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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 is,

$\mathbf{ln}\frac{\mathbf{P}\mathbf{2}}{\mathbf{P}\mathbf{1}}\mathbf{=}\frac{{\mathbf{Δ\pm á}}^{\mathbf{o}}\mathbf{vap}}{\mathbf{R}}\left[\frac{\mathbf{1}}{\mathbf{T}\mathbf{2}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{T}\mathbf{1}}\right]$

Here,

P is Pressure.

T is Temperature.

R is Gas Constant.

is Enthalpy of vaporisation.

PV=nRT

Ideal gas equation is a type of equation which defined the ideal gas. The ideal gas is not affected by the change in the pressure or volume of the gas at any temperature.

The equation for the ideal gas is:

$\mathbf{Number} \mathbf{of} \mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{MolarMass}}$

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

Vapour pressure${\text{P}}_{\text{1}}\text{= 10 atm}$,

Temperature,${\text{T}}_{\text{1}}{\text{= - 2.3}}^{\text{o}}\text{C = 270.85 k}$

Vapour pressure,${\text{P}}_{\text{2}}\text{= 40 atm}$

Temperature,${\text{T}}_{\text{2}}{\text{= 19}}^{\text{o}}\text{C = 292.15k}$

The Enthalpy of the vaporisation:

$$\begin{gathered} \ln \frac{\mathrm{P}1}{\mathrm{P}2}=\frac{-{\mathrm{Δ\pm á}}^{\mathrm{o}}\mathrm{vap}}{\mathrm{R}}\left[\frac{1}{\mathrm{T}2}-\frac{1}{\mathrm{T}1}\right] \\ \ln \frac{10}{40}=\frac{-{\mathrm{Δ\pm á}}^{\mathrm{o}}\mathrm{vap}}{8.314}\left[\frac{1}{270.85}-\frac{1}{292.15}\right] \\ \ln 0.25=\frac{-{\mathrm{Δ\pm á}}^{\mathrm{o}}\mathrm{vap}}{8.314}\left[\frac{292.15-270.85}{292.15×270.85}\right] \\ -1.38=\frac{-{\mathrm{Δ\pm á}}^{\mathrm{o}}\mathrm{vap}}{8.314}\left[\frac{21.3}{79129}\right] \end{gathered}$$

$$\begin{gathered} {\mathrm{Δ\pm á}}^{\mathrm{o}}\mathrm{vap}=\frac{1.38×8.314×79129}{21.3} \\ {\mathrm{Δ\pm á}}^{\mathrm{o}}\mathrm{vap}=42.3\mathrm{kJ}/\mathrm{mole} \end{gathered}$$

The enthalpy of vaporisation, ${\mathrm{Δ\pm á}}^{\mathrm{o}}\mathrm{vap}=42.6\mathrm{kJ}/\mathrm{mole}$

$$\begin{gathered} \ln \frac{\mathrm{P}1}{\mathrm{P}2}=\frac{-{\mathrm{Δ\pm á}}^{\mathrm{o}}\mathrm{vap}}{\mathrm{R}}\left[\frac{1}{\mathrm{T}2}-\frac{1}{\mathrm{T}1}\right] \\ \ln \frac{10}{\mathrm{P}2}=\frac{42.6}{8.314}\left[\frac{1}{262.15}-\frac{1}{270.85}\right] \\ \ln \frac{10}{\mathrm{P}2}=\frac{42.6}{8.314}\left[\frac{270.85-262.15}{262.15×270.85}\right] \\ \ln \frac{10}{\mathrm{P}2}=\frac{42.6}{8.314}\left[\frac{8.7}{71003}\right] \end{gathered}$$

$$\begin{gathered} \frac{10}{\mathrm{P}2}=\mathrm{Antiln}(0.00063) \\ \mathrm{P}=\frac{10}{1.88} \\ \mathrm{P}2=5.36\mathrm{torr} \end{gathered}$$

The vapour at temperature $\text{262.15k = 5.36torr}$.

As per ideal gas equation,

$$\begin{gathered} \mathrm{PV}=\mathrm{nRT} \\ 5.36×4.7=\frac{\mathrm{M}}{46}×8.314×262.15 \\ \mathrm{M}=\frac{5.36×4.7×46}{8.314×262.15} \\ \mathrm{M}=0.071\mathrm{g} \end{gathered}$$

Therefore, the mass of ethanol present in the vapor is 0.071g .

