Determine the pressure of the gases trapped in the closed end of a manometer app

Question

Determine the pressure of the gases trapped in the closed end of a manometer apparatus if the water column which separates the closed end from the open end is 10.55 cm higher on the closed end than the open end, and if the barometric pressure at the time of the analysis is found to be 767.0 mm Hg. What will be the partial pressure of the gases trapped in the closed end, other than the water vapor? Assume the temperature is 25oC. You will need to refer to a table of water vapor pressures at various temperatures in order to complete this problem (e.g., CRC tables). [ Pall gases = 759.2 mmHg or torr Pnon-water gases = 735.4 mmHg or torr]

Explanation / Answer

at closed end
total pressure, P = pressure due to gases + pressure due to water vapor + pressure due to water column = atmospheric pressure
=> P = Pnon-water gases + Pwater vapor + gh = Patm

=density of water = 1000 kg/m^3

g = acceleration due to gravity = 9.8 m/s^2

h = height of water column = 10.55 cm = 0.1055 m

Patm = 767 mm of Hg = 767*133.3224 Pa = 102258.2808 Pa

Pwater vapor = water vapor pressure at 25 C = 3.2 kPa = 3200 Pa

putting these values in equation => Pnon-water gases + 3200 + 1000*9.8*0.1055 = 102258.2808

=> Pnon-water gases = 102258.2808 - 3200 - 1033.9 = 98018.3808 Pa = pressure of trapped gases Ans

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