Question

The standard entropy of liquid methanol at 298K is 126.8 J/K-mol and its heat capacity is 81.6 J/K-mol. Methanol boils at 337K with an enthalpy of vaporization of 35.270 kJ/mol at that temperature. The heat capacity of the vapor is 43.9 J/K-mol.__Calculate the entropy of one mole of methanol vapor at 800 K.

Answer 1

Explanation:

First, we will calculate the entropies as follow.

 $\Delta S_{2} = C_{p_{2}} ln \frac{T_{2}}{T_{1}}$ J/K mol

As,   $T_{1}$ = 298 K,         $T_{2}$ = 373.8 K

Putting the given values into the above formula we get,

     $\Delta S_{2} = C_{p_{2}} ln \frac{T_{2}}{T_{1}}$ J/K mol

                 = $81.6 ln (\frac{373.8}{298})$

                 = 18.5 J/K mol

Now,

       $\Delta S_{3} = \frac{\Delta H_{vap}}{T_{b.p}}$

                  = $\frac{35270 J}{373.8}$

                  = 94.2 J/mol K

Also,

    $\Delta S_{4} = C_{p_{4}} ln \frac{T_{2}}{T_{1}}$

                = $43.89 \times ln (\frac{800}{373.8})$

                = 33.4 J/K mol

Now, we will calculate the entropy of one mole of methanol vapor at 800 K as follows.

   $\Delta S_{T} = \Delta S^{o}_{1} + \Delta S_{2} + \Delta S_{3} + \Delta S_{4}$

                 = (126.8 + 18.5 + 94.2 + 33.4) J/K mol

                 = 272.9 J/K mol

Thus, we can conclude that the entropy of one mole of methanol vapor at 800 K is 272.9 J/K mol.