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Question
A glass tube, sealed at both ends, is 100 cm long. It lies horizontally with the middle 10 cm containing mercury. The two ends of the tube contain air at 27°C and at a pressure 76 cm of mercury. The air column on one side is maintained at 0°C and the other side is maintained at 127°C. Calculate the length of the air column on the cooler side. Neglect the changes in the volume of mercury and of the glass.
Solution
Let CSA of the tube be A.
on the colder side :
P1 = 0.76 m Hg
T1 = 300K
V1 = V
T2 = 273K
V2 = Ax
`(P_1V)/T_1` = `(P_2Ax)/(T_2)`
⇒ `P_2= (P_1 VT_2)/(T_1Ax)`
on the hotter side :
P1= 0.76 m Hg
T1 = 300K
V1' = V
T2' = 400K
V2' = Ay
`(P_1'V)/T_1 = (P_2'Ay)/(T_2')`
⇒ `P_2' = (P_1VT_2')/(T_1Ay)`
In equilibrium , the pressures on both side will balance each other .
⇒ P2' = P2
⇒ `(P_1VT_2')/(T_1Ay) =( P_1VT_2)/(T_1Ax)`
⇒ `(T_2')/y = T_2/x`
From the length of the tube , we get
x + y + 0.1=1
⇒ y = 0.9-x
`400/((0.9-x)) = 273/x`
⇒ x = 0.365 m
⇒ x = 36.5 cm
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