A glass capillary tube sealed at both ends is 100cm long. It lies horizontally with the middle 10cm containing mercury. The two ends of the tube which are equal in length contain air at 27^(@)C at a pressure of 76cm of Hg. Now the air column at one end of the tube is kept at 0^(@)C and the other end is maintained at 127^(@)C. Calculate the length of the air column which is at 0^(@)C and its pressure. (Neglect the change in volume of Hg and glass).

A glass capillary tube sealed at both ends is 100cm long. It lies hori

1.	A glass capillary tube sealed at both ends is 100cm long. It lies horizontally with the middle 10cm containing mercury. The two ends of the tube which are equal in length contain air at 27^(@)C at a pressure of 76cm of Hg. Now the air column at one end of the tube is kept at 0^(@)C and the other end is maintained at 127^(@)C. Calculate the length of the air column which is at 0^(@)C and its pressure. (Neglect the change in volume of Hg and glass).
Answer» Solution :`2L+10=100," "i.e., L=45cm""...(1)` Now when end A is cooled while B is heated, the Hg COLUMN will be displaced towards A until the pressures on two sides of it again becomes equal, i.e., `P_(1)=P_(2)=P""...(2)` Now applying gas EQUATION to the air enclosed in side A, `(45AP_(0))/300=((45-x)AP_(1))/273,` i.e., `P_(1)=(273xx45xxP_(0))/(300(45-x))""...(3)` while for air on side B, `(45AP_(0))/300=((45+x)AP_(2))/400`, i.e., `P_(2)=(400xx45xxP_(0))/(300(45+x))""...(4)` SUBSTITUTING the values of `P_(1)andP_(2)` from equation (3) and (4) in (2), we get `(273xx45xxP_(0))/(300xx(45-x))=(400xx45xxP_(0))/(300xx(45+x))` Which on solving gives `x=(127xx45)//673=8.49cm` SO the length of air column which is at `0^(@)C=45-8.49=36.51` cm and from equation (3) with `P_(0)=76cm` of Hg (given), `P_(1)=(273xx45xx76)/(300xx36.51)=85.25cm` of Hg