Chapter 13, Problem 100A

Interpretation Introduction

(a)

Interpretation:

Density of tetrafluoroethane is to be calculated

Concept introduction:

The equation which helps to calculate density of a gas at given temperature and pressure is given by,

$\begin{array}{l}\text{D=}\frac{\text{PM}}{\text{RT}}\\ \text{where}\\ \text{D}\text{is}\text{the}\text{density}\text{of}\text{gas}\\ \text{M}\text{is}\text{the}\text{molecular}\text{mass}\text{of}\text{gas}\end{array}$

$\mathrm{P}$ is pressure of the gas

$\mathrm{R}$ is the ideal gas constant

$\mathrm{T}$ is the temperature of the gas

Answer to Problem 100A

The density of the gas is $\text{4}\text{.55}\text{g/L}$

Explanation of Solution

The formula to calculate density is given by:

$\text{D=}\frac{\text{PM}}{\text{RT}}$

According to the statement:

$\begin{array}{l}\text{P=1}\text{atm}\\ \text{M=2}\times \text{12+2}\times \text{1+4}\times 19\\ =102{\text{gmol}}^{\text{-1}}\\ \text{R=0}\text{.0821}{\text{LatmK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{T}}_{}\text{=273}\text{K}\end{array}$

Now substituting the values into formula,

$\begin{array}{c}\text{D=}\frac{\text{PM}}{\text{RT}}\\ =\frac{1\times 102}{0.0821\times 273}\\ \text{=4}\text{.55}\text{g/L}\end{array}$

Therefore density of the gas is $\text{4}\text{.55}\text{g/L}$

(b)

Interpretation Introduction

Interpretation:

Molecules per litre are to be find out for given condition of temperature and pressure

Concept introduction:

Ideal gas equation is used to calculate number of moles.

Answer to Problem 100A

$\text{1}\text{.51×}{\text{10}}^{\text{22}}\text{molecules}$ are the molecules per litre of tetrafluoroethane

Explanation of Solution

According to this law,

$\mathrm{PV}=nRT$

Or

$\mathrm{n}=\frac{PV}{RT}$

Given data is:

$\text{P=1}\text{atm}$

$\text{V=1}\text{L}$

$\text{R=0}\text{.0821}\text{Latm}{K}^{-1}mo{l}^{-1}$

$\begin{array}{c}\text{T=220}\text{°C}\\ \text{=220+273}\\ \text{=}493\text{K}\end{array}$

Now substituting the values into formula,

$\begin{array}{c}n=\frac{PV}{RT}\\ =\frac{1\times 1}{0.0821\times 493}\\ =0.025mol\end{array}$

Now one mole contains Avogadro number of molecules.

Therefore,

$0.025mol$ will contain $\begin{array}{l}\text{0}\text{.025×6}\text{.023×}{\text{10}}^{\text{23}}\text{molecules}\\ \text{=1}\text{.51×}{\text{10}}^{\text{22}}\text{molecules}\end{array}$