Respuesta :
Answer : The vapor pressure of propane at [tex]25^oC[/tex] is, 7067.24 torr
Solution : Given,
Initial temperature = [tex]-42^oC=273+(-42)=231K[/tex] [tex](0^oC=273K)[/tex]
Final temperature = [tex]25^oC=273+25=298K[/tex]
[tex]\Delta H_{vap}=19.04kj/mole=19040j/mole[/tex] (1 kj = 1000 j)
At normal boiling point, the vapor pressure is equal to 760 torr.
So, initial pressure = 760 torr
Using Clausius Clapeyron equation,
[tex]\ln(\frac{P_2}{P_1})=\frac{-\Delta H_{vap}}{R}[\frac{1}{T_2}-\frac{1}{T_1}][/tex]
where,
[tex]P_1[/tex] = vapor pressure at temperature [tex]T_1[/tex]
[tex]P_2[/tex] = vapor pressure at temperature [tex]T_2[/tex]
[tex]\Delta H_{vap}[/tex] = enthalpy of vaporization
R = Gas constant = 8.31 j/Kmole
Now put all the given values in this formula, w get
[tex]\ln(\frac{P_2}{760torr})=\frac{-(19040j/mole)}{8.31j/Kmole}[\frac{1}{298K}-\frac{1}{231K}][/tex]
[tex]\ln(\frac{P_2}{760torr})=2.23[/tex]
[tex]\frac{P_2}{760torr}=9.299[/tex]
[tex]P_2=7067.24torr[/tex]
Therefore, the vapor pressure of propane at [tex]25^oC[/tex] is, 7067.24 torr
The vapor pressure of propane at [tex]25{\text{ }}^\circ {\text{C}}[/tex] is [tex]\boxed{7042.369{\text{ torr}}}[/tex].
Further Explanation:
Clausius Clapeyron Equation:
The temperature and pressure of any substance are related to each other if two phases of the substances are present in equilibrium. Such a relationship is given by Clausius Clapeyron equation.
The mathematical expression of Clausius Clapeyron equation is as follows:
[tex]\ln \dfrac{{{{\text{P}}_{\text{1}}}}}{{{{\text{P}}_{\text{2}}}}} = \dfrac{{\Delta {H_{{\text{vap}}}}}}{{\text{R}}}\left( {\dfrac{1}{{{{\text{T}}_{\text{2}}}}} - \dfrac{1}{{{{\text{T}}_{\text{1}}}}}} \right)[/tex] ...... (1)
Here,
[tex]{{\text{P}}_{\text{1}}}[/tex] is the initial pressure.
[tex]{{\text{P}}_{\text{2}}}[/tex] is the final pressure.
[tex]{{\text{T}}_1}[/tex] is the initial temperature.
[tex]{{\text{T}}_{\text{2}}}[/tex] is the final temperature.
R is universal gas constant.
[tex]\Delta {H_{{\text{vap}}}}[/tex] is the heat of vaporization.
The temperatures are to be converted into K. The conversion factor for this is,
[tex]{\text{1 }}^\circ {\text{C}} = 273.1{\text{5 K}}[/tex]
Therefore the initial temperature can be calculated as follows:
[tex]\begin{aligned}{{\text{T}}_1} &= \left( { - 42 + 273.15} \right)\;{\text{K}}\\&= 231.{\text{15 K}}\\\end{aligned}[/tex]
Therefore the final temperature can be calculated as follows:
[tex]\begin{aligned}{{\text{T}}_2}&= \left( {25 + 273.15} \right)\;{\text{K}}\\&= 298.15{\text{ K}}\\\end{aligned}[/tex]
The value of [tex]\Delta {H_{{\text{vap}}}}[/tex] has to be converted into J. The conversion factor for this is,
[tex]1{\text{ kJ}} = {\text{1}}{{\text{0}}^3}{\text{ J}}[/tex]
Therefore [tex]\Delta {H_{{\text{vap}}}}[/tex] can be evaluated as follows:
[tex]\begin{aligned}\Delta {H_{{\text{vap}}}} &= \left( {\frac{{19.04{\text{ kJ}}}}{{1{\text{ mol}}}}} \right)\left( {\frac{{{{10}^3}{\text{ J}}}}{{1{\text{ kJ}}}}} \right)\\&= 19040{\text{ J/mol}}\\\end{aligned}[/tex]
The pressure corresponding to the normal boiling point of propane is 760 torr. So the value of [tex]{{\text{P}}_{\text{1}}}[/tex] becomes 760 torr.
Substitute 760 torr for [tex]{{\text{P}}_{\text{1}}}[/tex], 231.15 K for [tex]{{\text{T}}_{\text{1}}}[/tex], 298.15 K for [tex]{{\text{T}}_{\text{2}}}[/tex], 19040 J/mol for [tex]\Delta {H_{{\text{vap}}}}[/tex], [tex]8.324{\text{ J}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}[/tex] for R in equation (1).
[tex]\begin{aligned}\ln \left( {\frac{{{\text{760 torr}}}}{{{{\text{P}}_{\text{2}}}}}} \right) &= \left( {\frac{{19040{\text{ J}}}}{{{\text{8}}{\text{.314 J}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}}}} \right)\left( {\frac{1}{{{\text{298}}{\text{.15 K}}}} - \frac{1}{{231.15{\text{ K}}}}} \right)\\&= - 2.22639\\\end{aligned}[/tex]
Solve for [tex]{{\text{P}}_{\text{2}}}[/tex],
[tex]{{\text{P}}_{\text{2}}} = 7042.369{\text{ torr}}[/tex]
Therefore the vapor pressure of propane at [tex]25{\text{ }}^\circ {\text{C}}[/tex] is 7042.369 torr.
Learn more:
1. Calculate the enthalpy change using Hess’s Law: https://brainly.com/question/11293201
2. Find the enthalpy of decomposition of 1 mole of MgO: https://brainly.com/question/2416245
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Thermodynamics
Keywords: P1, P2, T1, T2, R, propane, 7042.369 torr, Clausius Clapeyron equation, normal boiling point, 760 torr, 298.15 K, 231.15 K.
