Question

Calculate the molar heat of vaporization of a fluid whose vapor pressure doubles when the temperature is raised from \[{75^ \circ }\]to $ {100^ \circ }C $ .

Answer 1

Hint: The molar heat of vaporization is the energy needed to vaporize one mole of a liquid. The units are usually kilojoules per mole. We solve this question by using the clausius-clapeyron equation. The Clausius-Clapeyron equation is an equation that displays the exponential relationship between vapor pressure and temperature. It is important because it gives a linear relationship between the natural logarithm of the vapor pressure and the inverse of the temperature.

Complete Step By Step Answer:
The variation of vapor pressure $ {p_i} $ with the desired boiling temperature $ {T_i} $ is given by the clausius-clapeyron equation:
Where, $ \Delta \overline {{H_{vap}}} $ is the molar enthalpy of vaporization and $ R = 8.314472\dfrac{J}{{mol.K}} $ is the universal gas constant.
Since we want to see the vapor pressure double due to raising the temperature from \[{75^ \circ }\]to $ {100^ \circ }C $ . Let:
 $ \Rightarrow {P_2} = 2{P_1} $
 $ \Rightarrow {T_1} = 75 + 273.15K = 348.15K $
 $ \Rightarrow {T_2} = 100 + 273.15K = 373.15K $
Therefore:
 $ \Rightarrow \ln (2) = - \dfrac{{\Delta \overline {{H_{vap}}} }}{{8.314\dfrac{j}{{mol.K}}}}\left[ {\dfrac{1}{{373.15k}} - \dfrac{1}{{348.15k}}} \right] $
 $ 2.315 \times {10^{ - 5}} \times \Delta \overline {{H_{vap}}} \dfrac{{mol}}{J} $
As a result:
 $ \Rightarrow \Delta \overline {{H_{vap}}} = \dfrac{{\ln 2}}{{2.315 \times {{10}^{ - 5}}}}\dfrac{J}{{mol}} $
 $ \Rightarrow 29946\dfrac{J}{{mol}} $
 $ \Rightarrow 29.95\dfrac{{kJ}}{{mol}} $ .

Additional Information:
Limitations of clausius-clapeyron equation:
-One cannot get something for nothing, because of the conservation of matter and energy.
-One cannot return to the same energy state, because entropy, or disorder, always increases.
-Absolute zero is unattainable because no perfectly pure substance exists.

Note:
Remember this kind of fluid would have moderate dipole-dipole forces, as water has $ \Delta \overline {{H_{vap}}} = 40.67\dfrac{{kJ}}{{mol}} $ at its boiling point. Dipole -dipole interactions occur when the partial charges formed within one molecule are attracted to an opposite partial charge in a nearby molecule. Polar molecules align so that the positive end of one molecule interacts with the negative end of another molecule.