Chapter 13.1, Problem 3RQ

Interpretation Introduction

Interpretation:

The interconversion table of unit of pressure have to be completed.

Concept Introduction:

The SI unit for pressure is the pascal (Pa) and the units such as atm, torr and mm Hg (millimeters of mercury) are most commonly used by chemists.

1 standard atmosphere = 1.00 atm = 760.0 mm Hg = 760.0 torr = 101,325 Pa

Answer to Problem 3RQ

torr	atm	pascal	mm Hg
459 torr	0.604 atm	61195 pascal	459 mm Hg
997 torr	1.31 atm	132,874 pascal	997 mm Hg
2386.4 torr	3.14 atm	318160 pascal	2386.4 mm Hg
842 torr	1.11 atm	112257 pascal	842 mm Hg

Explanation of Solution

1 standard atmosphere = 1.00 atm = 760.0 mm Hg = 760.0 torr = 101,325 Pa

1. $459torr=\frac{459}{760}atm=459mmHg=0.604atm=0.604\times 101325pascal=61195pascal$
2. $132874pascal=\frac{132874}{101325}atm=1.31atm=1.31\times 760torr=997torr=997mmHg$
3. $3.14atm=3.14\times 760.0torr=2386.4torr=2386.4mmHg=3.14\times 101325Pascal=318160pascal$
4. $842mmHg=842torr=\frac{842}{760}atm=1.11atm=1.11\times 101325pascal=112257pascal$