Chapter 13, Problem 23A

(a)

Interpretation Introduction

Interpretation:

The unknown quantity has to be calculated.

Concept Introduction:

The ideal gas law was derived by combining three important laws of the gases. The ideal gas law can be expressed as-

  $\begin{array}{l}\mathrm{V}\propto \frac{\mathrm{n}\mathrm{T}}{\mathrm{P}}\\ \mathrm{P}\mathrm{V}=\mathrm{n}\text{R}\mathrm{T}\end{array}$ Where, R is universal gas constant.

The ideal gas law consists of all the variables of a gas such as pressure ( $\mathrm{P}$ ), volume ( $\mathrm{V}$ ), number of moles ( $\mathrm{n}$ ) and temperature ( $\mathrm{T}$ ).

Answer to Problem 23A

  $\mathrm{V}=2.43\text{ L}$

Explanation of Solution

Given Information:

  $\begin{array}{l}\begin{array}{c}\mathrm{P}=782.4\text{ mm Hg }\\ =1.029\text{ atm}\end{array}\hfill \\ \begin{array}{c}\mathrm{T}=26.2°\text{C }\\ =299.2\text{ K}\end{array}\hfill \\ \mathrm{n}=0.1021\text{ mol}\hfill \\ \hfill \end{array}$

Calculation:

  $\begin{array}{l}\mathrm{P}\mathrm{V}=\mathrm{n}\text{R}\mathrm{T}\\ \mathrm{V}=\frac{\mathrm{n}\text{R}\mathrm{T}}{\mathrm{P}}\\ =\frac{0.1021\mathrm{m}\mathrm{o}\mathrm{l}\times 0.082\mathrm{L}\mathrm{a}\mathrm{t}\mathrm{m}/\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{K}\times 299.2\mathrm{K}}{1.029\mathrm{a}\mathrm{t}\mathrm{m}}\\ =2.43\mathrm{L}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The unknown quantity has to be calculated.

Concept Introduction:

The ideal gas law was derived by combining three important laws of the gases. The ideal gas law can be expressed as-

  $\begin{array}{l}\mathrm{V}\propto \frac{\mathrm{n}\mathrm{T}}{\mathrm{P}}\\ \mathrm{P}\mathrm{V}=\mathrm{n}\text{R}\mathrm{T}\end{array}$ Where, R is universal gas constant.

The ideal gas law consists of all the variables of a gas such as pressure ( $\mathrm{P}$ ), volume ( $\mathrm{V}$ ), number of moles ( $\mathrm{n}$ ) and temperature ( $\mathrm{T}$ ).

Answer to Problem 23A

  $\mathrm{P}=5126.9\text{ mm Hg}$

Explanation of Solution

Given Information:

  $\begin{array}{l}\begin{array}{c}\mathrm{V}=27.5\text{ mL }\\ \text{= 0}\text{.0275 L}\end{array}\hfill \\ \begin{array}{c}\mathrm{T}=16.6°\text{C }\\ \text{= 289}\text{.6 K}\end{array}\hfill \\ \mathrm{n}=0.007812\text{ mol}\hfill \end{array}$

Calculation:

  $\begin{array}{l}\mathrm{P}\mathrm{V}=\mathrm{n}\text{R}\mathrm{T}\\ \mathrm{P}=\frac{\mathrm{n}\text{R}\mathrm{T}}{\mathrm{V}}\\ =\frac{0.007812\mathrm{m}\mathrm{o}\mathrm{l}\times 0.082\mathrm{L}\mathrm{a}\mathrm{t}\mathrm{m}/\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{K}\times 289.6\mathrm{K}}{0.0275\mathrm{L}}\\ =6.746\mathrm{a}\mathrm{t}\mathrm{m}=6.746\times 760\mathrm{m}\mathrm{m}\mathrm{H}\mathrm{g}\\ =5126.9\mathrm{m}\mathrm{m}\mathrm{H}\mathrm{g}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The unknown quantity has to be calculated.

Concept Introduction:

The ideal gas law was derived by combining three important laws of the gases. The ideal gas law can be expressed as-

  $\begin{array}{l}\mathrm{V}\propto \frac{\mathrm{n}\mathrm{T}}{\mathrm{P}}\\ \mathrm{P}\mathrm{V}=\mathrm{n}\text{R}\mathrm{T}\end{array}$ Where, R is universal gas constant.

The ideal gas law consists of all the variables of a gas such as pressure ( $\mathrm{P}$ ), volume ( $\mathrm{V}$ ), number of moles ( $\mathrm{n}$ ) and temperature ( $\mathrm{T}$ ).

Answer to Problem 23A

Temperature $=-15.96°\text{C}$

Explanation of Solution

Given Information:

  $\begin{array}{l}\mathrm{P}=1.045\text{ atm}\hfill \\ \begin{array}{c}\mathrm{V}=45.2\text{ mL }\\ =0.0452\text{ L}\end{array}\hfill \\ \mathrm{n}=0.002241\text{ mol}\hfill \end{array}$

Calculation:

  $\begin{array}{c}\mathrm{P}\mathrm{V}=\mathrm{n}\text{R}\mathrm{T}\\ \mathrm{T}=\frac{\mathrm{P}\mathrm{V}}{\mathrm{n}\text{R}}\\ =\frac{1.045\mathrm{a}\mathrm{t}\mathrm{m}\times 0.0452\mathrm{L}}{0.002241\mathrm{m}\mathrm{o}\mathrm{l}\times 0.082\mathrm{L}\mathrm{a}\mathrm{t}\mathrm{m}/\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{K}}\\ =257.04\mathrm{K}=-15.96°\mathrm{C}\end{array}$