Question

A uniform capillary tube closed at one end contained dry air trapped by a thread of mercury 8.5×10^-2m long.When the tube was held horizontally the length of the air column was 5×10^-2m,when it was held vertically with the closed end downwards the length was 4.5×10^-2m.Determine the value of the atmospheric pressure.g=10m/s, density of mercury 1.36×10^4kg/m^-3

Answer 1

find the weight of the column of Hg (density*height*area).
pressure then from that pressure at base of column=density*height+1atm

so, the air trapped underneight it has the same pressure. Find the moles in the column vertically
n=PV/RT where v=length(area). Leave Area as a cunit, it will divide away soon.

now horizontally, same number of moles, but Pressure is now just 1 atm. What is the volume: Vnew=lengthnew*area

V=Length*area=nRT/1atm= you do it, solve for length

Answer 2

Do it again