0V0V1=20 (cm of Hg)

(76 cm of Hg)A(40cm)A(40x)cm76(cm of Hg)A40cm4(40+x)cm=20cm

76×2(140x140+x)=1

76×2×2x=1600x2

x2+304x1600=0

x=5.17cm


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