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", "licence": "Creative Commons Attribution-NonCommercial-NoDerivs 4.0 International"}, "statement": "", "advice": "Mathematically, the definition of entropy is:
\n\n$\\mathrm{dS=\\frac{dQ}{T}}$,
\n\nwhere $\\mathrm{dS}$ is an infinitesimal entropy change; $\\mathrm{dQ}$ is an infinitesimal heat transfer to the system being described (i.e. it has negative value if heat is removed from the system); and $\\mathrm{T}$ is the temperature of the system, in $\\mathrm{K}$.
\n\nThis equation works because the temperature of the system effectively remains constant during an infinitesimal heat transfer, whereas it generally does not work for any macroscopic heat transfer, as the temperature will change due to the heat transfer (the equation would effectively change as we added heat) - however, if the system is undergoing a phase change then its temperature will remain effectively constant, as the heat energy is being used to promote particles to the new phase instead of to increase their temperature. This allows us to integrate the formula whilst treating $\\mathrm{T}$ as a constant:
\n$\\mathrm{\\Delta S=\\int dS=\\int \\frac{dQ}{T}=\\frac{\\Delta Q}{T}}$ .
\n\nEven if you don't remember that explanation, it's important to remember that if a system is undergoing a phase change, then the entropy change corresponding to a given heat transfer to the system is given by:
\n$\\mathrm{\\Delta S=\\frac{\\Delta Q}{T}}$ .
\n\nAdditionally, if the heat supplied to the system is the total enthalpy change for that phase change, e.g. the enthalpy of vaporisation, then we will use that value in place of $\\mathrm{\\Delta Q}$:
\n$\\mathrm{\\Delta S=\\frac{{\\Delta H}_{vap}}{T}}$ .
\n\nWe are given the entropy change for the vaporisation of chloroform, $\\mathrm{{\\Delta S}_{vap}=\\var{sigformat(entro,5)}\\space J\\space K^{-1}\\space mol^{-1}}$, at a temparature of $\\mathrm{\\var{ktemp} \\space K}$, and asked to calculate the corresponding enthalpy change.
\n\nTo do this, we will rearrange our established equation, then simply substitute in the values we have been provided:
\n$\\mathrm{{\\Delta S}_{vap}=\\frac{{\\Delta H}_{vap}}{T}}$
\n\n$\\mathrm{\\Rightarrow\\space\\space\\space {\\Delta H}_{vap}=T{\\Delta S}_{vap}}$
\n\n$\\mathrm{\\Rightarrow\\space\\space\\space {\\Delta H}_{vap}=\\var{ktemp}\\times \\var{sigformat(entro,5)}=\\var{sigformat(trueheat,3)}\\space J\\space mol^{-1}}$ .
\n\nTo quote our answer in the units asked for, we divide by one thousand to obtain:
\n$\\mathrm{{\\Delta H}_{vap} = \\var{sigformat(heat,3)}\\space kJ\\space mol^{-1}}$ .
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\nThe enthalpy change of a system is equal to the amount of [[1]] between the system and [[2]] during a process that occurs at [[3]] pressure, if no other [[4]] occurs other than that due to expansion of the system.
\n\n\n\n\nThe entropy of vaporisation of chloroform is found to be $\\mathrm{\\var{sigformat(entro,5)}\\space J\\space K^{-1}\\space mol^{-1}}$ at its normal boiling point of $\\mathrm{\\var{ktemp}\\space K}$. Calculate the latent heat of vaporisation of chloroform at this temperature, in $\\mathrm{kJ\\space mol^{-1}}$ .
\n\n[[0]] $\\mathrm{kJ\\space mol^{-1}}$ .
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