If the pressure in a gas container that is connected to an open-end u-tube manometer is 116 kpa and the pressure of the atmosphere at the open end of the tube is 752 mm hg, the level of mercury in the tube will a. be 870 mm higher in the arm open to the atmosphere. b. be 870 mm higher in the arm connected to the gas cylinder. c. be 118 mm higher in the arm connected to the gas cylinder. d. be 118 mm higher in the arm open to the atmosphere.

P = absolute pressure in gas container connected to open-end U-tube = 116 kPa = 116000 Pa

we know that, 1 atm = 101325 Pa   and  1 atm = 760 mm of Hg , hence

P = 116000 Pa  (1 atm/101325 Pa) (760 mm of Hg/1 atm)

P = (116000 x 760/101325)  mm of Hg

P = 870.07 mm of Hg

P₀ = atmospheric pressure = 752 mm Hg

Level of mercury is given as

P' = P - P₀

inserting the values

P' = 870.07 - 752

P' = 118 mm

so correct choice is

d.be 118 mm higher in the arm open to the atmosphere.

answer: he got 23

explanation:

a cold ice-tea.

pour in room temperature or cold water. cover the pitcher and refrigerate. steep white or green tea for 6 to 8 hours; steep black or oolong tea for 8 to 12 hours.jun 28, 2015