Chapter 13.1, Problem 7LC

(a)

Interpretation Introduction

Interpretation: Pressure in unit kilopascal is 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 7LC

  $\text{96}\text{.25 kPa}$

Explanation of Solution

Relation between atm and kPa is,

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

Convert pressure in $\text{0}\text{.95 atm}$ to pressure in kPa as shown:

  $\begin{array}{c}\text{0}\text{.95 atm}=\text{0}\text{.95 }\bcancel{\text{atm}}\times \frac{\text{101}\text{.325 kPa}}{1\bcancel{\text{atm}}}\\ =\text{96}\text{.25 kPa}\end{array}$

Thus, pressure in kPa is $\text{96}\text{.25 kPa}$

(b)

Interpretation Introduction

Interpretation: Pressure in unit kilopascal is 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 7LC

  $\text{96}\text{.25 kPa}$

Explanation of Solution

Relation between mmHg and atmosphere is,

  $1\text{ atm}=760\text{ mmHg}$

Convert pressure in $45\text{ mmHg}$ to pressure in atm as shown:

  $\begin{array}{c}45\text{ mmHg}=45\bcancel{\text{mmHg}}\times \frac{1\text{ atm}}{760\bcancel{\text{mmHg}}}\\ =\text{0}\text{.05921 atm}\end{array}$

The 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:

  $\begin{array}{c}\text{0}\text{.05921 atm}=\text{0}\text{.05921 }\bcancel{\text{atm}}\times \frac{\text{101}\text{.325 kPa}}{1\bcancel{\text{atm}}}\\ =\text{5}\text{.999 kPa}\\ =\text{6 kPa}\end{array}$

Thus, pressure in kPa is $\text{6 kPa}$