Chapter 13.1, Problem 2SP

Interpretation Introduction

Interpretation:

Pressure in the unit atmosphere is to be calculated and to be compared with the given pressure.

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 2SP

  $\text{33}\text{.7 kPa}$ is greater than $0.25\text{ atm}$ .

Explanation of Solution

Relation between atm and kPa is,

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

Convert pressure $\text{33}\text{.7 kPa}$ to pressure in the atmosphere

  $\begin{array}{c}\text{33}\text{.7 kPa}=\text{33}\text{.7 }\bcancel{\text{kPa}}\times \frac{1\text{ atm}}{\text{101}\text{.325 }\bcancel{\text{kPa}}}\\ =\text{0}\text{.33 atm}\end{array}$

Thus, $\text{33}\text{.7 kPa}$ is greater than $0.25\text{ atm}$ .

Conclusion

Relation between various units is used as a conversion factor.