Q12.136CP - Chemistry: Molecular Nature Of Matter And Change

Question

Question: A 4.7-L sealed bottle containing 0.33 g of liquid ethanol, ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}\mathbf{O}$, is placed in a refrigerator and reaches equilibrium with its vapor at 11⁰C .

 

(a) What mass of ethanol is present in the vapor? 

(b) When the container is removed and warmed to room temperature, 20⁰C, will all the ethanol vaporize? 

(c) How much liquid ethanol would be present at -2.3⁰C? The vapor pressure of ethanol is 10. torr at  2.3⁰C and 40. torr at 19⁰C .

 

Step-by-Step Solution

Answer

Answer

 

1. The enthalpy of vaporization,   and the mass of ethanol present in the vapor is 0.071g.
2. Yes, all the ethanol will evaporate at temperature 293.15k which has a mass of 0.33g.
3. The vapor at temperature 293.15k is 41.74torr and the liquid ethanol that will evaporate at temperature 273.15k is 0.181g.

 

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 is,

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

 

 

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.

 

 

 

Step 2: Subpart (a) The mass of ethanol present in the vapor.

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: 

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

 $$\begin{gathered} {\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}=\frac{1.38\times 8.314\times 79129}{21.3} \\ {\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}=42.3\mathrm{kJ}/\mathrm{mole} \end{gathered}$$

The enthalpy of vaporisation, ${\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}=42.6\mathrm{kJ}/\mathrm{mole}$ 

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

 

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

 

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

 

As per ideal gas equation,

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

 

 

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

 

Step 2: Subpart (b) When the container is removed and warmed to room temperature,20 0 C .

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  ${\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}=42.6\mathrm{kJ}/\mathrm{mole}$

Therefore,

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

      

 

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

 

And,

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

 

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

 

Step 2: Subpart (c) The amount of liquid ethanol that would be present at 0.0 0 C

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 ${\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}=42.6\mathrm{kJ}/\mathrm{mole}$ .

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

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

 

Also,

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

 

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