Chapter 8, Problem 52QRT

(a)

Interpretation Introduction

Interpretation:

Partial pressure of $O_2$ should be determined.

Concept Introduction:

Dalton’s law of partial pressure:

According to this law, the total pressure exerted by each gas in a mixture is equal to the sum of the individual partial pressure of the gases.

  $\begin{array}{c}{P}_{total}\text{ = }{\displaystyle \sum p_i}\\ \\ P_{total}\text{ = }{p}_1{\text{ + p}}_2{\text{ + p}}_3\text{ + }\mathrm{....}\end{array}$

Answer to Problem 52QRT

Partial pressure of $O_2$ is $\text{154 mm Hg}.$

Explanation of Solution

Given information is shown below,

  $\begin{array}{c}{P}_{total}\text{ = 740 mm Hg}\\ \\ p_{N_2}\text{ = 575 mm Hg}\\ \\ p_{Ar}\text{ = 6}\text{.9 mm Hg}\\ \\ p_{C{O}_2}\text{ = 0}\text{.2 mm Hg}\\ \\ p_{H_{2}O}\text{ = 4}\text{.0 mm Hg}\end{array}$

According to this law, the total pressure exerted by each gas in a mixture is equal to the sum of the individual partial pressure of the gases at constant temperature and volume.

  ${\text{P}}_{\text{total}}{\text{ = p}}_{{\text{N}}_{\text{2}}}{\text{ + p}}_{{\text{O}}_{\text{2}}}{\text{ + p}}_{\text{Ar}}{\text{ + p}}_{{\text{CO}}_{\text{2}}}{\text{ + p}}_{{\text{H}}_{\text{2}}\text{O}}$

Partial pressure of $O_2$ can be determined as follows,

  $\begin{array}{c}{\text{ p}}_{{\text{O}}_{\text{2}}}{\text{ = P}}_{\text{total}}-\left[{\text{p}}_{{\text{N}}_{\text{2}}}{\text{+ p}}_{\text{Ar}}{\text{ + p}}_{{\text{CO}}_{\text{2}}}{\text{ + p}}_{{\text{H}}_{\text{2}}\text{O}}\right]\\ \\ \text{ = 740 mm Hg}-\left[\text{575 mm Hg + 6}\text{.9 mm Hg + 0}\text{.2 mm Hg + 4}\text{.0 mm Hg}\right]\\ \\ =\text{ 154 mm Hg}\end{array}$

Partial pressure of $O_2$ is $154mmHg.$

(b)

Interpretation Introduction

Interpretation:

Mole fraction of $N_2,\text{}{\text{ O}}_2,{\text{ Ar, CO}}_2,{\text{ H}}_{2}O$ should be determined.

Concept Introduction:

Mole fraction: Quantity which defines the number of moles of a substance in a mixture divided by the total number of moles of all substances present.

${\text{x}}_{\text{a}}\text{=}\frac{{\text{n}}_{\text{a}}}{{\text{n}}_{\text{total}}}$

Answer to Problem 52QRT

• Partial pressure of $N_2$ is $\text{0}\text{.777}.$
• Partial pressure of $O_2$ is $\text{0}\text{.208}\text{.}$
• Partial pressure of $Ar$ is $\text{0}\text{.0093}\text{.}$
• Partial pressure of $C{O}_2$ is $\text{0}\text{.0003}\text{.}$
• Partial pressure of $H_{2}O$ is $\text{0}\text{.0054}.$

Explanation of Solution

Given information is shown below,

  $\begin{array}{c}{P}_{total}\text{ = 740 mm Hg}\\ \\ p_{N_2}\text{ = 575 mm Hg}\\ \\ {\text{pO}}_2\text{ = 154 mm Hg}\\ \\ p_{Ar}\text{ = 6}\text{.9 mm Hg}\\ \\ p_{C{O}_2}\text{ = 0}\text{.2 mm Hg}\\ \\ p_{H_{2}O}\text{ = 4}\text{.0 mm Hg}\end{array}$

Given information is shown below,

  $\begin{array}{c}{\text{P}}_{\text{total}}\text{= 720 mmHg}\\ \\ \text{Percent}\text{of}\text{oxygen}\text{=}\text{21\%}\end{array}$

Mole fraction can be determined using the given formula,

  ${\text{X}}_{\text{a}}\text{=}\frac{{\text{n}}_{\text{a}}}{{\text{n}}_{\text{total}}}$

