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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.
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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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