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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

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

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.

$\mathbf{Number}\text{ }\mathbf{of}\text{ â¶Ä‰}\mathbf{Moles}\mathbf{=}\frac{\mathrm{Mass}}{\mathbf{Molar}\text{ }\mathbf{Mass}}$

As

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

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

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

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

Therefore,

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

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

As,

$\text{Number of â¶Ä‰Moles,n=}\frac{\text{Mass(M)}}{\text{MolarMass(M.M.)}}$

And,

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

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