Chapter 13.1, Problem 1SP

Interpretation Introduction

Interpretation:

Pressure in unit kilopascal and atmosphere are to be determined.

Concept Introduction :

The force of all gas particle/wall interactions divided by the surface area of the wall is the definition of pressure (P). Pressure is one of the fundamental measurable properties of gases.

Answer to Problem 1SP

  $\text{0}\text{.5065 atm}$

  $\text{51}\text{.321 kPa}$

Explanation of Solution

Relation between mmHg and atmosphere is,

  $1\text{ atm}=760\text{ mmHg}$

Convert pressure in $385\text{ mmHg}$ to pressure in atm as shown:

  $\begin{array}{c}385\text{ mmHg}=385\bcancel{\text{mmHg}}\times \frac{1\text{ atm}}{760\bcancel{\text{mmHg}}}\\ =\text{0}\text{.5065 atm}\end{array}$

Relation between atm and kPa is,

  $1\text{ atm}=\text{101}\text{.325 kPa}$

Convert pressure in $\text{0}\text{.5065 atm}$ to pressure in kPa as shown:

  $\begin{array}{c}\text{0}\text{.5065 atm}=\text{0}\text{.5065 }\bcancel{\text{atm}}\times \frac{\text{101}\text{.325 kPa}}{1\bcancel{\text{atm}}}\\ =\text{51}\text{.321 kPa}\end{array}$

Thus, pressure in atm is $\text{0}\text{.5065 atm}$ and in kPa is $\text{51}\text{.321 kPa}$

Conclusion

Relation between various units is used as a conversion factor.