Chapter 12, Problem 72AE

(a)

Interpretation Introduction

Interpretation:

Number of moles of air in ball has to be determined.

Concept Introduction:

Ideal gas law represents equation to relate its volume and pressure with its temperature. Expression for ideal gas equation is as follows:

  $PV=nRT$        (1)

Here,

$P$ is pressure of the gas.

$V$ is volume of gas.

$n$ denotes moles of gas.

$R$ is gas constant.

$T$ is temperature of gas.

Answer to Problem 72AE

Number of moles of air in ball is $0.0823\text{ mol}$.

Explanation of Solution

Rearrange expression (1) to calculate value of $n$ as follows:

  $n=\frac{PV}{RT}$        (2)

Conversion factor to convert $13\text{ lb/in}{\text{.}}^2$ to $\text{torr}$ is as follows:

  $\left(\frac{760\text{ torr}}{\text{14}\text{.7 lb/in}{\text{.}}^2}\right)$

Pressure $13\text{ lb/in}{\text{.}}^2$ can be converted to $\text{torr}$ as follows:

  $\begin{array}{c}\text{Pressure}\left(\text{torr}\right)=\left(13\text{ lb/in}{\text{.}}^2\right)\left(\frac{760\text{ torr}}{\text{14}\text{.7 lb/in}{\text{.}}^2}\right)\\ =6.72\times {10}^2\text{ torr}\end{array}$

Conversion of $20.0°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =20.0°\text{C}+273\\ =293\text{K}\end{array}$

Substitute $2.24\text{ L}$ for $V$, $6.72\times {10}^2\text{ torr}$ for $P$, $62.364\text{ L}\cdot \text{torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, and $293\text{K}$ for $T$ in equation (3).

  $\begin{array}{c}n=\frac{\left(6.72\times {10}^2\text{ torr}\right)\left(2.24\text{ L}\right)}{\left(62.364\text{ L}\cdot \text{torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(293\text{K}\right)}\\ =0.0823\text{ mol}\end{array}$

Hence number of moles of air in ball is $0.0823\text{ mol}$.

(b)

Interpretation Introduction

Interpretation:

Mass of air in ball has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 72AE

Mass of air in ball is $2.39\text{ g}$.

Explanation of Solution

Mass of $0.0823\text{ mol}$ air can be calculated as follows:

  $\begin{array}{c}\text{Mass}=\left(0.0823\text{ mol air}\right)\left(\frac{29\text{ g air}}{\text{1 mol air}}\right)\\ =2.39\text{ g air}\end{array}$

Hence mass of air in ball is $2.39\text{ g}$.

(c)

Interpretation Introduction

Interpretation:

Mass of air needed to remove to maintain constant pressure if temperature of ball increases from $\text{20}\text{.0 }°\text{C}$ to $\text{30}\text{.0 }°\text{C}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 72AE

Mass of air needed to remove to maintain constant pressure is $0.0783\text{ g}$.

Explanation of Solution

Expression to calculate final moles of air at constant pressure and volume is as follows:

  $n_2=\left(n_1\right)\left(\frac{T_1}{T_2}\right)$        (3)

Here,

$n_1$ is initial moles.

$n_2$ is initial moles.

$T_1$ is initial temperature.

$T_2$ is final temperature.

Conversion of $20.0°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =20.0°\text{C}+273\\ =293\text{K}\end{array}$

Conversion of $30.0°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =30.0°\text{C}+273\\ =303\text{K}\end{array}$

Substitute $0.0823\text{ mol}$ for $n_1$, $293\text{K}$ for $T_1$, and $303\text{K}$ for $T_2$ in equation (3).

  $\begin{array}{c}{n}_2=\left(0.0823\text{ mol}\right)\left(\frac{293\text{K}}{303\text{K}}\right)\\ =0.0796\text{ mol}\end{array}$

Moles of air needed to remove to maintain constant pressure can be calculated as follows:

  $\begin{array}{c}\text{Moles of air}=\left(0.0823-0.0796\right)\text{ mol}\\ =0.0027\text{ mol}\end{array}$

Mass of $0.0027\text{ mol}$ air can be calculated as follows:

  $\begin{array}{c}\text{Mass}=\left(0.0027\text{ mol air}\right)\left(\frac{29\text{ g air}}{\text{1 mol air}}\right)\\ =0.0783\text{ g air}\end{array}$

Hence mass of air needed to remove to maintain constant pressure is $0.0783\text{ g}$.