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 :

P_{1 }= 0.76 m Hg

T_{1} = 300K

V_{1} = V

T_{2 }= 273K

V_{2} = Ax

`(P_1V)/T_1` = `(P_2Ax)/(T_2)`

⇒ `P_2= (P_1 VT_2)/(T_1Ax)`

on the hotter side :

P_{1}= 0.76 m Hg

T_{1 }= 300K

V_{1}' = V

T_{2}' = 400K

V_{2}' = 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 .

⇒ P_{2}' = P_{2}

⇒ `(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