A glass capillary tube sealed at both ends is 100cm long. It lies horizontally with the middle 10 cm containing mercury. The two ends of the tube which are equal in length contain air at 27^(0) C at a pressure of 76 cm of Hg. Now the air column at one end of the tube is kept at 0^(0)C and the other end is maintained at 127^(0)C. Calculate the lellgth of the air column which is at 0^(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 10 cm containing mercury. The two ends of the tube which are equal in length contain air at 27^(0) C at a pressure of 76 cm of Hg. Now the air column at one end of the tube is kept at 0^(0)C and the other end is maintained at 127^(0)C. Calculate the lellgth of the air column which is at 0^(0)C and its pressure. (Neglect the change in volume of Hg and glass).
Answer» Solution : 2L + 10 = 100, i.e., L = 45 cm........ (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., Now applying gas equation to the AIR enclosed in side A, `(45 AP_(0))/(300)= ((45-x)AP_(1))/(273) , " i.e., " P_(1) = (273 xx 45 xx P_(0))/(300 (45 - x)) ` .... (3) while for air on side B, `(45 AP_(0))/(300 ) = ((45 + x) AP_(2))/(400)` , i.e., `P_(2) = (400 xx 45 xx P_(0))/(300 (45 + x)) ` ....... (4) Substituting the values of `P_(1) and P_(2)`from equation (3) and (4) in (2), we get `(273 xx 45 xx P_(0))/(300 xx (45 - x)) = (400 xx 45 xx P_(0))/(300 xx (45 + x))` Which on solving gives x = `(127 xx 45 ) // 673 ` = 8.49 cm So the length of air column which is at `0^(0)`C = 45 - 8.49 = 36.51 cm and from equation (3) with `P_(0)`= 76 cm of Hg (given), `p_(1) = (273 xx 45 xx 76)/(300 xx 36.51 )` = 85.25 cm of Hg