ΔThe heat of vaporization of water at 100^∘ C is 2.26 kJ / g ; at 37^∘ C (body temperature), it

step 1 So in this problem, we are going to calculate the heat of vaporization in kilojoules per mole. W

ΔThe heat of vaporization of water at 100^∘ C is 2.26 kJ /...

$\Delta$ The heat of vaporization of water at $100^{\circ} \mathrm{C}$ is $2.26 \mathrm{~kJ} / \mathrm{g} ;$ at $37^{\circ} \mathrm{C}$ (body temperature), it is $2.41 \mathrm{~kJ} / \mathrm{g} .$ (a) Convert the latter value to standard molar heat of vaporization, $\Delta H_{\text {vap }}^{\circ}$ at $37^{\circ} \mathrm{C}$. (b) Why is the heat of vaporization greater at $37^{\circ} \mathrm{C}$ than at $100^{\circ} \mathrm{C}$ ?

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Key Concepts

Unit Conversion in Thermodynamics

In thermodynamics, converting energy values from per gram to per mole units is a common practice, which involves using the molar mass of the substance. This conversion is crucial for standardizing measurements and comparing different substances, as it relates the microscopic quantities (mass) to macroscopic amounts (moles).

Temperature Dependence of Enthalpy of Vaporization

The heat of vaporization is temperature dependent, generally decreasing as the temperature increases. At lower temperatures, stronger intermolecular forces in the liquid state usually require more energy to overcome, resulting in a higher enthalpy of vaporization. This concept explains why the molar heat of vaporization of water is greater at 37°C than at 100°C, as additional energy is needed to counteract the stronger hydrogen bonding at the lower temperature.

Molar Heat of Vaporization

This concept refers to the amount of energy required to convert one mole of liquid into vapor at a constant temperature and pressure. It is a key thermodynamic property that allows chemists to compare the energies involved in phase transitions on a per-mole basis, rather than per gram, making it more universally applicable in chemical calculations.

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