Chapter 8, Problem 69QRT

Interpretation Introduction

Interpretation:

Volume occupied by $1\text{ mol}$ of each of the noble gas has to be determined and the trend has to be commented. Volume of one mole of the noble gas as an Ideal gas has to be calculated. The gas that shows the largest deviations from the ideality at room temperature has to be identified and reason for this has to be explained.

Answer to Problem 69QRT

Volume of $He$ is $32.0mL.$

Volume of $Ne$ is $16.8mL.$

Volume of $Ar$ is $28.5mL.$

Volume of $Kr$ is $34.6mL.$

Volume of $Xe$ is $44.5mL.$

The trend observed from the volume it that the liquids with larger atoms will occupy more volume in its liquid state.

Volume of $1\text{ mol}$ of  ideal gas at STP is $22.4\text{ L}\text{.}$

The gas showing the largest deviation from this value is $Xe.$ This is because of its larger size. Also the boiling point of $Xe$ is near to the room temperature.

Explanation of Solution

• Calculate the volume of $He$:

  $\begin{array}{c}mass\text{ = n}\times \text{Molar mass}\\ \\ \text{= }\left(1\text{ mol}\right)\left(4.003\text{ g/mol}\right)\text{ = }4.003\text{ g}\\ \\ density\text{ = }\frac{mass}{Volume}\\ \\ =\frac{4.003\text{ g}}{0.125\text{ g/mL}}\text{ = 32}\text{.0 mL He}\end{array}$

Volume of $He$ is $32.0mL.$

• Calculate the volume of $Ne$:

  $\begin{array}{c}mass\text{ = n}\times \text{Molar mass}\\ \\ \text{= }\left(1\text{ mol}\right)\left(20.18\text{ g/mol}\right)\text{ = }20.18\text{ g}\\ \\ density\text{ = }\frac{mass}{Volume}\\ \\ =\frac{20.18\text{ g}}{1.20\text{ g/mL}}\text{ = 16}\text{.8 mL Ne}\end{array}$

Volume of $Ne$ is $16.8mL.$

• Calculate the volume of $Ar$:

  $\begin{array}{c}mass\text{ = n}\times \text{Molar mass}\\ \\ \text{= }\left(1\text{ mol}\right)\left(\text{39}\text{.95 g/mol}\right)\text{ = 39}\text{.95 g}\\ \\ density\text{ = }\frac{mass}{Volume}\\ \\ =\frac{\text{39}\text{.95 g}}{1.40\text{ g/mL}}\text{ = 28}\text{.5 mL }Ar\end{array}$

Volume of $Ar$ is $28.5mL.$

• Calculate the volume of $Kr$:

  $\begin{array}{c}mass\text{ = n}\times \text{Molar mass}\\ \\ \text{= }\left(1\text{ mol}\right)\left(\text{83}\text{.80 g/mol}\right)\text{ = 83}\text{.80 g}\\ \\ density\text{ = }\frac{mass}{Volume}\\ \\ =\frac{\text{83}\text{.80 g}}{2.42\text{ g/mL}}\text{ = 34}\text{.6 mL }Kr\end{array}$

Volume of $Kr$ is $34.6mL.$

• Calculate the volume of $Xe$:

  $\begin{array}{c}mass\text{ = n}\times \text{Molar mass}\\ \\ \text{= }\left(1\text{ mol}\right)\left(\text{131}\text{.29 g/mol}\right)\text{ = 131}\text{.29 g}\\ \\ density\text{ = }\frac{mass}{Volume}\\ \\ =\frac{\text{131}\text{.29 g}}{\text{2}\text{.95 g/mL}}\text{ = 44}\text{.5 mL }Xe\end{array}$

Volume of $Xe$ is $44.5mL.$

Observing the volume occupied by $1\text{ mol}$ of each of the noble gas, it is clear that the liquids with larger atoms will occupy more volume in its liquid state.

• Volume of one mole of the noble gas as an Ideal gas can be calculated using Ideal gas equation as shown below,

  $\begin{array}{c}PV\text{ = nRT}\\ \\ \left(\text{1 atm}\right)\text{V= }\left(\text{1}\text{.0 mol}\right)\left(\text{0}{\text{.0821 L atm mol}}^{-\text{1}}{\text{K}}^{-\text{1}}\right)\left(\text{273}\text{.15 K}\right)\\ \\ V=\frac{\left(\text{1}\text{.0 mol}\right)\left(\text{0}{\text{.0821 L atm mol}}^{-\text{1}}{\text{K}}^{-\text{1}}\right)\left(\text{273}\text{.15 K}\right)}{\left(\text{1}\text{.0 atm}\right)}\\ \\ =\text{22}\text{.4 L}\end{array}$

Volume of $1\text{ mol}$ of ideal gas at STP is $22.4\text{ L}\text{.}$

The gas showing the largest deviation from this value is $Xe.$ This is because of its larger size. Also the boiling point of $Xe$ is near to the room temperature.