A glass tube sealed at both ends is 100cm long, It lies horizontally with the middle 10cm containing mercury. The two ends of the tube contain air at `27^(@)C`and at a pressure 76cm 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 on the cooler side. neglect the changes in the volume of mercury and of the glass.

Explanation

Here 2L + 10 = 100 cm
` rArr 2L = 90 cm `
` rArr L = 45 cm `
Applying combined gas equation to part 1
of the tube,
` ((45A)P_0/ 300) = ((45 - x) AP_1/ 273)`
` rArr (P_1)=(273 xx 45 xx (P_0)/ 300 (45 -x))`
Applying combined gas equation to part 2
of the tube,
` ((45A)P_0/ 300) = ((45 + x) AP_2/ 400)`
` rArr (P_2) = (400 xx 45 xx (P_0)/ 300(45 + x ))`
` (P_1)= (P_2)`
` rArr (273 xx 45 xx (P_0)/ 300 (45 - x)) = (400 xx 45 xx (P_0)/ 300(45 +x)`
` rArr (45 xx 273 / 45- x ) = (400 xx 45 / 45 + x)`
` rArr (45 - x )400 = (45 +x) 273 `
` rArr (18000 - 400x) = (12285 + 273 x )`
` rArr (400 + 273) x = 18000-12285`
` rArr x = 8.49`
` P_1 = (273 xx 45 xx 76/ 300 xx 36.51) `
` = 85.25 cm of Hg `
` Length of air column on the air cooler side
` = L-x`
` = 45 - 8.49 = 36.51 cm .`