- #1

- 466

- 18

## Homework Statement

The evaporation enthalpy of Hg is ##59.3 kJmol^-1## at its boiling temperature ##356.6ºC##. Calculate:

(a) the vaporization entropy of Hg at this temperature

(b) the change in entropy of the surroundings and universe

(c) the vaporization entropy of Hg at 400ºC.

I was also given heat capacities for liquid and gaseous states as ##27.983 JK^-1mol^-1## and ##20.786JK^-1mol^-1##.

## Homework Equations

##dS=dQ/T##

##Q=\Delta_{vap} H##

##Delta S_{univ}=\Delta S_{system} + \Delta S_{surroundings}##

## The Attempt at a Solution

(a)

##\Delta_{vap} S = \frac{\Delta_{vap} H }{T} = 59.3 / (356.6+273.15) = 0.094 kJ K^-1 mol^-1##

(b)

##\Delta S_{system} = \frac{-\Delta_{vap} H}{T_{surroundings}}=\frac{-59.3}{25+273.15}= -.1988 kJ K^-1 ## Here I've assumed that only one mole is being heated, because no other information has been given. I have also assumed that the temperature of the surroundings is 25ºC.

##Delta S_{universe} = -.1048 kJ/K##

(c)

Enthalpy doesn't change with temperature (approximation). Therefore

##\Delta_{vap} S = \frac{59.3}{400+273.15} = 0.088093 kJ K^-1 mol^-1##

**Few comments:**

This problem violates the second law of thermodynamics because the entropy of the universe has been calculated to be negative.

Also, I'm having doubts as to what the temperature of the surroundings is. Is it assumed to be 25ºC? Its not the same as the substance being heated? Why so?

Would you agree with me that these questions/procedures are assuming that things are happening reversibly? Doesn't the second law of thermodynamics say that the entropy of the universe is 0 for a reversible process?

What have I done wrong?

Many thanks.