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The liquid in the open-tube manometer in Fig. 12.8a is

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 16E Chapter 12

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 16E

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?

Step-by-Step Solution:

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

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Chapter 12, Problem 16E is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

The full step-by-step solution to problem: 16E from chapter: 12 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. The answer to “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?” is broken down into a number of easy to follow steps, and 74 words. This textbook survival guide was created for the textbook: University Physics, edition: 13. University Physics was written by and is associated to the ISBN: 9780321675460. Since the solution to 16E from 12 chapter was answered, more than 2647 students have viewed the full step-by-step answer. This full solution covers the following key subjects: pressure, absolute, tube, gas, open. This expansive textbook survival guide covers 26 chapters, and 2929 solutions.

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The liquid in the open-tube manometer in Fig. 12.8a is

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