Chapter 11, Problem 11.26QP

Interpretation Introduction

Interpretation:

The pressure in $\text{atm}$ and $\text{bars}$ exerted by a column of isopropanol has to be calculated.

Concept Introduction:

• $\text{760}\text{.0 mmHg}$ equals $\text{101}\text{.325 kPa,}$ as a result both values will be involved.  This location is slightly curious since most conversions involve a one, usually in the denominator.  The conversion examples above are examples of a one being involved.  For example, $\text{760}\text{.0 mmHg / 1}\text{.00 atm in examples 1 and 2}\text{. The 1}\text{.00}$ was assumed to be present.
• One $\text{atm}$ equals $\text{760}\text{.0 mmHg}$, thus there will be a multiplication or division based on the direction of the change.
• The gravitational constant is $\text{9}\text{.80665}{\text{m/s}}^{\text{2}}$
• One $\text{atm}$ equals $\text{101}\text{.325 kPa,}$ accordingly there will be a multiplication or division based on the direction of the change.