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.

Answer 1

Here , 

Initial pressure = Atmospheric pressure + pressure due to mercury 

⇒ P_{1 =} P₀ + P_Hg

Let the CSA of the tube be A.

P₁ = 0.76 + 0.2 = 0.96 m Hg

T₁ = T₂ = T

V_{1 =} 0.43 A

If the tube is slanted , then the atmospheric pressure P₀ remains the same . only the P_Hg  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