The liquid in the open-tube manometer in Fig. 12.8a is Mercury , and 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?

Solution 16E Step 1: Introduction : In this question, we have open tube manometer we need to find the absolute pressure at the bottom of the U tube In the second part we need to find the absolute pressure at the 4.00 cm depth below the free surface In the third part we need to find the absolute pressure of the gas in the container In the final part, we need to find the gauge pressure of the gas in pascals Data given Height of Mercury in tube 1 y = 3.00 cm 1 Height of Mercury in tube 2 y =27.00 cm Atmospheric pressure = 980 millibars = 98000 Pa Considering the figure given Step 2 : We need to find the absolute pressure at the bottom of the flask It is given by P = P +ogh Here the height of liquid at the bottom is given by y2 y 1 7.00 cm 3.00 cm h = y2 y 1 4.00 cm = 4.0 × 10 2 m 3 3 Density of mercury is = 13.6 × 10 kg/m 2 Gravitational constant g = 9.8 m/s Substituting values we get 3 3 2 2 P = 98000 Pa + 13.6 × 10 kg/m × 9.8 m/s × 4.0 × 10 m P = 98000 Pa + 5331.2 Pa P = 103331.2 Pa 5 2 P = 1.03 × 10 Pa = 14.94 lb/in 2 Hence we have absolute pressure at the bottom of the flask as 14.94 lb/in Step 3 : Here we need to find the absolute pressure at 4.00 cm below the free surface We have height as h = y 4 + (y y ) + y 2 2 2 1 1 h2= 7.00 cm 4.00 cm + (4.00 cm ) + 3.00 cm 2 h2= 10 cm = 10 × 10 m P = P o gh Substituting values we get 3 3 2 2 P = 98000 Pa + 13.6 × 10 kg/m × 9.8 m/s × 10 × 10 m m P = 98000 Pa + 13328 Pa P = 111328 Pa P = 1.13 × 10 Pa = 16.39 lb/in 2 Hence we have absolute pressure the absolute pressure at 4.00 cm below the free surface 2 as 16.39 lb/in