Chapter 13, Problem 25A

(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 25A

  $\mathrm{V}=0.201\text{ L}$

Explanation of Solution

Given Information:

  $\begin{array}{l}\mathrm{P}=1.01\text{ atm}\hfill \\ \begin{array}{c}\mathrm{T}=25°\text{C}\\ \text{=25+273}\\ \text{ = 298 K}\end{array}\hfill \\ \mathrm{n}=0.00831\text{ mol}\hfill \end{array}$

Calculation:

  $\begin{array}{l}\mathrm{P}\mathrm{V}=\mathrm{n}\text{R}\mathrm{T}\\ \mathrm{V}=\frac{\mathrm{n}\text{R}\mathrm{T}}{\mathrm{P}}\\ =\frac{0.00831\mathrm{m}\mathrm{o}\mathrm{l}\times 0.082\mathrm{L}\mathrm{a}\mathrm{t}\mathrm{m}/\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{K}\times 298\mathrm{K}}{1.01\mathrm{a}\mathrm{t}\mathrm{m}}\\ =0.201\mathrm{L}\end{array}$

(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 25A

  $\mathrm{P}=0.338\text{ atm}$

Explanation of Solution

Given Information:

  $\begin{array}{l}\begin{array}{c}\mathrm{V}=602\text{ mL }\\ =0.602\text{ L}\end{array}\hfill \\ \mathrm{T}=310\text{ K}\hfill \\ \mathrm{n}=8.01\times {10}^{-3}\text{mol}\hfill \end{array}$

Calculation:

  $\begin{array}{l}\mathrm{P}\mathrm{V}=\mathrm{n}\text{R}\mathrm{T}\\ \mathrm{P}=\frac{\mathrm{n}\text{R}\mathrm{T}}{\mathrm{V}}\\ =\frac{8.01\times 10^{-3}\mathrm{m}\mathrm{o}\mathrm{l}\times 0.082\mathrm{L}\mathrm{a}\mathrm{t}\mathrm{m}/\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{K}\times 310\mathrm{K}}{0.602\mathrm{L}}\\ =0.338\mathrm{a}\mathrm{t}\mathrm{m}\end{array}$

(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 25A

  $\mathrm{n}=0.0249\text{ mol}$

Explanation of Solution

Given Information:

  $\begin{array}{l}\mathrm{P}=0.998\text{ atm}\hfill \\ \begin{array}{c}\mathrm{V}=629\text{ mL }\\ \text{= 0}\text{.629 L}\end{array}\hfill \\ \begin{array}{c}\mathrm{T}=35°\text{C }\\ \text{= 308 K}\end{array}\hfill \end{array}$

Calculation:

  $\begin{array}{l}\mathrm{P}\mathrm{V}=\mathrm{n}\text{R}\mathrm{T}\\ \mathrm{n}=\frac{\mathrm{P}\mathrm{V}}{\text{R}\mathrm{T}}\\ =\frac{0.998\mathrm{a}\mathrm{t}\mathrm{m}\times 0.629\mathrm{L}}{0.082\mathrm{L}\mathrm{a}\mathrm{t}\mathrm{m}/\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{K}\times 308\mathrm{K}}=0.0249\mathrm{m}\mathrm{o}\mathrm{l}\end{array}$