Q12.132CP - Chemistry: Molecular Nature Of Matter And Change

Question

The boiling point of amphetamine, ${\mathbf{C}}_9{\mathbf{H}}_{13}\mathbf{N}$, is   ${\mathbf{201}}^o\mathbf{C}$at 760torr and ${\mathbf{83}}^o\mathbf{C}$   at 13torr. What is the concentration (in$\mathbf{g}\mathbf{/}{\mathbf{m}}^3$) of amphetamine when it is in contact with ${\mathbf{20}}^o\mathbf{C}$ air?

 

Step-by-Step Solution

Answer

The concentration of amphetamine when it is in contact with air at temperature ${\mathbf{20}}^o\mathbf{C}$ is .$\mathbf{42}\mathbf{.}\mathbf{12}\mathbf{g}\mathbf{/}{\mathbf{m}}^3$

 

Step 1: Definition

Enthalpy of vaporization may be defined as the energy required to convert the liquid into the vapor phase. It is also known as Enthalpy of evaporation.

 

Clausius- Clapeyron equation

 

 $\mathbf{ln}\frac{P2}{P1}\mathbf{=}\frac{{\mathbf{\Delta H}}^o\mathbf{vap}}{R}[\frac{1}{T2}\mathbf{-}\frac{1}{T1}]$

 

The ideal gas equation is a type of equation that defined the ideal gas. The ideal gas does not affect by the change in the pressure or volume of the gas at any temperature.The equation for the ideal gas is  .$\mathbf{PV}\mathbf{=}\mathbf{nRT}$

 

A number of moles can be defined as the ratio of the mass of the atom/molecule and the molar mass of the atom/molecule.

 $\mathbf{Number}\mathbf{of}\mathbf{Moles}\mathbf{=}\frac{\mathrm{Mass}}{\mathbf{Molar}\mathbf{Mass}}$

 

 

Step 2: The enthalpy of vaporization of dimethyl ether.

As

Initial Vapour pressure,${\text{P}}_{\text{1}}\text{ = 13 atm}$  

Initial Temperature,  ${\text{T}}_{\text{1}}{\text{= 83}}^{\text{o}}\text{C = 356 k}$

Final Vapour pressure,  ${\text{P}}_{\text{2}}\text{ = 760 atm}$

Final Temperature,  ${\text{T}}_{\text{2}}{\text{= 201}}^{\text{o}}\text{C = 474k}$

 

Therefore,

$\begin{array}{c}\mathbf{ln}\frac{P1}{P2}\mathbf{=}\frac{{\mathbf{\Delta H}}^o\mathbf{vap}}{R}[\frac{1}{T2}\mathbf{-}\frac{1}{T1}]\\ \Rightarrow \mathbf{ln}\frac{13}{760}\mathbf{=}\frac{{\mathbf{\Delta H}}^o\mathbf{vap}}{\mathbf{8}\mathbf{.}\mathbf{314}}[\frac{1}{474}\mathbf{-}\frac{1}{356}]\\ \Rightarrow \mathbf{ln}\mathbf{0}\mathbf{.}\mathbf{0171}\mathbf{=}\frac{{\mathbf{\Delta H}}^o\mathbf{vap}}{\mathbf{8}\mathbf{.}\mathbf{314}}\left[\frac{356-474}{474\times 356}\right]\\ \Rightarrow \mathbf{-}\mathbf{4}\mathbf{.}\mathbf{07}\mathbf{=}\frac{{\mathbf{\Delta H}}^o\mathbf{vap}}{\mathbf{8}\mathbf{.}\mathbf{314}}\left[\frac{-118}{168744}\right]\\ \Rightarrow {\mathbf{\Delta H}}^o\mathbf{vap}\mathbf{=}\frac{\mathbf{4}\mathbf{.}\mathbf{07}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{\times }\mathbf{168744}}{118}\\ \Rightarrow {\mathbf{\Delta H}}^o\mathbf{vap}\mathbf{=}\mathbf{48}\mathbf{.}\mathbf{4}\mathbf{kJ}\mathbf{/}\mathbf{mole}\end{array}$ 

 

 

The enthalpy of vaporisation,  ${\text{ΔH}}^{\text{o}}\text{vap = 48.4kJ/mole}$.

 

Step 3:

As, 

$\text{Number\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}Moles, n =}\frac{\text{Mass(M)}}{\text{MolarMass(M.M.)}}$ 

And,

 $\begin{array}{c}\mathrm{PV}=\mathrm{nRT}\\ \Rightarrow \mathbf{760}\mathbf{\times }\mathbf{V}\mathbf{=}\frac{M}{\mathbf{M}\mathbf{.}\mathbf{M}\mathbf{.}}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{\times }\mathbf{293}\\ \Rightarrow \mathbf{760}\mathbf{\times }\mathbf{V}\mathbf{=}\frac{M}{135}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{\times }\mathbf{293}\\ \Rightarrow \mathbf{760}\mathbf{\times }\mathbf{135}\mathbf{=}\frac{M}{V}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{\times }\mathbf{293}\\ \Rightarrow \frac{M}{V}\mathbf{=}\frac{760\times 135}{\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{\times }\mathbf{293}}\\ \Rightarrow \frac{M}{V}\mathbf{=}\mathbf{42}\mathbf{.}\mathbf{12}\mathbf{g}\mathbf{/}{\mathbf{m}}^3\end{array}$

 

 

Therefore, theconcentration of amphetamine when it is in contact with air at temperature ${\mathbf{20}}^o\mathbf{C}$  is$\mathbf{42}\mathbf{.}\mathbf{12}\mathbf{g}\mathbf{/}{\mathbf{m}}^3$  .