Q. 34

# 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 of the cooler side. Neglect the changes in the volume of mercury and of the glass.

Answer :

Let the curved surface area of tube A

Given

Length of mercury column=10cm=0.1m

Length of tube =100cm=1m

Pressure of mercury column P_{1}=76cm of Hg=0.76m of Hg

Temperature of mercury column T_{1}=27=300.15K

Temperature of air at cooler side T_{2}=0=273.15K

Temperature of air at hotter side T’_{2}=127=400.15K

Let the length of air column at cooler side be

The length of air column at hotter end be

Volume of cooler air =A

Volume of hotter air =A

Volume of mercury column = V

Pressure of cooler air =P_{2}

Pressure of hotter air =P’_{2}

We know that ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant =8.3JK^{-1}mol^{-1}

T=temperature

n=number of moles of gas

P=pressure of gas.

Applying ideal gas equation between cooler air and mercury column

Applying ideal gas equation between hotter air and mercury column

Under equilibrium condition the pressure P2 and P’2 will be same

Now length of entire tube

x+y+0.1=1

y=0.9-x

Substituting the value of y in equation (i)

(0.9-x)273.15=400.15x

So, the length of air column on the cooler side is 0.365m=36.5cm.

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