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

Step-by-step 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 * \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)`.