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The Clausius-Clapeyron equation is used to relate the equilibrium vapour pressure of a substance at one temperature with that at another temperature, as well as to the enthalpy of vaporisation of the substance:

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$\\mathrm{ln(p_2)-ln(p_1)=\\frac{{\\Delta H^\\circ}_{vap}}{R}[(\\frac{1}{T_1})-(\\frac{1}{T_2})]}$ ,

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where $\\mathrm{T_1}$ and $\\mathrm{T_2}$ are the initial and final temperatures, in $\\mathrm{K}$ ;   $\\mathrm{p_1}$ and $\\mathrm{p_2}$ are the corresponding initial and final equilibrium vapour pressures (they must have the same units as eachother);  $\\mathrm{{\\Delta H^\\circ}_{vap}}$ is the standard enthalpy of vaporisation of benzene; and $\\mathrm{R=8.314\\space J\\space K^{-1}\\space mol^{-1}}$ is the universal gas constant.

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We are given an initial equilibrium vapour pressure $\\mathrm{p_1 = \\var{p1}\\space bar}$ at a temperature of $\\mathrm{T_1=\\var{tc1} \\space ^\\circ C}$, as well as a new equilibrium vapour pressure, $\\mathrm{p_2 = \\var{sigformat(p2,3)}\\space bar}$, at temperature $\\mathrm{T_2=\\var{tc2} \\space ^\\circ C}$. We are asked to use these values to calculate the enthalpy of vaporisation, $\\mathrm{{\\Delta H^\\circ}_{vap}}$.

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We must remember to convert both of our temperatures to Kelvin for use in the equation, by adding $\\mathrm{273.15}$:

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$\\mathrm{T_1=\\var{tc1} \\space ^\\circ C=\\var{temp1}\\space K}$ ,

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$\\mathrm{T_2=\\var{tc2} \\space ^\\circ C=\\var{temp2}\\space K}$ .

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To calculate $\\mathrm{{\\Delta H^\\circ}_{vap}}$, we must first rearrange the Clausius-Clapeyron equation appropriately:

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$\\mathrm{ln(p_2)-ln(p_1)=\\frac{{\\Delta H^\\circ}_{vap}}{R}[(\\frac{1}{T_1})-(\\frac{1}{T_2})]}$

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$\\mathrm{\\Rightarrow\\space\\space\\space {\\Delta H^\\circ}_{vap}[(\\frac{1}{T_1})-(\\frac{1}{T_2})]=R[ln(p_2)-ln(p_1)]}$

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$\\mathrm{\\Rightarrow\\space\\space\\space {\\Delta H^\\circ}_{vap}=\\frac{R[ln(p_2)-ln(p_1)]}{[(\\frac{1}{T_1})-(\\frac{1}{T_2})]}}$ .

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We can now simply substitute in our values to calculate $\\mathrm{{\\Delta H^\\circ}_{vap}}$ for the reaction:

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$\\mathrm{{\\Delta H^\\circ}_{vap}=\\frac{8.314\\times[ln(\\var{sigformat(p2,3)})-ln(\\var{sigformat(p1,3)})]}{[(\\frac{1}{\\var{temp1}})-(\\frac{1}{\\var{temp2}})]}=\\var{enthalpy}\\space J\\space mol^{-1}}$ .

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Lastly, to quote our answer in the units asked for, we divide by one thousand to obtain:

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$\\mathrm{{\\Delta H^\\circ}_{vap} = \\var{siground(h,3)}\\space kJ\\space mol^{-1}}$ .

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The equilibrium vapour pressure of benzene at $\\mathrm{\\var{tc1}^\\circ C}$ is $\\mathrm{\\var{dpformat(p1,2)} \\space bar}$.

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When the temperature is decreased to $\\mathrm{\\var{tc2}^\\circ C}$, this vapour pressure decreases to $\\mathrm{\\var{sigformat(p2,3)} \\space bar}$.

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Calculate the enthalpy of vaporisation, $\\mathrm{{ΔH^\\circ}_{vap}}$, for benzene:

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[[0]] $\\mathrm{kJ \\space mol^{-1}}$

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