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Question
An ideal gas is trapped between a mercury column and the closed-end of a narrow vertical tube of uniform base containing the column. The upper end of the tube is open to the atmosphere. The atmospheric pressure equals 76 cm of mercury. The lengths of the mercury column and the trapped air column are 20 cm and 43 cm respectively. What will be the length of the air column when the tube is tilted slowly in a vertical plane through an angle of 60°? Assume the temperature to remain constant.
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Solution
Here ,
Initial pressure = Atmospheric pressure + pressure due to mercury
⇒ P1 = P0 + PHg
Let the CSA of the tube be A.
P1 = 0.76 + 0.2 = 0.96 m Hg
T1 = T2 = T
V1 = 0.43 A
If the tube is slanted , then the atmospheric pressure P0 remains the same . only the PHg changes
`P_2 = P_0 + P_Hg cos60^circ` = 0.76 + 0.2 × 0.5 = 0.86
`P_1V_1 = P_2V_2`
⇒ `V_2 = (P_1V_1)/P_2 = (0.96×0.43A)/0.86`
Let the length of the air column be l.
⇒ Al = `(P_1V_1)/P_2 = (0.96×0.43A)/0.86`
⇒ l = 0.48 m
⇒ l = 48 cm
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