Question

The heat of vaporization, ΔHvap of carbon tetrachloride (CCl4) is 43000 J/mol at 25 °C. 1 mol of liquid CCl4 has an entropy of 214 J/K. (a) What is the entropy of 1 mole of the vapor at 25 °C or 298 K? (b) How many intensive variables will be required to completely specify the vapor-liquid mixture of CCl4? (45 points)

Answer 1

Explanation:

(a)    The given data is as follows.

     Temperature (T) = $25^{o}C$ = (25 + 273) K = 298 K

      $\Delta H_{vap}$ = 43000 J/mol

Since, both the liquid and vapors are at equilibrium. Therefore, change in free energy will be calculated as follows.

      $\Delta G_{vap} = \Delta H_{vap} - T \Delta S_{vap}$ = 0        

            43000 - $(298 \times \Delta S_{vap})$ = 0

                      $\Delta S_{vap}$ = -144 J/mol K

Negative sign indicates an increase in entropy of the system.

Now, for 1 mole of $CCl_{4}$ is as follows.

     = 144 J/K

So,     $S_{vapor}$ - 214 = 144 J/k

                         = 358 J/K

Therefore, we can conclude that entropy of $CCl_{4}$ vapor is 358 J/K.

(b)  As we know that intensive variable are the variables which do not depend on the amount of a substance.

So, in the given situation only temperature will act as an intensive variable that will be required to completely specify the vapor-liquid mixture of $CCl_{4}$.