Q33 of 100 Page 34

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.

Let the curved surface area of tube be A.


Volume =areaheight


Given


Initial length of trapped air=20cm =0.2m


Length of mercury column=10cm=0.1m


So, Mercury column pressure =0.1g Pa


Initial volume of air trapped V1=0.2A


Atmospheric pressure =75cm of Hg=0.75m of Hg


1mm of Hg = hg Pa


Where h= height of mercury column =1mm=0.001m


= density of mercury


g= acceleration dur to gravity


So, atmospheric pressure= 0.75m of Hg= 0.75g Pa


Let the pressure of the trapped air when closed end of the tube is upward be P1. Now, pressure of the mercury and trapped air will then be equal to atmospheric pressure.


P1+0.1g=0.75g


P1=0.65g


When the tube is inverted such that closed end is downward then pressure of trapped air will be P2.


P2=atmospheric pressure + mercury column pressure


P2=0.75g+0.1g=0.85g


Let length of air column at P2=x


Volume of air trapped will be V2=A x


Now temperature in both cases will remain same, as no heat is being either added or abstracted from tube.


Applying boyle’s law,


P1V1=P2V2


0.65g0.2A=0.85gAx



The length of the air column trapped when the tube is inverted so that the closed-end goes down is 0.15m=15cm.


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31

Figure shows two vessels A and B with rigid walls containing ideal gases. The pressure, temperature and the volume are pA. TA, V in the vessel A and pB, TB, V in the vessel B. The vessels are now connected through a small tube. Show that the pressure p and the temperature T satisfy.


When equilibrium is achieved.



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.


 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.

 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.