Q12.124CP - Chemistry: Molecular Nature Of Matter And Change

Question

Methyl salicylate, ${\mathbf{C}}_8{\mathbf{H}}_8{\mathbf{O}}_3$, the odorous constituent of oil of wintergreen, has a vapor pressure of 1.00 torr at  $\mathbf{54}\mathbf{.}{\mathbf{3}}^o\mathbf{C}$  and 10.0 torr at  $\mathbf{95}\mathbf{.}{\mathbf{3}}^o\mathbf{C}$ . 

(a) What is its vapor pressure at ${\mathbf{25}}^o\mathbf{C}$ ? 

(b) What is the minimum number of liters of air that must pass over a sample of the compound at ${\mathbf{25}}^o\mathbf{C}$ to vaporize 1.0 mg of it?

Step-by-Step Solution

Answer

1. The vapor pressure at temperature  $\text{25°C}$  is 0.998 torr. There is a slight change in the pressure until a certain temperature but after that, there is a large of pressure.
2. The minimum number of liters of air to vaporize the 0.1mg that must pass over a sample of the compound at a given temperature $\text{25°C}$  is 0.14L.

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.

 

 $\mathbf{ln}\frac{P2}{P1}\mathbf{=}\frac{{\mathbf{\Delta H}}^o\mathbf{vap}}{R}\mathbf{[}\frac{\mathbf{1}}{\mathbf{T}\mathbf{2}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{T}\mathbf{1}}\mathbf{]}$

 

P is Pressure

T is Temperature

R is Gas Constant

$\text{ΔH}{°}_{\text{vap.}}^{}$  is Enthalpy of vaporisation.

 

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:

 $\mathbf{PV}\mathbf{=}\mathbf{nRT}$

 

 

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.

 

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

Step 2: Subpart (a) The vapor pressure at 25 o C .

Given,

 

Vapour pressure,  ${\mathbf{P}}_1\mathbf{=}\mathbf{1}\mathbf{torr}$

Temperature,  ${\mathbf{T}}_1\mathbf{=}\mathbf{54}\mathbf{.}{\mathbf{3}}^o\mathbf{C}\mathbf{=}\mathbf{327}\mathbf{.}\mathbf{3}\mathbf{k}$

Vapour pressure,  ${\mathbf{P}}_2\mathbf{=}\mathbf{10}\mathbf{torr}$ 

Temperature,  $\mathbf{=}\mathbf{95}\mathbf{.}{\mathbf{3}}^o\mathbf{C}\mathbf{=}\mathbf{368}\mathbf{.}\mathbf{3}\mathbf{k}$

 

Also,

 

The Enthalpy of the vaporisation: 

$\begin{array}{c}\ln \frac{\mathrm{P}1}{\mathrm{P}2}=\frac{-{\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}}{R}\left[\frac{1}{\mathrm{T}2}-\frac{1}{\mathrm{T}1}\right]\\ \ln \frac{1}{10}=\frac{-{\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}}{8.314}\left[\frac{1}{327.3}-\frac{1}{368.3}\right]\\ \ln 0.1=\frac{-{\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}}{8.314}\left[\frac{368.3-327.3}{327.3\times 368.3}\right]\\ -2.3=\frac{-{\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}}{8.314}\left[\frac{41}{120544.59}\right]\\ {\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}=\frac{2.3\times 8.314\times 120544.59}{41}\\ {\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}=56.2\mathrm{kJ}/\mathrm{mole}\end{array}$

The atomic radius of the body-centred unit cell  $\mathbf{=}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{-8}\mathbf{m}$ .

 

And,

  $\begin{array}{l}\ln \frac{\mathrm{P}1}{\mathrm{P}2}=\frac{{\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}}{}[\frac{1}{\mathrm{T}2}-\frac{1}{\mathrm{T}1}]\\ \ln \frac{}{}=\frac{52.2}{8.314}[\frac{1}{298}-\frac{1}{327.3}]\\ \ln \frac{}{}=\frac{52.2}{8.314}\left[\frac{327.3-298}{327.3\times 298}\right]\\ \ln \frac{}{}=\frac{52.2}{8.314}\left[\frac{29.3}{97535.4}\right]\end{array}$

 

 

Therefore,

$\begin{array}{l}\frac{1}{x}\mathbf{=}\mathbf{Antiln}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{002}\mathbf{)}\\ \frac{1}{x}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{002}\\ \mathbf{x}\mathbf{=}\frac{1}{\mathbf{1}\mathbf{.}\mathbf{002}}\\ \mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{998}\end{array}$

Step 3: Subpart (b) The minimum number of liters of air that must pass over a sample of the compound at 25 ° C to vaporize 1.0 mg of it.

As,

Temperature  $\mathbf{=}{\mathbf{25}}^o\mathbf{C}\mathbf{=}\mathbf{298}\mathbf{k}$ .

Pressure of gas at temperature ${\mathbf{25}}^o\mathbf{C}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{998}\mathbf{torr}$ .

Gas Constant $\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{L}\mathbf{/}\mathbf{mole}\mathbf{/}\mathbf{k}$ .

Mass of water given  $\mathbf{=}\mathbf{1}\mathbf{mg}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{001}\mathbf{g}$.

Molar mass of water  $\mathbf{=}\mathbf{18}\mathbf{g}\mathbf{/}\mathbf{mole}$.

 

And,

 

 $\begin{array}{c}\mathrm{PV}=\mathrm{nRT}\\ \mathbf{0}\mathbf{.}\mathbf{998}\mathbf{\times }\mathbf{V}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{001}}{18}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{\times }\mathbf{298}\end{array}$

So,

 

 $\begin{array}{c}\mathbf{V}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{001}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{\times }\mathbf{298}}{\mathbf{0}\mathbf{.}\mathbf{998}\mathbf{\times }\mathbf{18}}\\ \mathbf{V}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{14}\mathbf{L}\end{array}$