Q34 of 100 Page 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.

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 P1=76cm of Hg=0.76m of Hg


Temperature of mercury column T1=27=300.15K


Temperature of air at cooler side T2=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 =P2


Pressure of hotter air =P’2


We know that ideal gas equation


PV=nRT


Where V= volume of gas


R=gas constant =8.3JK-1mol-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.


More from this chapter

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32

A container of volume 50 cc contains air (mean molecular weight = 28.8 g) and is open to atmosphere where the pressure is 100 kPa. The container is kept in a bath containing melting ice (0°C).

(a) Find the mass of the air in the container when thermal equilibrium is reached.


(b) The container is now placed in another bath containing boiling water (100°C).


Find the mass of air in the container.


(c) The container is now closed and placed in the melting-ice bath. Find the pressure of the air when thermal equilibrium is reached.


 33

A uniform tube closed at one end, contains a pellet of mercury 10 cm long. When the tube is kept vertically with the closed-end upward, the length of the air column trapped is 20 cm. Find the length of the air column trapped when the tube is inverted so that the closed-end goes down. Atmospheric pressure = 75 cm of mercury.

 35

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 remains constant.

 36

Figure shows a cylindrical tube of length 30 cm which is partitioned by a tight-fitting separator. The separator is very weakly conducting and can freely slide along the tube. Ideal gases are filled in the two parts of the vessel. In the beginning, the temperatures in the parts A and B are 400 K and 100 K respectively. The separator slides to a momentary equilibrium position shown in the figure. Find the final equilibrium position of the separator, reached after a long time.