Q12.28P - Chemistry: Molecular Nature Of Matter And Change

Question

Butane is a common fuel used in cigarette lighters and camping stoves. Normally supplied in metal containers under pressure, the fuel exists as a mixture of liquid and gas, so high temperatures may cause the container to explode. At ,$\mathbf{25}\mathbf{.}{\mathbf{0}}^o\mathbf{C}$ the vapour pressure of butane is 2.3 atm. What is the pressure in the container at $\mathbf{135}\mathbf{.}{\mathbf{0}}^o\mathbf{C}$ vap ( =24.3 kJ/mol)?                

Step-by-Step Solution

Answer

The phase diagram of the  explained by the pressure vs temperature. The density of the solid is increases than the liquid even after the critical point where solid-liquid density are equal.

Step 1: Enthalpy of Vaporization

As Enthalpy means “heat” so enthalpy of the heat of vaporization can be defined as the heat which is a form of energy required to change the liquid phase into the vapour phase. 

 

$\mathbf{Log}\frac{P2}{P1}\mathbf{=}\frac{\mathbf{-}{{\mathbf{\Delta H}}^o}_{\mathrm{vap}}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{R}}\left(\frac{1}{T1}\mathbf{-}\frac{1}{T2}\right)$ 

Where

${\mathbf{P}}_1$  = Vapour pressure at a temperature${\mathbf{T}}_1$  

${\mathbf{P}}_2$  = Vapour pressure at a temperature  ${\mathbf{T}}_2$

R = Gas constant at the universal level.

 

Step 2: Numerical Explanation

Butane is used as igniting material in some lighter and camping stoves because of the moderate ignition temperature. If the ignition temperature is low it will burn easily whereas if the ignition temperature is high it will take time to burn.

Butane has enthalpy of Vaporisation,${{\mathbf{\Delta H}}^o}_{\mathrm{vap}}$  = 24.3kJ/mole

Initial Temperature =  $\mathbf{25}\mathbf{.}{\mathbf{0}}^o\mathbf{C}$

Final Temperature =  $\mathbf{135}\mathbf{.}{\mathbf{0}}^o\mathbf{C}$

Pressure = 2.3atm

 

Using the Clausius-Clapeyron equation we can calculate pressure:

$\begin{array}{l}Log\left(\frac{P2}{P1}\right)=\frac{-\Delta {H^o}_{vap}}{2.303\times R}\left(\frac{1}{T1}-\frac{1}{T2}\right)\\ \Rightarrow ln\frac{P2}{23atm}=\frac{-24.3kJ/mole}{8.314J/moleK}\left[\frac{1}{\left(273+25\right)k}-\frac{1}{(273+135)k}\right]\\ \Rightarrow ln\frac{P2}{23atm}=\frac{-24.3kJ/mole}{8.314J/moleK}\left[\frac{1}{298k}-\frac{1}{408k}\right]\\ \Rightarrow ln\frac{P2}{23atm}=\frac{-24.3kJ/mole}{8.314J/moleK}\left(\frac{408-298}{298\times 408}\right)\\ \Rightarrow ln\frac{P2}{23atm}=\frac{-24.3kJ/mole}{8.314J/moleK}\left(\frac{110}{121584}\right)\\ \Rightarrow ln\frac{P2}{23atm}=2.64\\ \Rightarrow \frac{P2}{23atm}=Anti-ln(2.64)\\ \Rightarrow P2=14.01\times 23\\ \Rightarrow P2=324.3atm\end{array}$

The vapour pressure at temperature $\mathbf{135}\mathbf{.}{\mathbf{0}}^o\mathbf{C}$  is 324.3atm.