Entropy change of mercury vaporization at different temperature

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1. Consider the system of 20g (0.1 mole) sample of liquid mercury at 1bar of pressure in an open beaker. Given the data below calculate the change in entropy of the universe to vaporize this sample at room temperature (25C). Normal boiling point of Hg is 356.7C. You may assume the heat capacities are constant over this temperature range.
Hg (lq) Hg (vap)

Heat capacities from: (http://courses.chem.indiana.edu/c360/documents/thermodynamicdata.pdf)
Heat of vaporization from (http://en.wikipedia.org/wiki/Mercury_%28element%29)

Solution.

We need to calculate change in entropy of our sample at the room temperature and we have thermodynamics data for the boiling temperature. So, we can imagine this process as the three part process:

1) Change in entropy caused by heating our liquid sample from to . Let's donate this as
2) Change in entropy caused by vaporizing our sample at the boiling point We have this data as
3) Change in entropy from cooling our vapor from to Let's this be

Sum of this three change in entropy will be our requested change in entropy when we vaporize sample at the room temperature.

Sign in subscript means that quantity is expressed per mole.

In order to calculate and , consider definition of the heat capacity:

If we divide last equation by , and we know that , we get

If we integrate this we get our equation to calculate and ,

So, for we have

For we have

It is logical that is negative because we are cooling our system and in that way we are decreasing disorderness (we know that entropy is measure of disorder or chaos in the system).

Ok, when we sum all three change of entropy, we get,

Because we have of sample, final change in entropy is

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