Answer:
(a) Absolute pressure at the bottom of the U-shaped tube is: 107339 Pa, (b) The absolute pressure in the open tube at a depth of 4.00 cm below the free surface is: 103337 Pa, (c) The absolute pressure of the gas in the container is: 103337 Pa and (d) The gauge pressure of the gas in pascals is: 5337 Pa.
Explanation:
We need to apply the Pascals' law ([tex]P=p*g*h[/tex]), where P is pressure, p is density, g is the gravity and h the height of the column, to determine the pressure at differents points in the open tube manometer. First we need to know the mercury density as: 13600 (Kg/m^3). Now remember that the absolute pressure is related with manometric pressure and atmospheric pressure like: [tex]P_{abs}=P_{gauge} +P_{atm}[/tex], so we can find the absolute pressure at the bottom of the U-shaped as: [tex]P_{abs(at the bottom)} = 13600*9.81*0.07+98000=103337(Pa)[/tex] (a). Using the same equation we can get the absolute pressure in the open tube at a depth of 4.00 cm as: [tex]P_{abs(4.00cm)} =13600*9.81*0.04+98000=103337(Pa)[/tex](b). Then to get the absolute pressure of the gas in the container we need to use the Pascal's law as:[tex]P_{abs(gas)}=13600*9.81*(0.07-0.03)+98000=103337(Pa)[/tex] (c). Finally to find the gauge pressure of the gas in pascals we solve for Pgauge as:[tex]P_{gauge}=P_{abs} -P_{atm}[/tex] so replacing the values we get:[tex]P_{gauge}=P_{abs(gas)}-P_{atm}=103337-98000=5337(Pa)[/tex] (d).
