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A jar contains a gas and a few drops of ...

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 temperatures are 30 and 25 mm of Hg. Calculate the new pressure in jar.

A

`792 mm" of "Hg`

B

`817 mm" of "Hg`

C

`800 mm" of "Hg`

D

`840 mm" of "Hg`

Text Solution

Verified by Experts

The correct Answer is:
B
`P_("gas")= P_("dry gas")+P_("moisture")" at TK or "P_("dry")= 830-30= 800 mm Hg`
`T_(2)= 0.99T_(1)`, At constant volume `(P_1)/(T_1)=(P_2)/(T_2)`
`P_("dry")= (800xx 0.99T)/(T)= 792mm therefore P_("gas")= P_("dry")+P_("moisture")= 792+25= 817mm`.
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