Question

con 6.13. Assume that the AvapH of an evaporating liquid comes from the liquid itself, which then changes temperature ac- cording to equation AvapH = -9 = mcAT. (a) Explain the presence of the negative sign in this equation. (b) If 1.00 g of H2O evaporates at 25.0°C in an adiabatic con- tainer initially containing 100.0 g of H,O, what is the final temperature of the water? Use c = 4.18 J/g.°C and AvapH = 43.99 kJ/mol at 25.0°C. The vapor pressure of H2O at 25.0°C is 23.77 torr; assume it remains constant throughout the tem- perature change of the liquid.

Answer 1

a) The negative sign has to do with the fact that the energy for the evaporation has to come from the water itself. This means that the energy needed to evaporate has to be "given" by the water in the form of heat, which is why it has a negative sign. (The negative sign is actually missing from the third term of the equation, where it should also be, we will see why):

b) To calculate this, we will use the expression given in item a), adding the missing negative sign and the number of moles multiplying the heat of vaporization:

$\Delta H_{vap}\cdot n=-m\cdot c\cdot \Delta T$

Using the fact that the number of moles is the quotient between the evaporated mass and the molar mass of water (18 g/mol) and assuming that the evaporated water doesn't "give" energy in the form of heat, we have:

$43990\frac{J}{mol}\cdot \frac{1g}{18g/mol}=-99g\cdot 4.18\frac{J}{g^{o}C}\cdot (T_{final}-25^{o}C)$

We can therefore rearrange and calculate the final temperature of the remaining 99 grams of water:

$T_{final}=-\frac{43990\frac{J}{mol}\cdot \frac{1g}{18g/mol}}{99g\cdot 4.18\frac{J}{g^{o}C}}+25^{o}C=19.1^{o}C$