Question

The liquid in the open-tube manometer in Fig. 12.8a is mercury, y1 = 3.00 cm, and y2 = 7.00 cm. Atmospheric pressure is 980 millibars. (a) What is the absolute pressure at the bottom of the U-shaped tube? (b) What is the absolute pressure in the open tube at a depth of 4.00 cm below the free surface? (c) What is the absolute pressure of the gas in the container? (d) What is the gauge pressure of the gas in pascals?

Answer 1

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 ($P=p*g*h$), 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: $P_{abs}=P_{gauge} +P_{atm}$, so we can find the absolute pressure at the bottom of the U-shaped as: $P_{abs(at the bottom)} = 13600*9.81*0.07+98000=103337(Pa)$ (a). Using the same equation we can get the absolute pressure in the open tube at a depth of 4.00 cm as: $P_{abs(4.00cm)} =13600*9.81*0.04+98000=103337(Pa)$(b). Then to get the absolute pressure of the gas in the container we need to use the Pascal's law as:$P_{abs(gas)}=13600*9.81*(0.07-0.03)+98000=103337(Pa)$ (c). Finally to find the gauge pressure of the gas in pascals we solve for Pgauge as:$P_{gauge}=P_{abs} -P_{atm}$ so replacing the values we get:$P_{gauge}=P_{abs(gas)}-P_{atm}=103337-98000=5337(Pa)$ (d).