Question

The liquid in the open-tube manometer in Figure 14.9a in thetextbook is mercury, $y_1$ = 3.00, and $y_2$ = 7.00. Atmospheric pressure is 980 millibars.

a) What is the absolute pressure at the bottom of the U-shapedtube?b) What is the absolute pressure in the open tube at a depthof 4.00 below the free surface?c) What is the absolute pressure of the gas in the tank?d) What is the gauge pressure of the gas in pascals?

Answer 1

The liquid in the open-tube manometer is mercury aty₁= 3.00 cm = 3.00 * 10^-2 m andy₂= 7.00 cm = 7.00 * 10^-2mThe atmospheric pressure is P = 980 millibars = 980 *10^-3 bars = 980 * 10^-3 * 10⁵ Pa =980 * 10² Paa)The absolute pressure at the bottom of the U-shaped tubeisP = P_o + ρ * g * y₂or P_o= P - ρ * g * y₂or P_o= 980 * 10² - 13.6 * 10³* 9.8 * 7.00 * 10^-2or P_o= 88670.4 Pab)The absolute pressure in the open tube at a depth of 4.00 cmbelow the free surface isP_o= P - ρ * g * yHere,y = 4.00 cm = 4.00 * 10^-2 mor P_o= 980 * 10² - 13.6 * 10³* 9.8 * 4.00 * 10^-2or P_o= 92668.8 Pac)The absolute pressure of the gas in the tank isP_o= ρ * g * yHere,ρ is the density of air.or P_o=1.225 * 9.8 * 7.00 * 10^-2or P_o= 0.84 Pad)The gauge pressure of the gas in pascals isP = ρ * g * (y₂ - y₁)or P = 1.225 * 9.8 * (7.00 - 3.00) * 10^-2or P = 0.48 Pa