A 100cm long glass capillary tube closed at both ends, has a mercury thread of length 10 cm.When the tube is horizontal the mercury thread stays at the middle of the tube with air columns of equal length on either side,at 76 cmHg pressure and 27^@C. Now the temperature of one side is changed to 0^@C,and of the other side to 127^@C. Find the length and the pressure of the air column kept at 0^@C. Neglect expansions of glass and mercury.

A 100cm long glass capillary tube closed at both ends, has a mercury t

1.	A 100cm long glass capillary tube closed at both ends, has a mercury thread of length 10 cm.When the tube is horizontal the mercury thread stays at the middle of the tube with air columns of equal length on either side,at 76 cmHg pressure and 27^@C. Now the temperature of one side is changed to 0^@C,and of the other side to 127^@C. Find the length and the pressure of the air column kept at 0^@C. Neglect expansions of glass and mercury.
Answer» Solution :Let the area of cross section of the capillary be `a cm^2` [Fig.6.8]. In the FIRST case,volume of air on either side of the mercury thread in the tube,`V_1=45a cm^3`,temperature `T_1=273+27=300K`,pressure of air `p_1=76cmHg`. In the second case let the left side of the Hg thread inside the tube be at `0^@C` or 273 K and the other side be at `127^@C` or 400 K.Let the pressure of the confined air on each side =PCMHG. Let L=LENGTH ofthe air column at `0^@C`,(90-l)=length of the other air column at `127^@C`. Using equation of state, `(p_1V_1)/T_1=(p_2V_2)/T_2.(76times45a)/300=(pla)/273`........(1) and `(76times45a)/300=(p(90-l)a)/400`........(2) From (1) and (2) we get, `(ptimes l times a)/273=(p(90-l)a)/400` or,`400l=273(90-l)or,673l=273times90` `thereforel=(90times273)/673=36.5cm` From equation (1) `(76times45a)/300=(ptimes36.5a)/273or,p=85.3cmHg` `therefore` The length of the air column at `0^@C` is 36.5 cn=m and its pressure is 85.3 cmHg.