Vapour pressure,${\text{P}}_{\text{1}}\text{= 10 atm}$

Temperature, ${\text{T}}_{\text{1}}{\text{= - 2.3}}^{\text{o}}\text{C = 270.85 k}$

Vapour pressure, P₂ = ?

Temperature, ${\text{T}}_{\text{2}}{\text{= 20}}^{\text{o}}\text{C = 293.15k}$

As the enthalpy of vaporisation is ${\mathrm{Δ\pm á}}^{\mathrm{o}}\mathrm{vap}=42.6\mathrm{kJ}/\mathrm{mole}$

Therefore,

$$\begin{gathered} \ln \frac{\mathrm{P}2}{\mathrm{P}1}=\frac{-{\mathrm{Δ\pm á}}^{\mathrm{o}}\mathrm{vap}}{\mathrm{R}}\left[\frac{1}{\mathrm{T}2}-\frac{1}{\mathrm{T}1}\right] \\ \ln \frac{\mathrm{P}2}{10}=\frac{42.6\mathrm{kJ}}{8.314}\left[\frac{1}{270.85}-\frac{1}{293.15}\right] \\ \ln \frac{\mathrm{P}2}{10}=\frac{42600\mathrm{J}}{8.314}\left[\frac{293.15-270.85}{293.15×270.85}\right] \\ \ln \frac{\mathrm{P}2}{10}=\frac{42600\mathrm{J}}{8.314}\left[\frac{22.3}{79400}\right] \\ \frac{\mathrm{P}2}{10}=\mathrm{Antiln}(1.43) \\ \mathrm{P}2=10×4.174 \\ \mathrm{P}2=41.74\mathrm{torr} \end{gathered}$$

The vapour at temperature$\text{293.15k = 41.74torr}$ .

And,

$$\begin{gathered} \mathrm{PV}=\mathrm{nRT} \\ 41.74×4.7=\frac{\mathrm{M}}{46}×8.314×262.15 \\ \mathrm{M}=\frac{41.74×4.7×46}{8.314×262.15} \\ \mathrm{M}=0.33\mathrm{g} \end{gathered}$$

Yes, all the ethanol will evaporate at temperature 293.15k which have mass 0.33g .

Vapour pressure,${\text{P}}_{\text{1}}\text{= 10 atm}$

Temperature,${\text{T}}_{\text{1}}{\text{= - 2.3}}^{\text{o}}\text{C = 270.85 k}$

Vapour pressure,${\text{P}}_{\text{2}}\text{= ?}$

Temperature,${\text{T}}_{\text{2}}{\text{=0}}^{\text{o}}\text{C = 273.15k}$

As the enthalpy of vaporisation is ${\mathrm{Δ\pm á}}^{\mathrm{o}}\mathrm{vap}=42.6\mathrm{kJ}/\mathrm{mole}$.

$$\begin{gathered} \ln \frac{\mathrm{P}2}{\mathrm{P}1}=\frac{-{\mathrm{Δ\pm á}}^{\mathrm{o}}\mathrm{vap}}{\mathrm{R}}\left[\frac{1}{\mathrm{T}2}-\frac{1}{\mathrm{T}1}\right] \\ \ln \frac{\mathrm{P}2}{10}=\frac{42.3\mathrm{kJ}}{8.314}\left[\frac{1}{270.85}-\frac{1}{273.15}\right] \\ \ln \frac{\mathrm{P}2}{10}=\frac{42300\mathrm{J}}{8.314}\left[\frac{273.15-270.85}{273.15×270.85}\right] \\ \ln \frac{\mathrm{P}2}{10}=\frac{42300\mathrm{J}}{8.314}\left[\frac{2.3}{73983}\right] \\ \frac{\mathrm{P}2}{10}=\mathrm{Antiln}(0.16) \\ \mathrm{P}2=10×1.1174 \\ \mathrm{P}2=11.174\mathrm{torr} \end{gathered}$$

The vapour at temperature $\text{273.15.15k = 11.174torr}$.

Also,

$$\begin{gathered} \mathrm{PV}=\mathrm{nRT} \\ 11.174×4.7=\frac{\mathrm{M}}{46}×8.314×262.15 \\ \mathrm{M}=\frac{11.174×4.7×46}{8.314×262.15} \\ \mathrm{M}=0.181\mathrm{g} \end{gathered}$$

Yes, the ethanol will evaporate at temperature 273.15k which have mass 0.181G .