## Calculating the work done when water freezes and expands

$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

Kira_Maszewski_1B
Posts: 17
Joined: Wed Sep 21, 2016 2:57 pm

### Calculating the work done when water freezes and expands

How do we calculate for the answer of Self-Test 8.1A in the the Chemical Principles 6e textbook? It is the example about finding the work that water does when it freezes and pushes back the metal wall of a pipe...

Jamie_Lin_3I
Posts: 25
Joined: Sat Jul 09, 2016 3:00 am
Been upvoted: 2 times

### Re: Calculating the work done when water freezes and expands

Hello Kira!
For this question, the answer key in the textbook does say -0.86 kJ, but from previous posts on chemistry community most students were getting -0.98 or -0.94 (that is if you rounded the answers until the end). However, I'm not sure which exactly is the correct answer...

To get an answer, however, you need to consider the change in volume between water as liquid and water as solid (when it freezes).

To find volume of water as liquid you must first convert grams into m^3 (SI unit of volume) using the density given. You do the same thing for ice, except you use a different density
100 g water x (1cm^3 / 1.00g) x (1m^3/10^6 cm^3) = volume of water as liquid ( in m^3)

Once you got the two volumes, find the change in volume (final minus initial* so volume of ice - volume of water)
Then using equation 3 in the textbook (since pressure is constant throughout expansion), find the work done by the system using the change in volume and the given pressure 1070 atm. (Remember to convert the given pressure [in atm] to pascal in order to get the answer in J then in kJ)