Mole fraction of $N_2,\text{}{\text{ O}}_2,{\text{ Ar, CO}}_2,{\text{ H}}_{2}O$ is obtained by substituting the values as follows,

  $\begin{array}{c}{X}_{N_2}\text{}\text{= }\frac{p_{N_2}}{P_{{}_{total}}}=\frac{\text{575 mmHg}}{\text{740 mmHg}}\text{ = 0}\text{.777}\\ \\ X_{O_2}\text{}\text{= }\frac{p_{O_2}}{P_{{}_{total}}}=\frac{\text{154 mmHg}}{\text{740 mmHg}}\text{ = 0}\text{.208}\\ \\ X_{Ar}\text{}\text{= }\frac{p_{Ar}}{P_{{}_{total}}}=\frac{\text{6}\text{.9 mmHg}}{\text{740 mmHg}}\text{ = 0}\text{.0093}\\ \\ X_{C{O}_2}\text{}\text{= }\frac{p_{C{O}_2}}{P_{{}_{total}}}=\frac{\text{0}\text{.2 mmHg}}{\text{740 mmHg}}\text{ = 0}\text{.0003}\\ \\ X_{H_{2}O}\text{}\text{= }\frac{p_{H_{2}O}}{P_{{}_{total}}}=\frac{\text{4}\text{.0 mmHg}}{\text{740 mmHg}}\text{ = 0}\text{.0054}\end{array}$

Therefore,

Partial pressure of $N_2$ is $0.777.$

Partial pressure of $O_2$ is $0.208.$

Partial pressure of $Ar$ is $0.0093.$

Partial pressure of $C{O}_2$ is $0.0003.$

Partial pressure of $H_{2}O$ is $0.0054.$

(c)

Interpretation Introduction

Interpretation:

Composition of the given sample in percentage by volume should be determined.

Answer to Problem 52QRT

Composition of the given sample in percentage by volume is $\text{77}{\text{.7\% N}}_2,\text{ 20}\text{.8\% }{O}_2,\text{ 0}\text{.93\% Ar, 0}{\text{.03\% CO}}_2\text{ and 0}{\text{.54\% H}}_{2}O.$

Composition of given sample slightly varies since the given sample is wet.

Explanation of Solution

Given information is shown below,

Partial pressure of $N_2$ is $\text{0}\text{.777}.$

Partial pressure of $O_2$ is $\text{0}\text{.208}\text{.}$

Partial pressure of $Ar$ is $\text{0}\text{.0093}\text{.}$

Partial pressure of $C{O}_2$ is $\text{0}\text{.0003}\text{.}$

Partial pressure of $H_{2}O$ is $\text{0}\text{.0054}.$

The composition of the given sample in percentage by volume can be determine d as follows,

  $\begin{array}{l}\%N_2\text{ = }\frac{p_{N_2}}{P_{total}}\times 100\%\text{ = 0}\text{.777}\times 100\%\text{ = 77}{\text{.7\% N}}_2\\ \\ \%O_2\text{ = }\frac{p_{O_2}}{P_{total}}\times 100\%\text{ = 0}\text{.208}\times 100\%\text{ = 20}{\text{.8\% O}}_2\\ \\ \%Ar\text{ = }\frac{p_{Ar}}{P_{total}}\times 100\%\text{ = 0}\text{.0093}\times 100\%\text{ = 0}\text{.93\% }Ar\\ \\ \%C{O}_2\text{ = }\frac{p_{C{O}_2}}{P_{total}}\times 100\%\text{ = 0}\text{.0003}\times 100\%\text{ = 0}{\text{.03\% CO}}_2\\ \\ \%H_2\text{O = }\frac{p_{H_2\text{O}}}{P_{total}}\times 100\%\text{ = 0}\text{.0054}\times 100\%\text{ = 0}\text{.54\% }{H}_2\text{O}\end{array}$

The composition of the given sample in percentage by volume is $\text{77}{\text{.7\% N}}_2,\text{ 20}\text{.8\% }{O}_2,\text{ 0}\text{.93\% Ar, 0}{\text{.03\% CO}}_2\text{ and 0}{\text{.54\% H}}_{2}O.$

Composition of samples in dry air is given below,

Composition of given sample slightly varies with the values in the table. This is because the given sample is wet whereas the composition values in table are for dry air.