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Question

A jar contains a gas and a few drops of water. The pressure in the jar is 830 mm of Hg. The temperature of the jar is reduced by 1%. The vapour pressure of water at two different temperatures is 30 and 25 mm Hg respectively. Calculate the new pressure in the jar.

A
792 mm of Hg
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B
817 mm of Hg
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C
800 mm of Hg
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D
840 mm of Hg
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Solution

The correct option is B 817 mm of Hg
Ptotal=Pgas+V.P=830 mm of Hg
Pgas=830V.P=83030=800
Since, V and n are constant for this gas, PT
P also reduces by 1%
Pgas=99100×800=792 mm
V.P @ lower T is 25 mm
Ptotal=792+25=817 mm

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