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

Step-by-step explanation:

First, we will calculate the entropies as follow.

`\Delta S_(2) = C_{p_(2)} ln (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 (T_(2))/(T_(1))` J/K mol

=
`81.6 ln ((373.8)/(298))`

= 18.5 J/K mol

Now,

`\Delta S_(3) = (\Delta H_(vap))/(T_(b.p))`

=
`(35270 J)/(373.8)`

= 94.2 J/mol K

Also,

`\Delta S_(4) = C_{p_(4)} ln (T_(2))/(T_(1))`

=
`43.89 * ln ((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.