Chapter 12, Problem 64AE

Interpretation Introduction

Interpretation:

Greatest volume in choices below has to be determined.

(a) $0.2\text{ mol}$ chlorine gas at $48°\text{C}$ and $80\text{ cm Hg}$.

(b) $4.2\text{ g}$ ammonia gas at $0.65\text{ atm}$ and $-11°\text{C}$.

(c) $21\text{ g}$ sulfur trioxide at $55°\text{C}$ and $110\text{ kPa}$.

Concept Introduction:

Ideal gas law represents an 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 64AE

Ammonia gas occupies greatest volume.

Explanation of Solution

Rearrange expression (1) to calculate volume of gas as follows:

  $V=\frac{nRT}{P}$        (2)

Conversion of temperature $48°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =48°\text{C}+273\\ =321\text{K}\end{array}$

Conversion factor to convert $\text{mm Hg}$ to $\text{atm}$ is as follows:

  $\left(\frac{1\text{ atm}}{760\text{ mm Hg}}\right)$

Pressure $80\text{ cm Hg}$ can be converted to $\text{atm}$ as follows:

  $\begin{array}{c}\text{Pressure}\left(\text{atm}\right)=\left(80\text{ cm Hg}\right)\left(\frac{1\text{ atm}}{760\text{ mm Hg}}\right)\left(\frac{10\text{ mm Hg}}{\text{1 cm Hg}}\right)\\ =1\text{ atm}\end{array}$

Substitute $0.2\text{ mol}$ for $n$, $\text{1 atm}$ for $P$, $8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, and $321\text{K}$ for $T$ in equation (2) to calculate volume of chlorine gas.

  $\begin{array}{c}V=\frac{\left(0.2\text{ mol}\right)\left(8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(321\text{K}\right)}{\text{1 atm}}\\ =5.3\text{ L}\end{array}$

Expression to calculate moles of $4.2{\text{ g NH}}_3$ is as follows:

  $\text{Moles}=\frac{\text{Given mass}}{\text{Molar mass}}$        (3)

Substitute $4.2\text{ g}$ for given mass and $17.03\text{ g/mol}$ for molar mass in equation (3) to calculate moles of ${\text{NH}}_3$.

  $\begin{array}{c}\text{Moles}=\frac{4.2\text{ g}}{17.03\text{ g/mol}}\\ =0.25\text{ mol}\end{array}$

Conversion of temperature $-11°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =-11°\text{C}+273\\ =262\text{K}\end{array}$

Substitute $0.25\text{ mol}$ for $n$, $\text{0}\text{.65 atm}$ for $P$, $8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, and $262\text{K}$ for $T$ in equation (2) to calculate volume of ammonia gas.

  $\begin{array}{c}V=\frac{\left(0.25\text{ mol}\right)\left(8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(262\text{K}\right)}{\text{0}\text{.65 atm}}\\ =8.3\text{ L}\end{array}$

Substitute $21\text{ g}$ for given mass and $80.07\text{ g/mol}$ for molar mass in equation (3) to calculate moles of ${\text{SO}}_3$.

  $\begin{array}{c}\text{Moles}=\frac{21\text{ g}}{80.07\text{ g/mol}}\\ =0.26\text{ mol}\end{array}$

Conversion of temperature $55°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =55°\text{C}+273\\ =328\text{K}\end{array}$

Conversion factor to convert $\text{kPa}$ to $\text{atm}$ is as follows:

  $\left(\frac{1\text{ atm}}{101.325\text{ kPa}}\right)$

Pressure $110\text{ kPa}$ can be converted to $\text{kPa}$ as follows:

  $\begin{array}{c}\text{Pressure}\left(\text{atm}\right)=\left(110\text{ kPa}\right)\left(\frac{1\text{ atm}}{101.325\text{ kPa}}\right)\\ =1.1\text{ atm}\end{array}$

Substitute $0.26\text{ mol}$ for $n$, $1.1\text{ atm}$ for $P$, $8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, and $328\text{K}$ for $T$ in equation (2) to calculate volume of ${\text{SO}}_3$ gas.

  $\begin{array}{c}V=\frac{\left(0.26\text{ mol}\right)\left(8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(328\text{K}\right)}{1.1\text{ atm}}\\ =6.4\text{ L}\end{array}$

Hence ${\text{NH}}_3$ occupies greatest volume.