4. -14 points My Notes Ask Your Teacher The liquid in the open-tube manometer in

Question

4. -14 points My Notes Ask Your Teacher The liquid in the open-tube manometer in the figure below is mercury, y1 = 3.75 cm, and y2 = 6.25 cm. Atmospheric pressure is 968 millibars. Open-tube manometer Po -Patm Pressure 2 yi The pressure is the same at the bottoms of the two tubes. (a) What is the absolute pressure at the bottom of the U-shaped tube? Pa (b) What is the absolute pressure in the open tube at a depth of 2.50 cm below the free surface? PSA (c) What is the absolute pressure of the gas in the tank? Pa (d) What is the gauge pressure of the gas in pascals? Pa Submit Answer Save Progress Practice Another Version

Explanation / Answer

The liquid in the open-tube manometer is mercury aty1= 3.75 cm = 3.75 * 10-2 m and y2= 6.25 cm = 6.25 * 10-2mThe atmospheric pressure is P = 968 millibars = 968 * 10^2 Pa

a)The absolute pressure at the bottom of the U-shaped tubeis

P = Po + ? * g * y2 or Po= P - ? * g * y2

Po= 968 * 10^2 - 13.6 * 10^3* 9.8 * 6.25 * 10-2

Po= 88470 Pa

b)The absolute pressure in the open tube at a depth of 2.50 cmbelow the free surface is

Po= P - ? * g * y Here,y = 2.50 * 10-2 m

Po= 968 * 10^2 - 13.6 * 10^3* 9.8 * 2.5 * 10-2

Po= 93468 Pa

c)The absolute pressure of the gas in the tank is

Po= ? * g * y Here,? is the density of air.

Po=1.225 * 9.8 * 6.25 * 10-2

Po= 0.75 Pa

d)The gauge pressure of the gas in pascals is

P = ? * g * (y2 - y1)

P = 1.225 * 9.8 * (6.25 - 3.75) * 10^-2

P = 0.3 Pa

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