[burntcookie]

Assuming that the osmotic pressure of a sample of sea water at 10 °C is 24.5 atm, calculate the work required under reversible conditions at this constant temperature to prepare 3 litre of pure water from this sea water. (Take the available volume of sea water as infinite and therefore unchanging in composition as 3 litre of pure water is extracted). Express your answer in J.

I tried w = 24.5 atm * 101.325 kPa/atm * 3L = 7447J, but my answer is wrong and I'm not sure how else to go about it. Help?

[Enthalpy]

Does sweet water have an osmotic pressure?

By the way, everyone would be happy to reach, or even approach, reversible conditions at water desalination. While reverse osmosis improves a lot over previous processes, it's still far from the theoretical limit.

[burntcookie]

Does sweet water have an osmotic pressure?

By the way, everyone would be happy to reach, or even approach, reversible conditions at water desalination. While reverse osmosis improves a lot over previous processes, it's still far from the theoretical limit.

Er...I think so? But it would be really small compared to the osmotic pressure of seawater...

[Enthalpy]

Or would the difference between seawater and sweet water be small?

[hahn1224]

 Because you have to push against not only the osmotic pressure, but the atmospheric pressure,
the P in w=PV should be 25.5atm.

So then, 25.5 x 3 x 101.325 = 7751.4J

[Enthalpy]

I dare to express a doubt.

Comparing the initial and final volumes often gives a hint about the role of atmospheric pressure.

[infinitefusion]

Because you have to push against not only the osmotic pressure, but the atmospheric pressure,
the P in w=PV should be 25.5atm.

So then, 25.5 x 3 x 101.325 = 7751.4J

I will confirm that this indeed produces the correct answer. I got the mark for this question.
w = (PI + 1)*V*(unit conversion factor)

[Enthalpy]

Pardon?

I was pleased with 24.5 atm *volume, and don't see what should be wrong there, except that this is a minimum, unattainable.

But why should atmospheric pressure work only against the sweet water side? The atmosphere also helps at the seawater side, so it must not be added to the osmotic pressure to compute the necessary work. Did I miss something?

[Enthalpy]

I checked at Wiki that the definition of osmotic pressure is absolutely normal
http://en.wikipedia.org/wiki/Osmotic_pressure
That is, the osmotic pressure of pure water is zero.

Then, as the atmospheric pressure pushes on the seawater side as well, the engine does not need to provide 1atm*volume. The expense is 24.5atm*volume.

Man, to make pure water from pure water, it doesn't cost any work!